{"id":3557,"date":"2015-05-06T03:51:00","date_gmt":"2015-05-06T03:51:00","guid":{"rendered":"https:\/\/courses.candelalearning.com\/oschemtemp\/?post_type=chapter&#038;p=3557"},"modified":"2016-10-27T15:40:07","modified_gmt":"2016-10-27T15:40:07","slug":"precipitation-and-dissolution-2","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/chapter\/precipitation-and-dissolution-2\/","title":{"raw":"Precipitation and Dissolution","rendered":"Precipitation and Dissolution"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\nBy the end of this module, you will be able to:\r\n<ul>\r\n \t<li>Write chemical equations and equilibrium expressions representing solubility equilibria<\/li>\r\n \t<li>Carry out equilibrium computations involving solubility, equilibrium expressions, and solute concentrations<\/li>\r\n<\/ul>\r\n<\/div>\r\nThe preservation of medical laboratory blood samples, mining of sea water for magnesium, formulation of over-the-counter medicines such as Milk of Magnesia and antacids, and treating the presence of hard water in your home\u2019s water supply are just a few of the many tasks that involve controlling the equilibrium between a slightly soluble ionic solid and an aqueous solution of its ions.\r\n\r\nIn some cases, we want to prevent\u00a0dissolution from occurring. Tooth decay, for example, occurs when the calcium hydroxylapatite, which has the formula Ca<sub>5<\/sub>(PO<sub>4<\/sub>)<sub>3<\/sub>(OH), in our teeth dissolves. The dissolution process is aided when bacteria in our mouths feast on the sugars in our diets to produce lactic acid, which reacts with the hydroxide ions in the calcium hydroxylapatite. Preventing the dissolution prevents the decay. On the other hand, sometimes we want a substance to dissolve. We want the calcium carbonate in a chewable antacid to dissolve because the CO<sub>3<\/sub><sup>2\u2212<\/sup> ions produced in this process help soothe an upset stomach.\r\n\r\nIn this section, we will find out how we can control the dissolution of a slightly soluble ionic solid by the application of Le Ch\u00e2telier\u2019s principle. We will also learn how to use the equilibrium constant of the reaction to determine the concentration of ions present in a solution.\r\n<h2>The Solubility Product Constant<\/h2>\r\nSilver chloride is what\u2019s known as a sparingly soluble ionic solid (Figure\u00a01). Recall from the solubility rules in an earlier chapter that halides of Ag<sup>+<\/sup> are not normally soluble. However, when we add an excess of solid AgCl to water, it dissolves to a small extent and produces a mixture consisting of a very dilute solution of Ag<sup>+<\/sup> and Cl<sup>\u2013<\/sup> ions in equilibrium with undissolved silver chloride:\r\n<p style=\"text-align: center;\">[latex]\\text{AgCl}\\left(s\\right){\\underset{\\text{precipitation}}{\\overset{\\text{dissolution}}{\\longrightleftharpoons}}}{\\text{Ag}}^{+}\\left(aq\\right)+{\\text{Cl}}^{-}\\left(aq\\right)[\/latex]<\/p>\r\nThis equilibrium, like other equilibria, is dynamic; some of the solid AgCl continues to dissolve, but at the same time, Ag<sup>+<\/sup> and Cl<sup>\u2013<\/sup> ions in the solution combine to produce an equal amount of the solid. At equilibrium, the opposing processes have equal rates.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"880\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213936\/CNX_Chem_15_01_AgCl.jpg\" alt=\"Two beakers are shown with a bidirectional arrow between them. Both beakers are just over half filled with a clear, colorless liquid. The beaker on the left shows a cubic structure composed of alternating green and slightly larger grey spheres. Evenly distributed in the region outside, 11 space filling models are shown. These are each composed of a central red sphere with two smaller white spheres attached in a bent arrangement. In the beaker on the right, the green and grey spheres are no longer connected in a cubic structure. Nine green spheres, 10 grey spheres, and 11 red and white molecules are evenly mixed and distributed throughout the liquid in the beaker.\" width=\"880\" height=\"354\" data-media-type=\"image\/jpeg\" \/> Figure\u00a01. Silver chloride is a sparingly soluble ionic solid. When it is added to water, it dissolves slightly and produces a mixture consisting of a very dilute solution of Ag+ and Cl\u2013 ions in equilibrium with undissolved silver chloride.[\/caption]\r\n\r\nThe equilibrium constant for the equilibrium between a slightly soluble ionic solid and a solution of its ions is called the <b>solubility product (<em>K<\/em><sub>sp<\/sub>)<\/b> of the solid. Recall from the chapter on solutions and colloids that we use an ion\u2019s concentration as an approximation of its activity in a dilute solution. For silver chloride, at equilibrium:\r\n<p style=\"text-align: center;\">[latex]\\text{AgCl}\\left(s\\right)\\rightleftharpoons {\\text{Ag}}^{\\text{+}}\\left(aq\\right)+{\\text{Cl}}^{-}\\left(aq\\right){K}_{\\text{sp}}=\\left[{\\text{Ag}}^{\\text{+}}\\left(aq\\right)\\right]\\left[{\\text{Cl}}^{-}\\left(aq\\right)\\right][\/latex]<\/p>\r\nWhen looking at dissolution reactions such as this, the solid is listed as a reactant, whereas the ions are listed as products. The solubility product constant, as with every equilibrium constant expression, is written as the product of the concentrations of each of the ions, raised to the power of their stoichiometric coefficients. Here, the solubility product constant is equal to Ag<sup>+<\/sup> and Cl<sup>\u2013<\/sup> when a solution of silver chloride is in equilibrium with undissolved AgCl. There is no denominator representing the reactants in this equilibrium expression since the reactant is a pure solid; therefore [AgCl] does not appear in the expression for <em>K<\/em><sub>sp<\/sub>.\r\n\r\nSome common solubility products are listed in Table\u00a01\u00a0according to their <em>K<\/em><sub>sp<\/sub> values, whereas a more extensive compilation of products appears in <a class=\"target-chapter\" href=\".\/chapter\/solubility-products\/\" target=\"_blank\">Solubility Products<\/a>. Each of these equilibrium constants is much smaller than 1 because the compounds listed are only slightly soluble. A small <em>K<\/em><sub>sp<\/sub> represents a system in which the equilibrium lies to the left, so that relatively few hydrated ions would be present in a saturated solution.\r\n<table id=\"fs-idp15839744\" class=\"span-all\" summary=\"No Summary Text\">\r\n<thead>\r\n<tr valign=\"middle\">\r\n<th colspan=\"2\">Table\u00a01. Common Solubility Products by Decreasing Equilibrium Constants<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"middle\">\r\n<td>Substance<\/td>\r\n<td><em>K<\/em><sub>sp<\/sub> at 25 \u00b0C<\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td>CuCl<\/td>\r\n<td>1.2 \u00d7 10<sup>\u20136<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td>CuBr<\/td>\r\n<td>6.27 \u00d7 10<sup>\u20139<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td>AgI<\/td>\r\n<td>1.5 \u00d7 10<sup>\u201316<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td>PbS<\/td>\r\n<td>7 \u00d7 10<sup>\u201329<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td>Al(OH)<sub>3<\/sub><\/td>\r\n<td>2 \u00d7 10<sup>\u201332<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td>Fe(OH)<sub>3<\/sub><\/td>\r\n<td>4 \u00d7 10<sup>\u201338<\/sup><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div class=\"textbox examples\">\r\n<h3>Example 1:\u00a0Writing Equations and Solubility Products<\/h3>\r\nWrite the ionic equation for the dissolution and the solubility product expression for each of the following slightly soluble ionic compounds:\r\n<ol>\r\n \t<li>AgI, silver iodide, a solid with antiseptic properties<\/li>\r\n \t<li>CaCO<sub>3<\/sub>, calcium carbonate, the active ingredient in many over-the-counter chewable antacids<\/li>\r\n \t<li>Mg(OH)<sub>2<\/sub>, magnesium hydroxide, the active ingredient in Milk of Magnesia<\/li>\r\n \t<li>Mg(NH<sub>4<\/sub>)PO<sub>4<\/sub>, magnesium ammonium phosphate, an essentially insoluble substance used in tests for magnesium<\/li>\r\n \t<li>Ca<sub>5<\/sub>(PO<sub>4<\/sub>)<sub>3<\/sub>OH, the mineral apatite, a source of phosphate for fertilizers<\/li>\r\n<\/ol>\r\n(Hint: When determining how to break 4 and 5 up into ions, refer to the list of polyatomic ions in the section on chemical nomenclature.)\r\n[reveal-answer q=\"50078\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"50078\"]\r\n<ol>\r\n \t<li>[latex]\\text{AgI}\\left(s\\right)\\rightleftharpoons\\text{Ag}^{+}\\left(aq\\right)+\\text{I}^{-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,{;}\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{sp}}=\\left[\\text{Ag}^{+}\\right]\\left[\\text{I}^{-}\\right][\/latex]<\/li>\r\n \t<li>[latex]\\text{CaCO}_{3}\\left(s\\right)\\rightleftharpoons\\text{Ca}^{2+}\\left(aq\\right)+\\text{CO}_{3}^{2-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,{;}\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{sp}}=\\left[\\text{Ca}^{2+}\\right]\\left[\\text{CO}_{3}^{2-}\\right][\/latex]<\/li>\r\n \t<li>[latex]\\text{Mg}\\left(\\text{OH}\\right)_{2}\\left(s\\right)\\rightleftharpoons\\text{Mg}^{2+}\\left(aq\\right)+2\\text{OH}^{-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,{;}\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{sp}}=\\left[\\text{Mg}^{2+}\\right]\\left[\\text{OH}^{-}\\right]^{2}[\/latex]<\/li>\r\n \t<li>[latex]\\text{Mg}\\left(\\text{NH}_{4}\\right)\\text{PO}_{4}\\left(s\\right)\\rightleftharpoons\\text{Mg}^{2+}\\left(aq\\right)+\\text{NH}_{4}^{+}\\left(aq\\right)+\\text{PO}_{4}^{3-}\\left(aq\\right)\r\n\\,\\,\\,\\,\\,\\,\\,{;}\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{sp}}=\\left[\\text{Mg}^{2+}\\right]\\left[\\text{NH}_{4}^\\text{+}\\right]\\left[\\text{PO}_{4}^{3-}\\right][\/latex]<\/li>\r\n \t<li>[latex]\\text{Ca}_{5}\\left(\\text{PO}_{4}\\right)3\\text{OH}\\left(\\text{s}\\right)\\rightleftharpoons5\\text{Ca}^{2+}\\left(aq\\right)+3\\text{PO}_{4}^{3-}\\left(aq\\right)+\\text{OH}-\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,{;}\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{sp}}=\\left[\\text{Ca}^{2+}\\right]^{5}\\left[\\text{PO}_{4}^{3-}\\right]^{3}\\left[\\text{OH}^{-}\\right][\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n<h4>Check Your Learning<\/h4>\r\nWrite the ionic equation for the dissolution and the solubility product for each of the following slightly soluble compounds:\r\n<ol>\r\n \t<li>BaSO<sub>4<\/sub><\/li>\r\n \t<li>Ag<sub>2<\/sub>SO<sub>4<\/sub><\/li>\r\n \t<li>Al(OH)<sub>3<\/sub><\/li>\r\n \t<li>Pb(OH)Cl<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"636225\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"636225\"]\r\n<ol>\r\n \t<li>[latex]\\text{BaSO}_{4}\\left(s\\right)\\rightleftharpoons\\text{Ba}^{2+}\\left(aq\\right)+\\text{SO}_{4}^{2-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,{;}\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{sp}}=\\left[\\text{Ba}^{2+}\\right]\\left[\\text{SO}_{4}^{2-}\\right][\/latex]<\/li>\r\n \t<li>[latex]{\\text{Ag}}_{2}{\\text{SO}}_{4}\\left(s\\right)\\rightleftharpoons {\\text{2Ag}}^{\\text{+}}\\left(aq\\right)+{\\text{SO}}_{4}{}^{2-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,{;}\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{sp}}={\\left[{\\text{Ag}}^{\\text{+}}\\right]}^{2}\\left[{\\text{SO}}_{4}{}^{2-}\\right][\/latex]<\/li>\r\n \t<li>[latex]\\text{Al}\\left(\\text{OH}\\right)_{3}\\left(s\\right)\\rightleftharpoons\\text{Al}^{2+}\\left(aq\\right)+3\\text{OH}^{-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,{;}\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{sp}}=\\left[\\text{Al}^{3+}\\right]\\left[\\text{OH}^{-}\\right]^{3}[\/latex]<\/li>\r\n \t<li>[latex]\\text{Pb}\\left(\\text{OH}\\right)\\text{Cl}\\left(s\\right)\\rightleftharpoons\\text{Pb}^{2+}\\left(aq\\right)+\\text{OH}^{-}\\left(aq\\right)+\\text{Cl}^{-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,{;}\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{sp}}=\\left[\\text{Pb}^{2+}\\right]\\left[\\text{OH}^{-}\\right]\\left[\\text{Cl}^{-}\\right][\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nNow we will extend the discussion of <em>K<\/em><sub>sp<\/sub> and show how the solubility product constant is determined from the solubility of its ions, as well as how <em>K<\/em><sub>sp<\/sub> can be used to determine the molar solubility of a substance.\r\n<h2><em>K<\/em><sub>sp<\/sub> and Solubility<\/h2>\r\nRecall that the definition of <em>solubility<\/em> is the maximum possible concentration of a solute in a solution at a given temperature and pressure. We can determine the solubility product of a slightly soluble solid from that measure of its solubility at a given temperature and pressure, provided that the only significant reaction that occurs when the solid dissolves is its dissociation into solvated ions, that is, the only equilibrium involved is:\r\n<p style=\"text-align: center;\">[latex]{\\text{M}}_{p}{\\text{X}}_{q}\\left(s\\right)\\rightleftharpoons p{\\text{M}}^{\\text{m+}}\\left(aq\\right)+q{\\text{X}}^{\\text{n}-}\\left(aq\\right)[\/latex]<\/p>\r\nIn this case, we calculate the solubility product by taking the solid\u2019s solubility expressed in units of moles per liter (mol\/L), known as its <strong>molar solubility<\/strong>.\r\n<div class=\"textbox examples\">\r\n<h3>Example 2:\u00a0Calculation of <em>K<\/em><sub>sp<\/sub> from Equilibrium Concentrations<\/h3>\r\nWe began the chapter with an informal discussion of how the mineral fluorite\u00a0is formed. Fluorite, CaF<sub>2<\/sub>, is a slightly soluble solid that dissolves according to the equation:\r\n<p style=\"text-align: center;\">[latex]{\\text{CaF}}_{2}\\left(s\\right)\\rightleftharpoons {\\text{Ca}}^{\\text{2+}}\\left(aq\\right)+{\\text{2F}}^{-}\\left(aq\\right)[\/latex]<\/p>\r\nThe concentration of Ca<sup>2+<\/sup> in a saturated solution of CaF<sub>2<\/sub> is 2.1 \u00d7 10<sup>\u20134<\/sup><em>M<\/em>; therefore, that of F<sup>\u2013<\/sup> is 4.2 \u00d7 10<sup>\u20134<\/sup><em>M<\/em>, that is, twice the concentration of Ca<sup>2+<\/sup>. What is the solubility product of fluorite?\r\n[reveal-answer q=\"15268\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"15268\"]\r\n\r\nFirst, write out the <em>K<\/em><sub>sp<\/sub> expression, then substitute in concentrations and solve for <em>K<\/em><sub>sp<\/sub>:\r\n<p style=\"text-align: center;\">[latex]{\\text{CaF}}_{2}\\left(s\\right)\\rightleftharpoons {\\text{Ca}}^{\\text{2+}}\\left(aq\\right)+{\\text{2F}}^{-}\\left(aq\\right)[\/latex]<\/p>\r\nA saturated solution is a solution at equilibrium with the solid. Thus:\r\n<p style=\"text-align: center;\">[latex]{K}_{\\text{sp}}={\\text{[Ca}}^{\\text{2+}}\\text{]}\\text{[}{\\text{F}}^{-}{]}^{2}=\\text{(2.1}\\times {10}^{-4}\\big){\\left(4.2\\times {10}^{-4}\\right)}^{2}=\\text{3.7}\\times {10}^{-11}[\/latex]<\/p>\r\nAs with other equilibrium constants, we do not include units with <em>K<\/em><sub>sp<\/sub>.\r\n\r\n[\/hidden-answer]\r\n<h4>Check Your Learning<\/h4>\r\nIn a saturated solution that is in contact with solid Mg(OH)<sub>2<\/sub>, the concentration of Mg<sup>2+<\/sup> is 3.7 \u00d7 10<sup>\u20135<\/sup><em>M<\/em>. What is the solubility product for Mg(OH)<sub>2<\/sub>?\r\n<p style=\"text-align: center;\">[latex]{\\text{Mg(OH)}}_{2}\\left(s\\right)\\rightleftharpoons {\\text{Mg}}^{\\text{2+}}\\left(aq\\right)+{\\text{2OH}}^{-}\\left(aq\\right)[\/latex]<\/p>\r\n[reveal-answer q=\"918992\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"918992\"]2.0 \u00d7 10<sup>\u201313<\/sup>[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 3:\u00a0Determination of Molar Solubility from <em>K<\/em><sub>sp<\/sub><\/h3>\r\nThe <em>K<\/em><sub>sp<\/sub> of copper(I) bromide, CuBr, is 6.3 \u00d7 10<sup>\u20139<\/sup>. Calculate the molar solubility of copper bromide.\r\n[reveal-answer q=\"315393\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"315393\"]\r\n\r\nThe solubility product constant of copper(I) bromide is 6.3 \u00d7 10<sup>\u20139<\/sup>.\u00a0The reaction is:\r\n<p style=\"text-align: center;\">[latex]\\text{CuBr}\\left(s\\right)\\rightleftharpoons {\\text{Cu}}^{\\text{+}}\\left(aq\\right)+{\\text{Br}}^{-}\\left(aq\\right)[\/latex]<\/p>\r\nFirst, write out the solubility product equilibrium constant expression:\r\n<p style=\"text-align: center;\">[latex]{K}_{\\text{sp}}=\\text{[}{\\text{Cu}}^{\\text{+}}{]\\text{[}\\text{Br}}^{-}\\text{]}[\/latex]<\/p>\r\nCreate an ICE table (as introduced in the chapter on fundamental equilibrium concepts), leaving the CuBr column empty as it is a solid and does not contribute to the <em>K<\/em><sub>sp<\/sub>:\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213937\/CNX_Chem_15_01_ICETable1_img.jpg\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), and Equilibrium concentration ( M ). The second column has the header of, \u201cC u B r equilibrium arrow C u superscript positive sign plus B r superscript negative sign.\u201d Under the second column is a subgroup of three rows and three columns. The first column is blank. The second column has the following: 0, x, 0 plus x equals x. The third column has the following 0, x, 0 plus x equals x.\" width=\"879\" height=\"238\" data-media-type=\"image\/jpeg\" \/>\r\n\r\nAt equilibrium:\r\n<p style=\"text-align: center;\">[latex]{K}_{\\text{sp}}=\\left[\\text{Cu}^{+}\\right]\\left[\\text{Br}^{-}\\right][\/latex]\r\n[latex]6.3\\times {10}^{-9}=\\left(x\\right)\\left(x\\right)={x}^{2}[\/latex]\r\n[latex]x=\\sqrt{\\left(6.3\\times {10}^{-9}\\right)}=\\text{7.9}\\times {10}^{-5}[\/latex]<\/p>\r\nTherefore, the molar solubility of CuBr is 7.9 \u00d7 10<sup>\u20135<\/sup><em>M<\/em>.\r\n\r\n[\/hidden-answer]\r\n<h4>Check Your Learning<\/h4>\r\nThe <em>K<\/em><sub>sp<\/sub> of AgI is 1.5 \u00d7 10<sup>\u201316<\/sup>. Calculate the molar solubility of silver iodide.\r\n[reveal-answer q=\"528472\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"528472\"]1.2 \u00d7 10<sup>\u20138<\/sup><em>M<\/em>[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 4:\u00a0Determination of Molar Solubility from <em>K<\/em><sub>sp<\/sub>, Part II<\/h3>\r\nThe <em>K<\/em><sub>sp<\/sub> of calcium hydroxide, Ca(OH)<sub>2<\/sub>, is 8.0 \u00d7 10<sup>\u20136<\/sup>. Calculate the molar solubility of calcium hydroxide.\r\n[reveal-answer q=\"17713\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"17713\"]\r\n\r\nThe solubility product constant of calcium hydroxide is 8.0 \u00d7 10<sup>\u20136<\/sup>.\u00a0The reaction is:\r\n<p style=\"text-align: center;\">[latex]{\\text{Ca(OH)}}_{2}\\left(s\\right)\\rightleftharpoons {\\text{Ca}}^{\\text{2+}}\\left(aq\\right)+{\\text{2OH}}^{-}\\left(aq\\right)[\/latex]<\/p>\r\nFirst, write out the solubility product equilibrium constant expression:\r\n<p style=\"text-align: center;\">[latex]{K}_{\\text{sp}}=\\left[\\text{Ca}^{2+}\\right]\\left[\\text{OH}^{-}\\right]^{2}[\/latex]<\/p>\r\nCreate an ICE table, leaving the Ca(OH)<sub>2<\/sub> column empty as it is a solid and does not contribute to the <em>K<\/em><sub>sp<\/sub>:\r\n\r\n<img class=\"wp-image-4942 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/08\/23214454\/CNX_Chem_15_01_ICETable7_img.jpg\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), and Equilibrium concentration ( M ). The second column has the header of, \u201cC a ( O H ) subscript 2 equilibrium arrow C a superscript 2 positive sign plus 2 O H superscript negative sign.\u201d Under the second column is a subgroup of three rows and three columns. The first column is blank. The second column has the following: 0, x, and 0 plus x equals x. The third column has the following 0, 2 x, and 0 plus 2 x equals 2 x.\" width=\"880\" height=\"238\" \/>\r\n\r\nAt equilibrium:\r\n<p style=\"text-align: center;\">[latex]{K}_{\\text{sp}}=\\left[\\text{Ca}^{2+}\\right]\\left[\\text{OH}^{-}\\right]^{2}[\/latex]\r\n[latex]8.0\\times{10}^{-6}=\\left(x\\right)\\left(2x\\right)^{2}=\\left(x\\right)\\left(4x^{2}\\right)=4{x}^{3}[\/latex]\r\n[latex]x=\\sqrt[3]{\\frac{8.0\\times {10}^{-6}}{4}}=\\text{1.3}\\times {10}^{-2}[\/latex]<\/p>\r\nTherefore, the molar solubility of Ca(OH)<sub>2<\/sub> is 1.3 \u00d7 10<sup>\u20132<\/sup><em>M<\/em>.\r\n\r\n[\/hidden-answer]\r\n<h4>Check Your Learning<\/h4>\r\nThe <em>K<\/em><sub>sp<\/sub> of PbI<sub>2<\/sub> is 1.4 \u00d7 10<sup>\u20138<\/sup>. Calculate the molar solubility of lead(II) iodide.\r\n[reveal-answer q=\"194233\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"194233\"]1.5 \u00d7 10<sup>\u20133<\/sup><em>M<\/em>[\/hidden-answer]\r\n\r\n<\/div>\r\nNote that solubility is not always given as a molar value. When the solubility of a compound is given in some unit other than moles per liter, we must convert the solubility into moles per liter (i.e., molarity) in order to use it in the solubility product constant expression. Example 5\u00a0shows how to perform those unit conversions before determining the solubility product equilibrium.\r\n<div class=\"textbox examples\">\r\n<h3>Example 5:\u00a0Determination of <em>K<\/em><sub>sp<\/sub> from Gram Solubility<\/h3>\r\nMany of the pigments used by artists in oil-based paints (Figure\u00a02) are sparingly soluble in water. For example, the solubility of the artist\u2019s pigment chrome yellow, PbCrO<sub>4<\/sub>, is 4.3 \u00d7 10<sup>\u20135<\/sup> g\/L. Determine the solubility product equilibrium constant for PbCrO<sub>4<\/sub>.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"501\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213940\/CNX_Chem_15_01_OilPaints.jpg\" alt=\"A photograph is shown of a portion of an oil painting which reveals colors of orange, brown, yellow, green, blue, and purple colors in its strokes. A few water droplets rest on the surface.\" width=\"501\" height=\"334\" data-media-type=\"image\/jpeg\" \/> Figure 2. Oil paints contain pigments that are very slightly soluble in water. In addition to chrome yellow (PbCrO<sub>4<\/sub>), examples include Prussian blue (Fe<sub>7<\/sub>(CN)<sub>18<\/sub>), the reddish-orange color vermilion (HgS), and green color veridian (Cr<sub>2<\/sub>O<sub>3<\/sub>). (credit: Sonny Abesamis)[\/caption]\r\n\r\n[reveal-answer q=\"586402\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"586402\"]\r\n\r\nWe are given the solubility of PbCrO<sub>4<\/sub> in grams per liter. If we convert this solubility into moles per liter, we can find the equilibrium concentrations of Pb<sup>2+<\/sup> and [latex]{\\text{CrO}}_{4}^{2-}[\/latex], then <em>K<\/em><sub>sp<\/sub>:\r\n\r\n<img class=\"alignnone wp-image-4943\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213942\/CNX_Chem_15_01_PbCrO4_img.jpg\" alt=\"This figure shows four horizontally oriented rectangles. The first three from the left are shaded green and the last one at the right is shaded white. Right pointing arrows between the rectangles are labeled \u201c1,\u201d \u201c2,\u201d and \u201c3\u201d moving left to right across the diagram. The first rectangle is labeled \u201cSolubility of P b C r O subscript 4, in g divdided by L.\u201d The second rectangle is labeled \u201c[ P b C r O subscript 4 ], in m o l divided by L.\u201d The third is labeled \u201c[ P b superscript 2 plus] and [ C r O subscript 4 superscript 2 negative ].\u201d The fourth rectangle is labeled \u201cK subscript s p.\u201d\" width=\"880\" height=\"156\" \/>\r\n<ol>\r\n \t<li><em>Use the molar mass of PbCrO<sub>4<\/sub><\/em> [latex]\\left(\\frac{323.2\\text{g}}{1\\text{mol}}\\right)[\/latex] <em>to convert the solubility of PbCrO<sub>4<\/sub> in grams per liter into moles per liter:\r\n<\/em>[latex]\\begin{array}{l}{ }\\left[{\\text{PbCrO}}_{4}\\right]&amp;=&amp;\\frac{4.3\\times {10}^{-5}{\\text{g PbCrO}}_{4}}{1\\text{L}}\\times \\frac{1{\\text{mol PbCrO}}_{4}}{323.2{\\text{g PbCrO}}_{4}}\\\\&amp; =&amp;\\frac{1.3\\times {10}^{-7}{\\text{mol PbCrO}}_{4}}{1\\text{L}}\\\\&amp; =&amp;\\text{}1.3\\times {10}^{-7}M\\end{array}[\/latex]<\/li>\r\n \t<li><em>The chemical equation for the dissolution indicates that 1 mol of PbCrO<sub>4<\/sub> gives 1 mol of Pb<sup>2+<\/sup>(aq) and 1 mol of<\/em> [latex]{\\text{CrO}}_{4}{}^{\\text{2-}}\\text{(}aq\\text{)}:[\/latex][latex]{\\text{PbCrO}}_{4}\\left(s\\right)\\rightleftharpoons {\\text{Pb}}^{2+}\\left(aq\\right)+{\\text{CrO}}_{4}{}^{2-}\\left(aq\\right)[\/latex]Thus, both [Pb<sup>2+<\/sup>] and [latex]\\left[{\\text{CrO}}_{4}{}^{\\text{2-}}\\right][\/latex] are equal to the molar solubility of PbCrO<sub>4<\/sub>:[latex]\\left[\\text{Pb}^{2+}\\right]=\\left[\\text{CrO}_{4}^{2-}\\right]=1.3\\times{10}^{-7}M[\/latex]<\/li>\r\n \t<li><em>Solve. K<\/em><sub>sp<\/sub> = [Pb<sup>2+<\/sup>] [latex]\\left[{\\text{CrO}}_{4}{}^{2-}\\right][\/latex] = (1.3 \u00d7 10<sup>\u20137<\/sup>)(1.3 \u00d7 10<sup>\u20137<\/sup>) = 1.7 \u00d7 10<sup>\u201314<\/sup><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n<h4>Check Your Learning<\/h4>\r\nThe solubility of TlCl [thallium(I) chloride], an intermediate formed when thallium is being isolated from ores, is 3.2 grams per liter at 20 \u00b0C. What is its solubility product?\r\n[reveal-answer q=\"95775\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"95775\"]1.4 \u00d7 10<sup>\u20134<\/sup> (1.5 \u00d7 10<sup>\u20134<\/sup> if we round the solubility to two digits before calculating <em>K<\/em><sub>sp<\/sub>)[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 6:\u00a0Calculating the Solubility of Hg<sub>2<\/sub>Cl<sub>2<\/sub><\/h3>\r\nCalomel, Hg<sub>2<\/sub>Cl<sub>2<\/sub>, is a compound composed of the diatomic ion of mercury(I), [latex]{\\text{Hg}}_{2}{}^{\\text{2+}}[\/latex], and chloride ions, Cl<sup>\u2013<\/sup>. Although most mercury compounds are now known to be poisonous, eighteenth-century physicians used calomel as a medication. Their patients rarely suffered any mercury poisoning from the treatments because calomel is quite insoluble:\r\n<p style=\"text-align: center;\">[latex]{\\text{Hg}}_{2}{\\text{Cl}}_{2}\\left(s\\right)\\rightleftharpoons {\\text{Hg}}_{2}{}^{\\text{2+}}\\left(aq\\right)+{\\text{2Cl}}^{-}\\left(aq\\right){K}_{\\text{sp}}=\\text{1.1}\\times {10}^{-18}[\/latex]<\/p>\r\nCalculate the molar solubility of Hg<sub>2<\/sub>Cl<sub>2<\/sub>.\r\n[reveal-answer q=\"561036\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"561036\"]\r\n\r\nThe molar solubility of Hg<sub>2<\/sub>Cl<sub>2<\/sub> is equal to the concentration of [latex]{\\text{Hg}}_{2}{}^{\\text{2+}}[\/latex] ions because for each 1 mol of Hg<sub>2<\/sub>Cl<sub>2<\/sub> that dissolves, 1 mol of [latex]{\\text{Hg}}_{2}{}^{\\text{2+}}[\/latex] forms:\r\n\r\n<img class=\"alignnone wp-image-4944\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213944\/CNX_Chem_15_01_Format_img.jpg\" alt=\"This figure shows four horizontally oriented light green rectangles. Right pointing arrows are placed between them. The first rectangle is labeled \u201cDetermine the direction of change.\u201d The second rectangle is labeled \u201cDetermine x and the equilibrium concentrations.\u201d The third is labeled \u201cSolve for x and the equilibrium concentrations.\u201d The fourth rectangle is labeled \u201cCheck the math.\u201d\" width=\"880\" height=\"156\" \/>\r\n<ol>\r\n \t<li><em>Determine the direction of change.<\/em> Before any Hg<sub>2<\/sub>Cl<sub>2<\/sub> dissolves, <em>Q<\/em> is zero, and the reaction will shift to the right to reach equilibrium.<\/li>\r\n \t<li><em>Determine<\/em> x<em> and equilibrium concentrations.<\/em> Concentrations and changes are given in the following ICE table:<img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213945\/CNX_Chem_15_01_ICETable2_img.jpg\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header of, \u201cH g subscript 2 C l subscript 2 equilibrium arrow H g subscript 2 superscript 2 positive sign plus 2 C l superscript negative sign.\u201d Under the second column is a subgroup of three rows and three columns. The first column is blank. The second column has the following: 0, x, 0 plus x equals x. The third column has the following: 0, 2 x, 0 plus 2 x equals 2 x.\" width=\"838\" height=\"227\" data-media-type=\"image\/jpeg\" \/>Note that the change in the concentration of Cl<sup>\u2013<\/sup> (2<em>x<\/em>) is twice as large as the change in the concentration of [latex]{\\text{Hg}}_{2}{}^{\\text{2+}}[\/latex] (<em>x<\/em>) because 2 mol of Cl<sup>\u2013<\/sup> forms for each 1 mol of [latex]{\\text{Hg}}_{2}{}^{\\text{2+}}[\/latex] that forms. Hg<sub>2<\/sub>Cl<sub>2<\/sub> is a pure solid, so it does not appear in the calculation.<\/li>\r\n \t<li><em>Solve for<\/em> x<em> and the equilibrium concentrations.<\/em> We substitute the equilibrium concentrations into the expression for <em>K<\/em><sub>sp<\/sub> and calculate the value of <em>x<\/em>: [latex]{K}_{\\text{sp}}=\\left[\\text{Hg}_{2}^{2+}\\right]\\left[\\text{Cl}^{-}\\right]^{2}[\/latex][latex]1.1\\times{10}^{-18}=\\left(x\\right)\\left(2x\\right)^{2}[\/latex][latex]4{x}^{3}=\\text{1.1}\\times {10}^{-18}[\/latex]\r\n\r\n[latex]x=\\sqrt[3]{\\left(\\frac{1.1\\times {10}^{-18}}{4}\\right)}=\\text{6.5}\\times {10}^{-7}\\text{}M[\/latex]\r\n\r\n[latex]\\left[{\\text{Hg}}_{2}{}^{\\text{2+}}\\right]=\\text{6.5}\\times {10}^{-7}\\text{}M=\\text{6.5}\\times {10}^{-7}\\text{}M[\/latex]\r\n\r\n[latex]\\left[\\text{Cl}^{-}\\right]=2x=2\\left(6.5\\times{10}^{-7}\\right)=1.3\\times {10}^{-6}M[\/latex]\r\n\r\nThe molar solubility of Hg<sub>2<\/sub>Cl<sub>2<\/sub> is equal to [latex]{\\text{[Hg}}_{2}{}^{\\text{2+}}][\/latex], or 6.5 \u00d7 10<sup>\u20137<\/sup><em>M<\/em>.<\/li>\r\n \t<li><em>Check the work.<\/em> At equilibrium, <em>Q<\/em> = <em>K<\/em><sub>sp<\/sub>: [latex]Q=\\left[\\text{Hg}_{2}^{2+}\\right]\\left[\\text{Cl}^{-}\\right]^{2}=\\left(6.5\\times{10}^{-7}\\right)\\left(1.3\\times{10}^{-6}\\right)^{2}=1.1\\times{10}^{-18}[\/latex]\r\nThe calculations check.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n<h4>Check Your Learning<\/h4>\r\nDetermine the molar solubility of MgF<sub>2<\/sub> from its solubility product: <em>K<\/em><sub>sp<\/sub> = 6.4 \u00d7 10<sup>\u20139<\/sup>.\r\n[reveal-answer q=\"24056\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"24056\"]1.2 \u00d7 10<sup>\u20133<\/sup><em>M<\/em>[\/hidden-answer]\r\n\r\n<\/div>\r\nTabulated <em>K<\/em><sub>sp<\/sub> values can also be compared to reaction quotients calculated from experimental data to tell whether a solid will precipitate in a reaction under specific conditions: <em>Q<\/em> equals <em>K<\/em><sub>sp<\/sub> at equilibrium; if <em>Q<\/em> is less than <em>K<\/em><sub>sp<\/sub>, the solid will dissolve until <em>Q<\/em> equals <em>K<\/em><sub>sp<\/sub>; if <em>Q<\/em> is greater than <em>K<\/em><sub>sp<\/sub>, precipitation will occur at a given temperature until <em>Q<\/em> equals <em>K<\/em><sub>sp<\/sub>.\r\n<div class=\"textbox shaded\">\r\n<h3 data-type=\"title\">Using Barium Sulfate for Medical Imaging<\/h3>\r\nVarious types of medical imaging techniques are used to aid diagnoses of illnesses in a noninvasive manner. One such technique utilizes the ingestion of a barium compound before taking an X-ray image. A suspension of barium sulfate, a chalky powder, is ingested by the patient. Since the <em>K<\/em><sub>sp<\/sub> of barium sulfate is 1.1 \u00d7 10<sup>\u201310<\/sup>, very little of it dissolves as it coats the lining of the patient\u2019s intestinal tract. Barium-coated areas of the digestive tract then appear on an X-ray as white, allowing for greater visual detail than a traditional X-ray (Figure\u00a03).\r\n\r\n[caption id=\"attachment_4945\" align=\"aligncenter\" width=\"601\"]<img class=\"wp-image-4945\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213946\/CNX_Chem_15_01_BariumXray.jpg\" alt=\"This figure contains one image. A black and white abdominal x-ray image is shown in which the intestinal tract of a person is clearly visible in white.\" width=\"601\" height=\"404\" \/> Figure\u00a03. The suspension of barium sulfate coats the intestinal tract, which allows for greater visual detail than a traditional X-ray. (credit modification of work by \u201cglitzy queen00\u201d\/Wikimedia Commons)[\/caption]\r\n\r\nFurther diagnostic testing can be done using barium sulfate and fluoroscopy. In fluoroscopy, a continuous X-ray is passed through the body so the doctor can monitor, on a TV or computer screen, the barium sulfate\u2019s movement as it passes through the digestive tract. Medical imaging using barium sulfate can be used to diagnose acid reflux disease, Crohn\u2019s disease, and ulcers in addition to other conditions.\r\n\r\nVisit <a href=\"http:\/\/www.hopkinsmedicine.org\/healthlibrary\/conditions\/adult\/digestive_disorders\/barium_x-rays_upper_and_lower_gi_85,p01275\/\" target=\"_blank\">this website for more information on how barium is used in medical diagnoses<\/a> and which conditions it is used to diagnose.\r\n\r\n<\/div>\r\n<h2>Predicting Precipitation<\/h2>\r\nThe equation that describes the equilibrium between solid calcium carbonate and its solvated ions is:\r\n<p style=\"text-align: center;\">[latex]{\\text{CaCO}}_{3}\\left(s\\right)\\rightleftharpoons {\\text{Ca}}^{\\text{2+}}\\left(aq\\right)+{\\text{CO}}_{3}^{2-}\\left(aq\\right)[\/latex]<\/p>\r\nWe can establish this equilibrium either by adding solid calcium carbonate to water or by mixing a solution that contains calcium ions with a solution that contains carbonate ions. If we add calcium carbonate to water, the solid will dissolve until the concentrations are such that the value of the reaction quotient (<em>Q<\/em> = [Ca<sup>2+<\/sup>] [latex]\\left[{\\text{CO}}_{3}^{2-}\\right][\/latex] ) is equal to the solubility product (<em>K<\/em><sub>sp<\/sub> = 4.8 \u00d7 10<sup>\u20139<\/sup>). If we mix a solution of calcium nitrate, which contains Ca<sup>2+<\/sup> ions, with a solution of sodium carbonate, which contains [latex]{\\text{CO}}_{3}^{2-}[\/latex] ions, the slightly soluble ionic solid CaCO<sub>3<\/sub> will precipitate, provided that the concentrations of Ca<sup>2+<\/sup> and [latex]{\\text{CO}}_{3}^{2-}[\/latex] ions are such that <em>Q<\/em> is greater than <em>K<\/em><sub>sp<\/sub> for the mixture. The reaction shifts to the left and the concentrations of the ions are reduced by formation of the solid until the value of <em>Q<\/em> equals <em>K<\/em><sub>sp<\/sub>. A saturated solution in equilibrium with the undissolved solid will result. If the concentrations are such that <em>Q<\/em> is less than <em>K<\/em><sub>sp<\/sub>, then the solution is not saturated and no precipitate will form.\r\n\r\nWe can compare numerical values of <em>Q<\/em> with <em>K<\/em><sub>sp<\/sub> to predict whether precipitation will occur, as Example 7\u00a0shows. (Note: Since all forms of equilibrium constants are temperature dependent, we will assume a room temperature environment going forward in this chapter unless a different temperature value is explicitly specified.)\r\n<div class=\"textbox examples\">\r\n<h3>Example 7:\u00a0Precipitation of Mg(OH)<sub>2<\/sub><\/h3>\r\nThe first step in the preparation of magnesium metal is the precipitation of Mg(OH)<sub>2<\/sub> from sea water by the addition of lime, Ca(OH)<sub>2<\/sub>, a readily available inexpensive source of OH<sup>\u2013<\/sup> ion:\r\n<p style=\"text-align: center;\">[latex]{\\text{Mg(OH)}}_{2}\\left(s\\right)\\rightleftharpoons {\\text{Mg}}^{\\text{2+}}\\left(aq\\right)+{\\text{2OH}}^{-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,{;}\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{sp}}=\\text{2.1}\\times {10}^{-13}[\/latex]<\/p>\r\nThe concentration of Mg<sup>2+<\/sup>(<em>aq<\/em>) in sea water is 0.0537 <em>M<\/em>. Will Mg(OH)<sub>2<\/sub> precipitate when enough Ca(OH)<sub>2<\/sub> is added to give a [OH<sup>\u2013<\/sup>] of 0.0010 <em>M<\/em>?\r\n[reveal-answer q=\"543435\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"543435\"]\r\n\r\nThis problem asks whether the reaction:\r\n<p style=\"text-align: center;\">[latex]{\\text{Mg(OH)}}_{2}\\left(s\\right)\\rightleftharpoons {\\text{Mg}}^{\\text{2+}}\\left(aq\\right)+{\\text{2OH}}^{-}\\left(aq\\right)[\/latex]<\/p>\r\nshifts to the left and forms solid Mg(OH)<sub>2<\/sub> when [Mg<sup>2+<\/sup>] = 0.0537 <em>M<\/em> and [OH<sup>\u2013<\/sup>] = 0.0010 <em>M<\/em>. The reaction shifts to the left if <em>Q<\/em> is greater than <em>K<\/em><sub>sp<\/sub>. Calculation of the reaction quotient under these conditions is shown here:\r\n<p style=\"text-align: center;\">[latex]Q=\\left[\\text{Mg}^{2+}\\right]\\left[\\text{OH}^{-}\\right]^{2}=\\left(0.0537\\right)\\left(0.0010\\right)^{2}=5.4\\times{10}^{-8}[\/latex]<\/p>\r\nBecause <em>Q<\/em> is greater than <em>K<\/em><sub>sp<\/sub> (<em>Q<\/em> = 5.4 \u00d7 10<sup>\u20138<\/sup> is larger than <em>K<\/em><sub>sp<\/sub> = 2.1 \u00d7 10<sup>\u201313<\/sup>), we can expect the reaction to shift to the left and form solid magnesium hydroxide. Mg(OH)<sub>2<\/sub>(<em>s<\/em>) forms until the concentrations of magnesium ion and hydroxide ion are reduced sufficiently so that the value of <em>Q<\/em> is equal to <em>K<\/em><sub>sp<\/sub>.\r\n\r\n[\/hidden-answer]\r\n<h4>Check Your Learning<\/h4>\r\nUse the solubility product in\u00a0<a class=\"target-chapter\" href=\".\/chapter\/solubility-products\/\" target=\"_blank\">Solubility Products<\/a>\u00a0to determine whether CaHPO<sub>4<\/sub> will precipitate from a solution with [Ca<sup>2+<\/sup>] = 0.0001 <em>M<\/em> and [latex]\\left[{\\text{HPO}}_{4}{}^{2-}\\right][\/latex] = 0.001 <em>M<\/em>.\r\n[reveal-answer q=\"41133\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"41133\"]No precipitation of CaHPO<sub>4<\/sub>; <em>Q<\/em> = 1 \u00d7 10<sup>\u20137<\/sup>, which is less than <em>K<\/em><sub>sp<\/sub>[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 8:\u00a0Precipitation of AgCl upon Mixing Solutions<\/h3>\r\nDoes silver chloride precipitate when equal volumes of a 2.0 \u00d7 10<sup>\u20134<\/sup>-<em>M<\/em> solution of AgNO<sub>3<\/sub> and a 2.0 \u00d7 10<sup>\u20134<\/sup>-<em>M<\/em> solution of NaCl are mixed?\r\n\r\n(Note: The solution also contains Na<sup>+<\/sup> and [latex]{\\text{NO}}_{3}^{-}[\/latex] ions, but when referring to solubility rules, one can see that sodium nitrate is very soluble and cannot form a precipitate.)\r\n[reveal-answer q=\"81096\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"81096\"]\r\n\r\nThe equation for the equilibrium between solid silver chloride, silver ion, and chloride ion is:\r\n<p style=\"text-align: center;\">[latex]\\text{AgCl}\\left(s\\right)\\rightleftharpoons {\\text{Ag}}^{\\text{+}}\\left(aq\\right)+{\\text{Cl}}^{-}\\left(aq\\right)[\/latex]<\/p>\r\nThe solubility product is 1.8 \u00d7 10<sup>\u201310<\/sup> (see\u00a0<a class=\"target-chapter\" href=\".\/chapter\/solubility-products\/\" target=\"_blank\">Solubility Products<\/a>).\r\n\r\nAgCl will precipitate if the reaction quotient calculated from the concentrations in the mixture of AgNO<sub>3<\/sub> and NaCl is greater than <em>K<\/em><sub>sp<\/sub>. The volume doubles when we mix equal volumes of AgNO<sub>3<\/sub> and NaCl solutions, so each concentration is reduced to half its initial value. Consequently, immediately upon mixing, [Ag<sup>+<\/sup>] and [Cl<sup>\u2013<\/sup>] are both equal to:\r\n<p style=\"text-align: center;\">[latex]\\frac{1}{2}\\left(2.0\\times {10}^{-4}\\right)\\text{}M=\\text{1.0}\\times {10}^{-4}\\text{}M[\/latex].<\/p>\r\nThe reaction quotient, <em>Q<\/em>, is <em>momentarily<\/em> greater than <em>K<\/em><sub>sp<\/sub> for AgCl, so a supersaturated solution is formed:\r\n<p style=\"text-align: center;\">[latex]Q=\\left[\\text{Ag}^{+}\\right]\\left[\\text{Cl}^{-}\\right]=\\left(1.0\\times{10}^{-4}\\right)\\left(1.0\\times {10}^{-4}\\right)=1.0\\times{10}^{-8}\\gt{K}_{\\text{sp}}[\/latex]<\/p>\r\nSince supersaturated solutions are unstable, AgCl will precipitate from the mixture until the solution returns to equilibrium, with <em>Q<\/em> equal to <em>K<\/em><sub>sp<\/sub>.\r\n\r\n[\/hidden-answer]\r\n<h4>Check Your Learning<\/h4>\r\nWill KClO<sub>4<\/sub> precipitate when 20 mL of a 0.050-<em>M<\/em> solution of K<sup>+<\/sup> is added to 80 mL of a 0.50-<em>M<\/em> solution of [latex]{\\text{ClO}}_{4}{}^{-}[\/latex]? (Remember to calculate the new concentration of each ion after mixing the solutions before plugging into the reaction quotient expression.)\r\n[reveal-answer q=\"920440\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"920440\"]No, <em>Q<\/em> = 4.0 \u00d7 10<sup>\u20133<\/sup>, which is less than <em>K<\/em><sub>sp<\/sub> = 1.07 \u00d7 10<sup>\u20132<\/sup>[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the previous two examples, we have seen that Mg(OH)<sub>2<\/sub> or AgCl precipitate when <em>Q<\/em> is greater than <em>K<\/em><sub>sp<\/sub>. In general, when a solution of a soluble salt of the M<sup>m+<\/sup> ion is mixed with a solution of a soluble salt of the X<sup>n\u2013<\/sup> ion, the solid, M<sub>p<\/sub>X<sub>q<\/sub> precipitates if the value of <em>Q<\/em> for the mixture of M<sup>m+<\/sup> and X<sup>n\u2013<\/sup> is greater than <em>K<\/em><sub>sp<\/sub> for M<sub>p<\/sub>X<sub>q<\/sub>. Thus, if we know the concentration of one of the ions of a slightly soluble ionic solid and the value for the solubility product of the solid, then we can calculate the concentration that the other ion must exceed for precipitation to begin. To simplify the calculation, we will assume that precipitation begins when the reaction quotient becomes equal to the solubility product constant.\r\n<div class=\"textbox examples\">\r\n<h3>Example 9:\u00a0Precipitation of Calcium Oxalate<\/h3>\r\nBlood will not clot if calcium ions are removed from its plasma. Some blood collection tubes contain salts of the oxalate ion, [latex]{\\text{C}}_{2}{\\text{O}}_{4}^{2-}[\/latex], for this purpose (Figure\u00a04). At sufficiently high concentrations, the calcium and oxalate ions form solid, CaC<sub>2<\/sub>O<sub>4<\/sub>\u00b7H<sub>2<\/sub>O (which also contains water bound in the solid). The concentration of Ca<sup>2+<\/sup> in a sample of blood serum is 2.2 \u00d7 10<sup>\u20133<\/sup><em>M<\/em>. What concentration of [latex]{\\text{C}}_{2}{\\text{O}}_{4}^{2-}[\/latex] ion must be established before CaC<sub>2<\/sub>O<sub>4<\/sub>\u00b7H<sub>2<\/sub>O begins to precipitate?\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"500\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213948\/CNX_Chem_15_01_Blood.jpg\" alt=\"A photograph is shown of 6 vials of blood resting on and near a black and white document. Two of the vials have purple caps, three have tan caps, and one has a red cap. Each has a label and the vials with tan caps have a small amount of an off-white material present in a layer at the base of the vial.\" width=\"500\" height=\"487\" data-media-type=\"image\/jpeg\" \/> Figure\u00a04. Anticoagulants can be added to blood that will combine with the Ca<sup>2+<\/sup> ions in blood serum and prevent the blood from clotting. (credit: modification of work by Neeta Lind)[\/caption]\r\n\r\n[reveal-answer q=\"639745\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"639745\"]\r\n\r\nThe equilibrium expression is:\r\n<p style=\"text-align: center;\">[latex]{\\text{CaC}}_{2}{\\text{O}}_{4}\\left(s\\right)\\rightleftharpoons {\\text{Ca}}^{\\text{2+}}\\left(aq\\right)+{\\text{C}}_{2}{\\text{O}}_{4}{}^{2-}\\left(aq\\right)[\/latex]<\/p>\r\nFor this reaction:\r\n<p style=\"text-align: center;\">[latex]{K}_{\\text{sp}}=\\left[{\\text{Ca}}^{\\text{2+}}\\right]\\left[{\\text{C}}_{2}{\\text{O}}_{4}^{2-}\\right]=2.27\\times {10}^{-9}[\/latex] (see\u00a0<a class=\"target-chapter\" href=\".\/chapter\/solubility-products\/\" target=\"_blank\">Solubility Products<\/a>)<\/p>\r\nCaC<sub>2<\/sub>O<sub>4<\/sub> does not appear in this expression because it is a solid. Water does not appear because it is the solvent.\r\n\r\nSolid CaC<sub>2<\/sub>O<sub>4<\/sub> does not begin to form until <em>Q<\/em> equals <em>K<\/em><sub>sp<\/sub>. Because we know <em>K<\/em><sub>sp<\/sub> and [Ca<sup>2+<\/sup>], we can solve for the concentration of [latex]{\\text{C}}_{2}{\\text{O}}_{4}^{2-}[\/latex] that is necessary to produce the first trace of solid:\r\n<p style=\"text-align: center;\">[latex]Q={K}_{\\text{sp}}=\\left[\\text{Ca}^{2+}\\right]\\left[\\text{C}_{2}\\text{O}_{4}^{2-}\\right]=2.27\\times{10}^{-9}[\/latex]\r\n[latex]\\left(2.2\\times {10}^{-3}\\right){\\text{[C}}_{2}{\\text{O}}_{4}{}^{\\text{2-}}]=\\text{2.27}\\times {10}^{-9}[\/latex]\r\n[latex]\\left[{\\text{C}}_{2}{\\text{O}}_{4}{}^{2-}\\right]=\\frac{2.27\\times {10}^{-9}}{2.2\\times {10}^{-3}}=1.0\\times {10}^{-6}[\/latex]<\/p>\r\nA concentration of [latex]\\left[{\\text{C}}_{2}{\\text{O}}_{4}^{2-}\\right][\/latex] = 1.0 \u00d7 10<sup>\u20136<\/sup><em>M<\/em> is necessary to initiate the precipitation of CaC<sub>2<\/sub>O<sub>4<\/sub> under these conditions.\r\n\r\n[\/hidden-answer]\r\n<h4>Check Your Learning<\/h4>\r\nIf a solution contains 0.0020 mol of [latex]{\\text{CrO}}_{4}^{2-}[\/latex] per liter, what concentration of Ag<sup>+<\/sup> ion must be reached by adding solid AgNO<sub>3<\/sub> before Ag<sub>2<\/sub>CrO<sub>4<\/sub> begins to precipitate? Neglect any increase in volume upon adding the solid silver nitrate.\r\n[reveal-answer q=\"24336\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"24336\"]7.0 \u00d7 10<sup>\u20135<\/sup><em>M<\/em>[\/hidden-answer]\r\n\r\n<\/div>\r\nIt is sometimes useful to know the concentration of an ion that remains in solution after precipitation. We can use the solubility product for this calculation too: If we know the value of <em>K<\/em><sub>sp<\/sub> and the concentration of one ion in solution, we can calculate the concentration of the second ion remaining in solution. The calculation is of the same type as that in Example 9\u2014calculation of the concentration of a species in an equilibrium mixture from the concentrations of the other species and the equilibrium constant. However, the concentrations are different; we are calculating concentrations after precipitation is complete, rather than at the start of precipitation.\r\n<div class=\"textbox examples\">\r\n<h3>Example 10:\u00a0Concentrations Following Precipitation<\/h3>\r\nClothing washed in water that has a manganese [Mn<sup>2+<\/sup>(<em>aq<\/em>)] concentration exceeding 0.1 mg\/L (1.8 \u00d7 10<sup>\u20136<\/sup><em>M<\/em>) may be stained by the manganese upon oxidation, but the amount of Mn<sup>2+<\/sup> in the water can be reduced by adding a base. If a person doing laundry wishes to add a buffer to keep the pH high enough to precipitate the manganese as the hydroxide, Mn(OH)<sub>2<\/sub>, what pH is required to keep [Mn<sup>2+<\/sup>] equal to 1.8 \u00d7 10<sup>\u20136<\/sup><em>M<\/em>?\r\n[reveal-answer q=\"943057\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"943057\"]\r\n\r\nThe dissolution of Mn(OH)<sub>2<\/sub> is described by the equation:\r\n<p style=\"text-align: center;\">[latex]{\\text{Mn(OH)}}_{2}\\left(s\\right)\\rightleftharpoons {\\text{Mn}}^{\\text{2+}}\\left(aq\\right)+{\\text{2OH}}^{-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,{;}\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{sp}}=\\text{4.5}\\times {10}^{-14}[\/latex]<\/p>\r\nWe need to calculate the concentration of OH<sup>\u2013<\/sup> when the concentration of Mn<sup>2+<\/sup> is 1.8 \u00d7 10<sup>\u20136<\/sup><em>M<\/em>. From that, we calculate the pH. At equilibrium:\r\n<p style=\"text-align: center;\">[latex]{K}_{\\text{sp}}=\\left[\\text{Mn}^{2+}\\right]\\left[\\text{OH}^{-}\\right]^{2}[\/latex]\u00a0or\u00a0[latex]\\left(1.8\\times {10}^{-6}\\right){\\left[{\\text{OH}}^{-}\\right]}^{2}=\\text{4.5}\\times {10}^{-14}[\/latex]<\/p>\r\nso\u00a0[latex]\\left[{\\text{OH}}^{-}\\right]=\\text{1.6}\\times {10}^{-4}M[\/latex].\r\n\r\nNow we calculate the pH from the pOH:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\text{pOH}=-\\text{log}\\left[{\\text{OH}}^{-}\\right]=-\\text{log}\\left(1.6\\times 10 - 4\\right)=\\text{3.80}\\\\ \\text{pH}=\\text{14.00}-\\text{pOH}=\\text{14.00}-\\text{3.80}=\\text{10.20}\\end{array}[\/latex]<\/p>\r\nIf the person doing laundry adds a base, such as the sodium silicate (Na<sub>4<\/sub>SiO<sub>4<\/sub>) in some detergents, to the wash water until the pH is raised to 10.20, the manganese ion will be reduced to a concentration of 1.8 \u00d7 10<sup>\u20136<\/sup><em>M<\/em>; at that concentration or less, the ion will not stain clothing.\r\n\r\n[\/hidden-answer]\r\n<h4>Check Your Learning<\/h4>\r\nThe first step in the preparation of magnesium metal is the precipitation of Mg(OH)<sub>2<\/sub> from sea water by the addition of Ca(OH)<sub>2<\/sub>. The concentration of Mg<sup>2+<\/sup>(<em>aq<\/em>) in sea water is 5.37 \u00d7 10<sup>\u20132<\/sup><em>M<\/em>. Calculate the pH at which [Mg<sup>2+<\/sup>] is diminished to 1.0 \u00d7 10<sup>\u20135<\/sup><em>M<\/em> by the addition of Ca(OH)<sub>2<\/sub>.\r\n[reveal-answer q=\"138585\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"138585\"]11.09[\/hidden-answer]\r\n\r\n<\/div>\r\nDue to their light sensitivity, mixtures of silver halides are used in fiber optics for medical lasers, in photochromic eyeglass lenses (glass lenses that automatically darken when exposed to sunlight), and\u2014before the advent of digital photography\u2014in photographic film. Even though AgCl (<em>K<\/em><sub>sp<\/sub> = 1.6 \u00d7 10<sup>\u201310<\/sup>), AgBr (<em>K<\/em><sub>sp<\/sub> = 7.7 \u00d7 10<sup>\u201313<\/sup>), and AgI (<em>K<\/em><sub>sp<\/sub> = 8.3 \u00d7 10<sup>\u201317<\/sup>) are each quite insoluble, we cannot prepare a homogeneous solid mixture of them by adding Ag<sup>+<\/sup> to a solution of Cl<sup>\u2013<\/sup>, Br<sup>\u2013<\/sup>, and I<sup>\u2013<\/sup>; essentially all of the AgI will precipitate before any of the other solid halides form because of its smaller value for <em>K<\/em><sub>sp<\/sub>. However, we can prepare a homogeneous mixture of the solids by slowly adding a solution of Cl<sup>\u2013<\/sup>, Br<sup>\u2013<\/sup>, and I<sup>\u2013<\/sup> to a solution of Ag<sup>+<\/sup>.\r\n\r\nWhen two anions form slightly soluble compounds with the same cation, or when two cations form slightly soluble compounds with the same anion, the less soluble compound (usually, the compound with the smaller <em>K<\/em><sub>sp<\/sub>) generally precipitates first when we add a precipitating agent to a solution containing both anions (or both cations). When the <em>K<\/em><sub>sp<\/sub> values of the two compounds differ by two orders of magnitude or more (e.g., 10<sup>\u20132<\/sup> vs. 10<sup>\u20134<\/sup>), almost all of the less soluble compound precipitates before any of the more soluble one does. This is an example of <b>selective precipitation<\/b>, where a reagent is added to a solution of dissolved ions causing one of the ions to precipitate out before the rest.\r\n<div class=\"textbox shaded\">\r\n<h3 data-type=\"title\">The Role of Precipitation in Wastewater Treatment<\/h3>\r\nSolubility equilibria are useful tools in the treatment of wastewater carried out in facilities that may treat the municipal water in your city or town (Figure\u00a05). Specifically, selective precipitation is used to remove contaminants from wastewater before it is released back into natural bodies of water. For example, phosphate ions [latex]{\\text{(PO}}_{4}^{2-})[\/latex] are often present in the water discharged from manufacturing facilities. An abundance of phosphate causes excess algae to grow, which impacts the amount of oxygen available for marine life as well as making water unsuitable for human consumption.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"500\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213949\/CNX_Chem_15_01_Wastewater.jpg\" alt=\"A color photograph is shown of a high volume wastewater treatment facility. Nineteen large circular pools of water undergoing treatment are visible across the center of the photograph. A building and parking lot are visible in the foreground.\" width=\"500\" height=\"338\" data-media-type=\"image\/jpeg\" \/> Figure\u00a05. Wastewater treatment facilities, such as this one, remove contaminants from wastewater before the water is released back into the natural environment. (credit: \u201ceutrophication&amp;hypoxia\u201d\/Wikimedia Commons)[\/caption]\r\n\r\nOne common way to remove phosphates from water is by the addition of calcium hydroxide, known as lime, Ca(OH)<sub>2<\/sub>. The lime is converted into calcium carbonate, a strong base, in the water. As the water is made more basic, the calcium ions react with phosphate ions to produce hydroxylapatite, Ca<sub>5<\/sub>(PO4)<sub>3<\/sub>(OH), which then precipitates out of the solution:\r\n<p style=\"text-align: center;\">[latex]5{\\text{Ca}}^{\\text{2+}}+{\\text{3PO}}_{4}{}^{3-}+{\\text{OH}}^{-}\\rightleftharpoons {\\text{Ca}}_{10}{{\\text{(PO}}_{4})}_{6}\\cdot {\\text{(OH)}}_{2}\\left(s\\right)[\/latex]<\/p>\r\nThe precipitate is then removed by filtration and the water is brought back to a neutral pH by the addition of CO<sub>2<\/sub> in a recarbonation process. Other chemicals can also be used for the removal of phosphates by precipitation, including iron(III) chloride and aluminum sulfate.\r\n\r\n<a href=\"http:\/\/science.jrank.org\/pages\/5146\/Phosphorus-Removal.html\" target=\"_blank\">View this site for more information on how phosphorus is removed from wastewater.<\/a>\r\n\r\n<\/div>\r\nSelective precipitation can also be used in qualitative analysis. In this method, reagents are added to an unknown chemical mixture in order to induce precipitation. Certain reagents cause specific ions to precipitate out; therefore, the addition of the reagent can be used to determine whether the ion is present in the solution.\r\n<div class=\"textbox\">View <a href=\"http:\/\/employees.oneonta.edu\/viningwj\/sims\/common_ion_effect_s1.html\" target=\"_blank\">this simulation to study the process of salts dissolving and forming saturated solutions<\/a> and precipitates for specific compounds, or compounds for which you select the charges on the ions and the <em>K<\/em><sub>sp<\/sub><\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 11:\u00a0Precipitation of Silver Halides<\/h3>\r\nA solution contains 0.0010 mol of KI and 0.10 mol of KCl per liter. AgNO<sub>3<\/sub> is gradually added to this solution. Which forms first, solid AgI or solid AgCl?\r\n[reveal-answer q=\"657940\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"657940\"]\r\n\r\nThe two equilibria involved are:\r\n<p style=\"text-align: center;\">[latex]\\text{AgCl}\\left(s\\right)\\rightleftharpoons {\\text{Ag}}^{\\text{+}}\\left(aq\\right)+{\\text{Cl}}^{-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,{;}\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{sp}}=\\text{1.8}\\times {10}^{-10}[\/latex]\r\n[latex]\\text{AgI}\\left(s\\right)\\rightleftharpoons {\\text{Ag}}^{\\text{+}}\\left(aq\\right)+{\\text{I}}^{-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,{;}\\,\\,\\,\\,\\,\\,\\,K_{\\text{sp}}=\\text{1.5}\\times {10}^{-16}[\/latex]<\/p>\r\nIf the solution contained about <em>equal<\/em> concentrations of Cl<sup>\u2013<\/sup> and I<sup>\u2013<\/sup>, then the silver salt with the smallest <em>K<\/em><sub>sp<\/sub> (AgI) would precipitate first. The concentrations are not equal, however, so we should find the [Ag<sup>+<\/sup>] at which AgCl begins to precipitate and the [Ag<sup>+<\/sup>] at which AgI begins to precipitate. The salt that forms at the lower [Ag<sup>+<\/sup>] precipitates first.\r\n\r\nFor AgI: AgI precipitates when <em>Q<\/em> equals <em>K<\/em><sub>sp<\/sub> for AgI (1.5 \u00d7 10<sup>\u201316<\/sup>). When [I<sup>\u2013<\/sup>] = 0.0010 <em>M<\/em>:\r\n<p style=\"text-align: center;\">[latex]Q=\\left[\\text{Ag}^{+}\\right]\\left[\\text{I}^{-}\\right]=\\left[\\text{Ag}^{+}\\right]\\left(0.0010\\right)=1.5\\times {10}^{-16}[\/latex]\r\n[latex][{\\text{Ag}}^{\\text{+}}\\text{]}=\\frac{\\text{1.8}\\times {10}^{-10}}{0.10}=\\text{1.8}\\times {10}^{-9}[\/latex]<\/p>\r\nAgI begins to precipitate when [Ag<sup>+<\/sup>] is 1.5 \u00d7 10<sup>\u201313<\/sup><em>M<\/em>.\r\n\r\nFor AgCl: AgCl precipitates when <em>Q<\/em> equals <em>K<\/em><sub>sp<\/sub> for AgCl (1.8 \u00d7 10<sup>\u201310<\/sup>). When [Cl<sup>\u2013<\/sup>] = 0.10 <em>M<\/em>:\r\n<p style=\"text-align: center;\">[latex]{Q}_{\\text{sp}}=\\left[\\text{Ag}^{+}\\right]\\left[\\text{Cl}^{-}\\right]=\\left[\\text{Ag}^{+}\\right]\\left(0.10\\right)=1.8\\times{10}^{-10}[\/latex]\r\n[latex][{\\text{Ag}}^{\\text{+}}\\text{]}=\\frac{1.8\\times {10}^{-10}}{0.10}\\text{=1.8}\\times {10}^{-9}M[\/latex]<\/p>\r\nAgCl begins to precipitate when [Ag<sup>+<\/sup>] is 1.8 \u00d7 10<sup>\u20139<\/sup><em>M<\/em>.\r\n\r\nAgI begins to precipitate at a lower [Ag<sup>+<\/sup>] than AgCl, so AgI begins to precipitate first.\r\n\r\n[\/hidden-answer]\r\n<h4>Check Your Learning<\/h4>\r\nIf silver nitrate solution is added to a solution which is 0.050 <em>M<\/em> in both Cl<sup>\u2013<\/sup> and Br<sup>\u2013<\/sup> ions, at what [Ag<sup>+<\/sup>] would precipitation begin, and what would be the formula of the precipitate?\r\n[reveal-answer q=\"671573\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"671573\"][Ag<sup>+<\/sup>] = 1.5 \u00d7 10<sup>\u201311<\/sup><em>M<\/em>; AgBr precipitates first[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Common Ion Effect<\/h2>\r\nAs we saw when we discussed buffer solutions, the hydronium ion concentration of an aqueous solution of acetic acid decreases when the strong electrolyte sodium acetate, NaCH<sub>3<\/sub>CO<sub>2<\/sub>, is added. We can explain this effect using Le Ch\u00e2telier\u2019s principle. The addition of acetate ions causes the equilibrium to shift to the left, decreasing the concentration of [latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}[\/latex] to compensate for the increased acetate ion concentration. This increases the concentration of CH<sub>3<\/sub>CO<sub>2<\/sub>H:\r\n<p style=\"text-align: center;\">[latex]{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}+{\\text{H}}_{2}\\text{O}\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}+{\\text{CH}}_{3}{\\text{CO}}_{2}^{-}[\/latex]<\/p>\r\nBecause sodium acetate and acetic acid have the acetate ion in common, the influence on the equilibrium is called the <b>common ion effect<\/b>.\r\n\r\nThe common ion effect can also have a direct effect on solubility equilibria. Suppose we are looking at the reaction where silver iodide is dissolved:\r\n<p style=\"text-align: center;\">[latex]\\text{AgI}\\left(s\\right)\\rightleftharpoons {\\text{Ag}}^{\\text{+}}\\left(aq\\right)+{\\text{I}}^{-}\\left(aq\\right)[\/latex]<\/p>\r\nIf we were to add potassium iodide (KI) to this solution, we would be adding a substance that shares a common ion with silver iodide. Le Ch\u00e2telier\u2019s principle tells us that when a change is made to a system at equilibrium, the reaction will shift to counteract that change. In this example, there would be an excess of iodide ions, so the reaction would shift toward the left, causing more silver iodide to precipitate out of solution.\r\n<div class=\"textbox examples\">\r\n<h3>Example 12:\u00a0Common Ion Effect<\/h3>\r\nCalculate the molar solubility of cadmium sulfide (CdS) in a 0.010-<em>M<\/em> solution of cadmium bromide (CdBr<sub>2<\/sub>). The <em>K<\/em><sub>sp<\/sub> of CdS is 1.0 \u00d7 10<sup>\u201328<\/sup>.\r\n[reveal-answer q=\"848161\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"848161\"]\r\n\r\nThe first thing you should notice is that the cadmium sulfide is dissolved in a solution that contains cadmium ions. We need to use an ICE table to set up this problem and include the CdBr<sub>2<\/sub> concentration as a contributor of cadmium ions:\r\n<p style=\"text-align: center;\">[latex]\\text{CdS}\\left(s\\right)\\rightleftharpoons {\\text{Cd}}^{2+}\\left(aq\\right)+{\\text{S}}^{2-}\\left(aq\\right)[\/latex]<\/p>\r\n<p style=\"text-align: center;\"><img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213951\/CNX_Chem_15_01_ICETable3_img.jpg\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), and Equilibrium concentration ( M ). The second column has the header, \u201cC d S equilibrium arrow C d to the second power plus S to the second power superscript negative sign.\u201d Under the second column is a subgroup of three rows and three columns. The first column is blank. The second column has the following: 0.010, x, 0.010 plus x. The third column has the following: 0, x, 0 plus x equals x.\" width=\"879\" height=\"238\" data-media-type=\"image\/jpeg\" \/><\/p>\r\n<p style=\"text-align: center;\">[latex]{K}_{\\text{sp}}=\\left[{\\text{Cd}}^{\\text{2+}}\\right]\\left[{\\text{S}}^{2-}\\right]=1.0\\times {10}^{-28}[\/latex]\r\n[latex]\\left(0.010+x\\right)\\left(x\\right)=\\text{1.0}\\times {10}^{-28}[\/latex]\r\n[latex]{x}^{2}+\\text{0.010}x-\\text{1.0}\\times {10}^{-28}=0[\/latex]<\/p>\r\nWe can solve this equation using the quadratic formula, but we can also make an assumption to make this calculation much simpler. Since the <em>K<\/em><sub>sp<\/sub> value is so small compared with the cadmium concentration, we can assume that the change between the initial concentration and the equilibrium concentration is negligible, so that 0.010 + <em>x<\/em> ~ 0.010. Going back to our <em>K<\/em><sub>sp<\/sub> expression, we would now get:\r\n<p style=\"text-align: center;\">[latex]{K}_{\\text{sp}}=\\left[\\text{Cd}^{2+}\\right]\\left[\\text{S}^{2-}\\right]=1.0\\times{10}^{-28}[\/latex]\r\n[latex]\\left(0.010\\right)\\left(x\\right)=\\text{1.0}\\times {10}^{-28}[\/latex]\r\n[latex]x=\\text{1.0}\\times {10}^{-26}[\/latex]<\/p>\r\nTherefore, the molar solubility of CdS in this solution is 1.0 \u00d7 10<sup>\u201326<\/sup><em>M<\/em>.\r\n\r\n[\/hidden-answer]\r\n<h4>Check Your Learning<\/h4>\r\nCalculate the molar solubility of aluminum hydroxide, Al(OH)<sub>3<\/sub>, in a 0.015-<em>M<\/em> solution of aluminum nitrate, Al(NO<sub>3<\/sub>)<sub>3<\/sub>. The <em>K<\/em><sub>sp<\/sub> of Al(OH)<sub>3<\/sub> is 2 \u00d7 10<sup>\u201332<\/sup>.\r\n[reveal-answer q=\"176865\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"176865\"]1 \u00d7 10<sup>\u201310<\/sup><em>M<\/em>[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Key Concepts and Summary<\/h3>\r\nThe equilibrium constant for an equilibrium involving the precipitation or dissolution of a slightly soluble ionic solid is called the solubility product, <em>K<\/em><sub>sp<\/sub>, of the solid. When we have a heterogeneous equilibrium involving the slightly soluble solid M<sub>p<\/sub>X<sub>q<\/sub> and its ions M<sup>m+<\/sup> and X<sup>n\u2013<\/sup>:\r\n<p style=\"text-align: center;\">[latex]\\text{M}_{\\text{p}}{\\text{X}}_{\\text{q}}\\left(s\\right)\\rightleftharpoons p{\\text{M}}^{\\text{m+}}\\left(aq\\right)+q{\\text{X}}^{\\text{n}-}\\left(aq\\right)[\/latex]<\/p>\r\nWe write the solubility product expression as:\r\n<p style=\"text-align: center;\">[latex]{K}_{\\text{sp}}={{\\text{[M}}^{\\text{m+}}\\text{]}}^{\\text{p}}{{\\text{[X}}^{\\text{n}-}]}^{\\text{q}}[\/latex]<\/p>\r\nThe solubility product of a slightly soluble electrolyte can be calculated from its solubility; conversely, its solubility can be calculated from its <em>K<\/em><sub>sp<\/sub>, provided the only significant reaction that occurs when the solid dissolves is the formation of its ions.\r\n\r\nA slightly soluble electrolyte begins to precipitate when the magnitude of the reaction quotient for the dissolution reaction exceeds the magnitude of the solubility product. Precipitation continues until the reaction quotient equals the solubility product.\r\n\r\nA reagent can be added to a solution of ions to allow one ion to selectively precipitate out of solution. The common ion effect can also play a role in precipitation reactions. In the presence of an ion in common with one of the ions in the solution, Le Ch\u00e2telier\u2019s principle applies and more precipitate comes out of solution so that the molar solubility is reduced.\r\n<h4>Key Equations<\/h4>\r\n<ul>\r\n \t<li>[latex]\\text{Mp}_{\\text{Xq}}\\left(s\\right)\\rightleftharpoons p{\\text{M}}^{\\text{m+}}\\left(aq\\right)+q{\\text{X}}^{\\text{n}-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,{;}\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{sp}}=\\left[\\text{M}^{\\text{m+}}\\right]^{\\text{p}}\\left[\\text{X}^{\\text{n}-}\\right]^\\text{q}[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Exercises<\/h3>\r\n<ol>\r\n \t<li>Complete the changes in concentrations for each of the following reactions:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\begin{array}{ccc}\\text{AgI}\\left(s\\right)\\longrightarrow &amp; {\\text{Ag}}^{\\text{+}}\\left(aq\\right)&amp; +{\\text{I}}^{-}\\left(aq\\right)\\\\ &amp; x&amp;\\text{ _____}\\end{array}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{array}{ccc}{\\text{CaCO}}_{3}\\left(s\\right)\\longrightarrow &amp; {\\text{Ca}}^{\\text{2+}}\\left(aq\\right)+&amp; {\\text{CO}}_{3}{}^{2-}\\left(aq\\right)\\\\ &amp; \\text{____}&amp; x\\end{array}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{array}{ccc}\\text{Mg}{\\left(\\text{OH}\\right)}_{2}\\left(s\\right)\\longrightarrow &amp; {\\text{Mg}}^{\\text{2+}}\\left(aq\\right)+&amp; 2{\\text{OH}}^{-}\\left(aq\\right)\\\\ &amp; x&amp;\\text{ _____}\\end{array}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{array}{ccc}{\\text{Mg}}_{3}{\\left({\\text{PO}}_{4}\\right)}_{2}\\left(s\\right)\\longrightarrow &amp; 3{\\text{Mg}}^{\\text{2+}}\\left(aq\\right)+&amp; 2{\\text{PO}}_{4}{}^{3-}\\left(aq\\right)\\\\ &amp; &amp; x\\text{_____}\\end{array}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{array}{cccc}{\\text{Ca}}_{5}{\\left({\\text{PO}}_{4}\\right)}_{3}\\text{OH}\\left(s\\right)\\longrightarrow &amp; 5{\\text{Ca}}^{\\text{2+}}\\left(aq\\right)+&amp; 3{\\text{PO}}_{4}^{3-}\\left(aq\\right)+&amp; {\\text{OH}}^{-}\\left(aq\\right)\\\\ &amp; \\text{_____}&amp; \\text{_____}&amp; x\\end{array}[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Complete the changes in concentrations for each of the following reactions:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\begin{array}{ccc}{\\text{BaSO}}_{4}\\left(s\\right)\\longrightarrow &amp; {\\text{Ba}}^{\\text{2+}}\\left(aq\\right)+&amp; {\\text{SO}}_{4}{}^{2-}\\left(aq\\right)\\\\ &amp; x&amp; \\text{_____}\\end{array}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{array}{ccc}{\\text{Ag}}_{2}{\\text{SO}}_{4}\\left(s\\right)\\longrightarrow &amp; 2{\\text{Ag}}^{\\text{+}}\\left(aq\\right)+&amp; {\\text{SO}}_{4}^{2-}\\left(aq\\right)\\\\ &amp; \\text{_____}&amp; x\\end{array}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{array}{ccc}\\text{Al}{\\left(\\text{OH}\\right)}_{3}\\left(s\\right)\\longrightarrow &amp; {\\text{Al}}^{\\text{3+}}\\left(aq\\right)+&amp; 3{\\text{OH}}^{-}\\left(aq\\right)\\\\ &amp; x&amp; \\text{_____}\\end{array}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{array}{cccc}\\text{Pb}\\left(\\text{OH}\\right)\\text{Cl}\\left(s\\right)\\longrightarrow &amp; {\\text{Pb}}^{\\text{2+}}\\left(aq\\right)+&amp; {\\text{OH}}^{-}\\left(aq\\right)+&amp; {\\text{Cl}}^{-}\\left(aq\\right)\\\\ &amp; \\text{_____}\\text{}&amp; x&amp; \\text{_____}\\end{array}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{array}{ccc}{\\text{Ca}}_{3}{\\left({\\text{AsO}}_{4}\\right)}_{2}\\left(s\\right)\\longrightarrow &amp; 3{\\text{Ca}}^{\\text{2+}}\\left(aq\\right)+&amp; 2{\\text{AsO}}_{4}^{3-}\\left(aq\\right)\\\\ &amp; 3x&amp; \\text{_____}\\end{array}[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>How do the concentrations of Ag<sup>+<\/sup> and [latex]{\\text{CrO}}_{4}^{2-}[\/latex] in a saturated solution above 1.0 g of solid Ag<sub>2<\/sub>CrO<sub>4<\/sub> change when 100 g of solid Ag<sub>2<\/sub>CrO<sub>4<\/sub> is added to the system? Explain.<\/li>\r\n \t<li>How do the concentrations of Pb<sup>2+<\/sup> and S<sup>2\u2013<\/sup> change when K<sub>2<\/sub>S is added to a saturated solution of PbS?<\/li>\r\n \t<li>What additional information do we need to answer the following question: How is the equilibrium of solid silver bromide with a saturated solution of its ions affected when the temperature is raised?<\/li>\r\n \t<li>Which of the following slightly soluble compounds has a solubility greater than that calculated from its solubility product because of hydrolysis of the anion present: CoSO<sub>3<\/sub>, CuI, PbCO<sub>3<\/sub>, PbCl<sub>2<\/sub>, Tl<sub>2<\/sub>S, KClO<sub>4<\/sub>?<\/li>\r\n \t<li>Which of the following slightly soluble compounds has a solubility greater than that calculated from its solubility product because of hydrolysis of the anion present: AgCl, BaSO<sub>4<\/sub>, CaF<sub>2<\/sub>, Hg<sub>2<\/sub>I<sub>2<\/sub>, MnCO<sub>3<\/sub>, ZnS, PbS?<\/li>\r\n \t<li>Write the ionic equation for dissolution and the solubility product (<em>K<\/em><sub>sp<\/sub>) expression for each of the following slightly soluble ionic compounds:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>PbCl<sub>2<\/sub><\/li>\r\n \t<li>Ag<sub>2<\/sub>S<\/li>\r\n \t<li>Sr<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub><\/li>\r\n \t<li>SrSO<sub>4<\/sub><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Write the ionic equation for the dissolution and the <em>K<\/em><sub>sp<\/sub> expression for each of the following slightly soluble ionic compounds:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>LaF<sub>3<\/sub><\/li>\r\n \t<li>CaCO<sub>3<\/sub><\/li>\r\n \t<li>Ag<sub>2<\/sub>SO<sub>4<\/sub><\/li>\r\n \t<li>Pb(OH)<sub>2<\/sub><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>The <a href=\"http:\/\/hbcponline.com\/faces\/contents\/ContentsSearch.xhtml\" target=\"_blank\"><em>Handbook of Chemistry and Physics<\/em><\/a> gives solubilities of the following compounds in grams per 100 mL of water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>BaSiF<sub>6<\/sub>, 0.026 g\/100 mL (contains [latex]{\\text{SiF}}_{6}^{2-}[\/latex] ions)<\/li>\r\n \t<li>Ce(IO<sub>3<\/sub>)<sub>4<\/sub>, 1.5 \u00d7 10<sup>\u20132<\/sup> g\/100 mL<\/li>\r\n \t<li>Gd<sub>2<\/sub>(SO<sub>4<\/sub>)<sub>3<\/sub>, 3.98 g\/100 mL<\/li>\r\n \t<li>(NH<sub>4<\/sub>)<sub>2<\/sub>PtBr<sub>6<\/sub>, 0.59 g\/100 mL (contains [latex]{\\text{PtBr}}_{6}^{2-}[\/latex] ions)<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>The <a href=\"http:\/\/hbcponline.com\/faces\/contents\/ContentsSearch.xhtml\" target=\"_blank\"><em>Handbook of Chemistry and Physics<\/em><\/a> gives solubilities of the following compounds in grams per 100 mL of water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>BaSeO<sub>4<\/sub>, 0.0118 g\/100 mL<\/li>\r\n \t<li>Ba(BrO<sub>3<\/sub>)<sub>2<\/sub>\u00b7H<sub>2<\/sub>O, 0.30 g\/100 mL<\/li>\r\n \t<li>NH<sub>4<\/sub>MgAsO<sub>4<\/sub>\u00b76H<sub>2<\/sub>O, 0.038 g\/100 mL<\/li>\r\n \t<li>La<sub>2<\/sub>(MoO<sub>4<\/sub>)<sub>3<\/sub>, 0.00179 g\/100 mL<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Use solubility products and predict which of the following salts is the most soluble, in terms of moles per liter, in pure water: CaF<sub>2<\/sub>, Hg<sub>2<\/sub>Cl<sub>2<\/sub>, PbI<sub>2<\/sub>, or Sn(OH)<sub>2<\/sub>.<\/li>\r\n \t<li>Assuming that no equilibria other than dissolution are involved, calculate the molar solubility of each of the following from its solubility product:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>KHC<sub>4<\/sub>H<sub>4<\/sub>O<sub>6<\/sub><\/li>\r\n \t<li>PbI<sub>2<\/sub><\/li>\r\n \t<li>Ag<sub>4<\/sub>[Fe(CN)<sub>6<\/sub>], a salt containing the [latex]{\\text{Fe(CN)}}_{4}^{-}[\/latex] ion.<\/li>\r\n \t<li>Hg<sub>2<\/sub>I<sub>2<\/sub><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Assuming that no equilibria other than dissolution are involved, calculate the molar solubility of each of the following from its solubility product:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Ag<sub>2<\/sub>SO<sub>4<\/sub><\/li>\r\n \t<li>PbBr<sub>2<\/sub><\/li>\r\n \t<li>AgI<\/li>\r\n \t<li>CaC<sub>2<\/sub>O<sub>4<\/sub>\u00b7H<sub>2<\/sub>O<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that changes in the initial concentrations of the common ions can be neglected.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>AgCl(<em>s<\/em>) in 0.025 <em>M<\/em> NaCl<\/li>\r\n \t<li>CaF<sub>2<\/sub>(<em>s<\/em>) in 0.00133 <em>M<\/em> KF<\/li>\r\n \t<li>Ag<sub>2<\/sub>SO<sub>4<\/sub>(<em>s<\/em>) in 0.500 L of a solution containing 19.50 g of K<sub>2<\/sub>SO<sub>4<\/sub><\/li>\r\n \t<li>Zn(OH)<sub>2<\/sub>(<em>s<\/em>) in a solution buffered at a pH of 11.45<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that changes in the initial concentrations of the common ions can be neglected.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>TlCl(<em>s<\/em>) in 1.250 <em>M<\/em> HCl<\/li>\r\n \t<li>PbI<sub>2<\/sub>(<em>s<\/em>) in 0.0355 <em>M<\/em> CaI<sub>2<\/sub><\/li>\r\n \t<li>Ag<sub>2<\/sub>CrO<sub>4<\/sub>(<em>s<\/em>) in 0.225 L of a solution containing 0.856 g of K<sub>2<\/sub>CrO<sub>4<\/sub><\/li>\r\n \t<li>Cd(OH)<sub>2<\/sub>(<em>s<\/em>) in a solution buffered at a pH of 10.995<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that it is not appropriate to neglect the changes in the initial concentrations of the common ions.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>TlCl(<em>s<\/em>) in 0.025 <em>M<\/em> TlNO<sub>3<\/sub><\/li>\r\n \t<li>BaF<sub>2<\/sub>(<em>s<\/em>) in 0.0313 <em>M<\/em> KF<\/li>\r\n \t<li>MgC<sub>2<\/sub>O<sub>4<\/sub> in 2.250 L of a solution containing 8.156 g of Mg(NO<sub>3<\/sub>)<sub>2<\/sub><\/li>\r\n \t<li>Ca(OH)<sub>2<\/sub>(<em>s<\/em>) in an unbuffered solution initially with a pH of 12.700<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Explain why the changes in concentrations of the common ions in Question\u00a017 can be neglected.<\/li>\r\n \t<li>Explain why the changes in concentrations of the common ions in Question\u00a018 cannot be neglected.<\/li>\r\n \t<li>Calculate the solubility of aluminum hydroxide, Al(OH)<sub>3<\/sub>, in a solution buffered at pH 11.00.<\/li>\r\n \t<li>Refer to\u00a0<a class=\"target-chapter\" href=\".\/chapter\/solubility-products\/\" target=\"_blank\">Solubility Products<\/a>\u00a0for solubility products for calcium salts. Determine which of the calcium salts listed is most soluble in moles per liter and which is most soluble in grams per liter.<\/li>\r\n \t<li>Most barium compounds are very poisonous; however, barium sulfate is often administered internally as an aid in the X-ray examination of the lower intestinal tract (Figure\u00a03). This use of BaSO<sub>4<\/sub> is possible because of its low solubility. Calculate the molar solubility of BaSO<sub>4<\/sub> and the mass of barium present in 1.00 L of water saturated with BaSO<sub>4<\/sub>.<\/li>\r\n \t<li>Public Health Service standards for drinking water set a maximum of 250 mg\/L (2.60 \u00d7 10<sup>\u20133<\/sup> M) of [latex]{\\text{SO}}_{4}^{2-}[\/latex] because of its cathartic action (it is a laxative). Does natural water that is saturated with CaSO<sub>4<\/sub> (\u201cgyp\u201d water) as a result or passing through soil containing gypsum, CaSO<sub>4<\/sub>\\cdot 2H<sub>2<\/sub>O, meet these standards? What is [latex]{\\text{SO}}_{4}^{2-}[\/latex] in such water?<\/li>\r\n \t<li>Perform the following calculations:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Calculate [Ag<sup>+<\/sup>] in a saturated aqueous solution of AgBr.<\/li>\r\n \t<li>What will [Ag<sup>+<\/sup>] be when enough KBr has been added to make [Br<sup>\u2013<\/sup>] = 0.050 <em>M<\/em>?<\/li>\r\n \t<li>What will [Br<sup>\u2013<\/sup>] be when enough AgNO<sub>3<\/sub> has been added to make [Ag<sup>+<\/sup>] = 0.020 <em>M<\/em>?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>The solubility product of CaSO<sub>4<\/sub>\u00b72H<sub>2<\/sub>O is 2.4 \u00d7 10<sup>\u20135<\/sup>. What mass of this salt will dissolve in 1.0 L of 0.010 <em>M<\/em> [latex]{\\text{SO}}_{4}^{2-}?[\/latex]<\/li>\r\n \t<li>Assuming that no equilibria other than dissolution are involved, calculate the concentrations of ions in a saturated solution of each of the following (see\u00a0<a class=\"target-chapter\" href=\".\/chapter\/solubility-products\/\" target=\"_blank\">Solubility Products<\/a>).\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>TlCl<\/li>\r\n \t<li>BaF<sub>2<\/sub><\/li>\r\n \t<li>Ag<sub>2<\/sub>CrO<sub>4<\/sub><\/li>\r\n \t<li>CaC<sub>2<\/sub>O<sub>4<\/sub>\u00b7H<sub>2<\/sub>O<\/li>\r\n \t<li>the mineral anglesite, PbSO<sub>4<\/sub><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Assuming that no equilibria other than dissolution are involved, calculate the concentrations of ions in a saturated solution of each of the following (see\u00a0<a class=\"target-chapter\" href=\".\/chapter\/solubility-products\/\" target=\"_blank\">Solubility Products<\/a>):\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>AgI<\/li>\r\n \t<li>Ag<sub>2<\/sub>SO<sub>4<\/sub><\/li>\r\n \t<li>Mn(OH)<sub>2<\/sub><\/li>\r\n \t<li>\u00a0Sr(OH)<sub>2<\/sub>\u00b78H<sub>2<\/sub>O<\/li>\r\n \t<li>the mineral brucite, Mg(OH)<sub>2<\/sub><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>The following concentrations are found in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, calculate <em>K<\/em><sub>sp<\/sub> for each of the slightly soluble solids indicated:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>AgBr: [Ag<sup>+<\/sup>] = 5.7 \u00d7 10<sup>\u20137<\/sup><em>M<\/em>, [Br<sup>\u2013<\/sup>] = 5.7 \u00d7 10<sup>\u20137<\/sup><em>M<\/em><\/li>\r\n \t<li>CaCO<sub>3<\/sub>: [Ca<sup>2+<\/sup>] = 5.3 \u00d7 10<sup>\u20133<\/sup><em>M<\/em>, [latex]\\left[\\text{CO}_{3}^{2-}\\right][\/latex] = 9.0 \u00d7 10<sup>\u20137<\/sup><em>M<\/em><\/li>\r\n \t<li>PbF<sub>2<\/sub>: [Pb<sup>2+<\/sup>] = 2.1 \u00d7 10<sup>\u20133<\/sup><em>M<\/em>, [F<sup>\u2013<\/sup>] = 4.2 \u00d7 10<sup>\u20133<\/sup><em>M<\/em><\/li>\r\n \t<li>Ag<sub>2<\/sub>CrO<sub>4<\/sub>: [Ag<sup>+<\/sup>] = 5.3 \u00d7 10<sup>\u20135<\/sup><em>M<\/em>, 3.2 \u00d7 10<sup>\u20133<\/sup><em>M<\/em><\/li>\r\n \t<li>InF<sub>3<\/sub>: [In<sup>3+<\/sup>] = 2.3 \u00d7 10<sup>\u20133<\/sup><em>M<\/em>, [F<sup>\u2013<\/sup>] = 7.0 \u00d7 10<sup>\u20133<\/sup><em>M<\/em><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>The following concentrations are found in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, calculate <em>K<\/em><sub>sp<\/sub> for each of the slightly soluble solids indicated:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>TlCl: [Tl<sup>+<\/sup>] = 1.21 \u00d7 10<sup>\u20132<\/sup><em>M<\/em>, [Cl<sup>\u2013<\/sup>] = 1.2 \u00d7 10<sup>\u20132<\/sup><em>M<\/em><\/li>\r\n \t<li>Ce(IO<sub>3<\/sub>)<sub>4<\/sub>: [Ce<sup>4+<\/sup>] = 1.8 \u00d7 10<sup>\u20134<\/sup><em>M<\/em>, [latex]\\left[\\text{IO}_{3}^{-}\\right][\/latex] = 2.6 \u00d7 10<sup>\u201313<\/sup><em>M<\/em><\/li>\r\n \t<li>Gd<sub>2<\/sub>(SO<sub>4<\/sub>)<sub>3<\/sub>: [Gd<sup>3+<\/sup>] = 0.132 <em>M<\/em>, [latex]\\left[\\text{SO}_{4}^{2-}\\right][\/latex] = 0.198 <em>M<\/em><\/li>\r\n \t<li>Ag<sub>2<\/sub>SO<sub>4<\/sub>: [Ag<sup>+<\/sup>] = 2.40 \u00d7 10<sup>\u20132<\/sup><em>M<\/em>, [latex]\\left[{\\text{SO}}_{4}^{2-}\\right][\/latex] = 2.05 \u00d7 10<sup>\u20132<\/sup><em>M<\/em><\/li>\r\n \t<li>BaSO<sub>4<\/sub>: [Ba<sup>2+<\/sup>] = 0.500 <em>M<\/em>, [latex]\\left[{\\text{SO}}_{4}^{2-}\\right][\/latex] = 2.16 \u00d7 10<sup>\u201310<\/sup><em>M<\/em><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Which of the following compounds precipitates from a solution that has the concentrations indicated? (See\u00a0<a class=\"target-chapter\" href=\".\/chapter\/solubility-products\/\" target=\"_blank\">Solubility Products<\/a>\u00a0for <em>K<\/em><sub>sp<\/sub> values.)\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>KClO<sub>4<\/sub>: [K<sup>+<\/sup>] = 0.01 <em>M<\/em>, [latex]\\left[{\\text{ClO}}_{4}^{-}\\right][\/latex] = 0.01 <em>M<\/em><\/li>\r\n \t<li>K<sub>2<\/sub>PtCl<sub>6<\/sub>: [K<sup>+<\/sup>] = 0.01 <em>M<\/em>, [latex]{\\text{[PtCl}}_{6}^{2-}][\/latex] = 0.01 <em>M<\/em><\/li>\r\n \t<li>PbI<sub>2<\/sub>: [Pb<sup>2+<\/sup>] = 0.003 <em>M<\/em>, [I<sup>\u2013<\/sup>] = 1.3 \u00d7 10<sup>\u20133<\/sup><em>M<\/em><\/li>\r\n \t<li>Ag<sub>2<\/sub>S: [Ag<sup>+<\/sup>] = 1 \u00d7 10<sup>\u201310<\/sup><em>M<\/em>, [S<sup>2\u2013<\/sup>] = 1 \u00d7 10<sup>\u201313<\/sup><em>M<\/em><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Which of the following compounds precipitates from a solution that has the concentrations indicated? (See\u00a0<a class=\"target-chapter\" href=\".\/chapter\/solubility-products\/\" target=\"_blank\">Solubility Products<\/a>\u00a0for <em>K<\/em><sub>sp<\/sub> values.)\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>CaCO<sub>3<\/sub>: [Ca<sup>2+<\/sup>] = 0.003 <em>M<\/em>, [latex]{\\text{[CO}}_{3}^{2-}][\/latex] = 0.003 <em>M<\/em><\/li>\r\n \t<li>Co(OH)<sub>2<\/sub>: [Co<sup>2+<\/sup>] = 0.01 <em>M<\/em>, [OH<sup>\u2013<\/sup>] = 1 \u00d7 10<sup>\u20137<\/sup><em>M<\/em><\/li>\r\n \t<li>CaHPO<sub>4<\/sub>: [Ca<sup>2+<\/sup>] = 0.01 <em>M<\/em>, [latex]{\\text{[HPO}}_{4}^{2-}][\/latex] = 2 \u00d7 10<sup>\u20136<\/sup><em>M<\/em><\/li>\r\n \t<li>Pb<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub>: [Pb<sup>2+<\/sup>] = 0.01 <em>M<\/em>, [latex]{\\text{[PO}}_{4}^{3-}][\/latex] 1 \u00d7 10<sup>\u201313<\/sup><em>M<\/em><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Calculate the concentration of Tl<sup>+<\/sup> when TlCl just begins to precipitate from a solution that is 0.0250 <em>M<\/em> in Cl<sup>\u2013<\/sup>.<\/li>\r\n \t<li>Calculate the concentration of sulfate ion when BaSO<sub>4<\/sub> just begins to precipitate from a solution that is 0.0758 <em>M<\/em> in Ba<sup>2+<\/sup>.<\/li>\r\n \t<li>Calculate the concentration of Sr<sup>2+<\/sup> when SrF<sub>2<\/sub> starts to precipitate from a solution that is 0.0025 <em>M<\/em> in F<sup>\u2013<\/sup>.<\/li>\r\n \t<li>Calculate the concentration of [latex]{\\text{PO}}_{4}^{3-}[\/latex] when Ag<sub>3<\/sub>PO<sub>4<\/sub> starts to precipitate from a solution that is 0.0125 <em>M<\/em> in Ag<sup>+<\/sup>.<\/li>\r\n \t<li>Calculate the concentration of F<sup>\u2013<\/sup> required to begin precipitation of CaF<sub>2<\/sub> in a solution that is 0.010 <em>M<\/em> in Ca<sup>2+<\/sup>.<\/li>\r\n \t<li>Calculate the concentration of Ag<sup>+<\/sup> required to begin precipitation of Ag<sub>2<\/sub>CO<sub>3<\/sub> in a solution that is 2.50 \u00d7 10<sup>\u20136<\/sup><em>M<\/em> in [latex]{\\text{CO}}_{3}^{2-}[\/latex].<\/li>\r\n \t<li>What [Ag<sup>+<\/sup>] is required to reduce [latex]\\left[{\\text{CO}}_{3}^{2-}\\right][\/latex] to 8.2 \u00d7 10<sup>\u20134<\/sup><em>M<\/em> by precipitation of Ag<sub>2<\/sub>CO<sub>3<\/sub>?<\/li>\r\n \t<li>What [F<sup>\u2013<\/sup>] is required to reduce [Ca<sup>2+<\/sup>] to 1.0 \u00d7 10<sup>\u20134<\/sup><em>M<\/em> by precipitation of CaF<sub>2<\/sub>?<\/li>\r\n \t<li>A volume of 0.800 L of a 2 \u00d7 10<sup>\u20134<\/sup>-<em>M<\/em> Ba(NO<sub>3<\/sub>)<sub>2<\/sub> solution is added to 0.200 L of 5 \u00d7 10<sup>\u20134<\/sup><em>M<\/em> Li<sub>2<\/sub>SO<sub>4<\/sub>. Does BaSO<sub>4<\/sub> precipitate? Explain your answer.<\/li>\r\n \t<li>Perform these calculations for nickel(II) carbonate.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>With what volume of water must a precipitate containing NiCO<sub>3<\/sub> be washed to dissolve 0.100 g of this compound? Assume that the wash water becomes saturated with NiCO<sub>3<\/sub> (<em>K<\/em><sub>sp<\/sub> = 1.36 \u00d7 10<sup>\u20137<\/sup>).<\/li>\r\n \t<li>If the NiCO<sub>3<\/sub> were a contaminant in a sample of CoCO<sub>3<\/sub> (<em>K<\/em><sub>sp<\/sub> = 1.0 \u00d7 10<sup>\u201312<\/sup>), what mass of CoCO<sub>3<\/sub> would have been lost? Keep in mind that both NiCO<sub>3<\/sub> and CoCO<sub>3<\/sub> dissolve in the same solution.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Iron concentrations greater than 5.4 \u00d7 10<sup>\u20136<\/sup><em>M<\/em> in water used for laundry purposes can cause staining. What [OH<sup>\u2013<\/sup>] is required to reduce [Fe<sup>2+<\/sup>] to this level by precipitation of Fe(OH)<sub>2<\/sub>?<\/li>\r\n \t<li>A solution is 0.010 <em>M<\/em> in both Cu<sup>2+<\/sup> and Cd<sup>2+<\/sup>. What percentage of Cd<sup>2+<\/sup> remains in the solution when 99.9% of the Cu<sup>2+<\/sup> has been precipitated as CuS by adding sulfide?<\/li>\r\n \t<li>A solution is 0.15 <em>M<\/em> in both Pb<sup>2+<\/sup> and Ag<sup>+<\/sup>. If Cl<sup>\u2013<\/sup> is added to this solution, what is [Ag<sup>+<\/sup>] when PbCl<sub>2<\/sub> begins to precipitate?<\/li>\r\n \t<li>What reagent might be used to separate the ions in each of the following mixtures, which are 0.1 <em>M<\/em> with respect to each ion? In some cases it may be necessary to control the pH. (Hint: Consider the <em>K<\/em><sub>sp<\/sub> values given in\u00a0<a class=\"target-chapter\" href=\".\/chapter\/solubility-products\/\" target=\"_blank\">Solubility Products<\/a>.)\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]{\\text{Hg}}_{2}^{2+}[\/latex] and Cu<sup>2+<\/sup><\/li>\r\n \t<li>[latex]{\\text{SO}}_{4}^{2-}[\/latex] and Cl<sup>\u2013<\/sup><\/li>\r\n \t<li>Hg<sup>2+<\/sup> and Co<sup>2+<\/sup><\/li>\r\n \t<li>Zn<sup>2+<\/sup> and Sr<sup>2+<\/sup><\/li>\r\n \t<li>Ba<sup>2+<\/sup> and Mg<sup>2+<\/sup><\/li>\r\n \t<li>[latex]{\\text{CO}}_{3}^{2-}[\/latex] and OH<sup>\u2013<\/sup><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>A solution contains 1.0 \u00d7 10<sup>\u20135<\/sup> mol of KBr and 0.10 mol of KCl per liter. AgNO<sub>3<\/sub> is gradually added to this solution. Which forms first, solid AgBr or solid AgCl?<\/li>\r\n \t<li>A solution contains 1.0 \u00d7 10<sup>\u20132<\/sup> mol of KI and 0.10 mol of KCl per liter. AgNO<sub>3<\/sub> is gradually added to this solution. Which forms first, solid AgI or solid AgCl?<\/li>\r\n \t<li>The calcium ions in human blood serum are necessary for coagulation (Figure\u00a04). Potassium oxalate, K<sub>2<\/sub>C<sub>2<\/sub>O<sub>4<\/sub>, is used as an anticoagulant when a blood sample is drawn for laboratory tests because it removes the calcium as a precipitate of CaC<sub>2<\/sub>O<sub>4<\/sub>\u00b7H<sub>2<\/sub>O. It is necessary to remove all but 1.0% of the Ca<sup>2+<\/sup> in serum in order to prevent coagulation. If normal blood serum with a buffered pH of 7.40 contains 9.5 mg of Ca<sup>2+<\/sup> per 100 mL of serum, what mass of K<sub>2<\/sub>C<sub>2<\/sub>O<sub>4<\/sub> is required to prevent the coagulation of a 10 mL blood sample that is 55% serum by volume? (All volumes are accurate to two significant figures. Note that the volume of serum in a 10-mL blood sample is 5.5 mL. Assume that the K<sub>sp<\/sub> value for CaC<sub>2<\/sub>O<sub>4<\/sub> in serum is the same as in water.)<\/li>\r\n \t<li>About 50% of urinary calculi (kidney stones) consist of calcium phosphate, Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub>. The normal mid range calcium content excreted in the urine is 0.10 g of Ca<sup>2+<\/sup> per day. The normal mid range amount of urine passed may be taken as 1.4 L per day. What is the maximum concentration of phosphate ion that urine can contain before a calculus begins to form?<\/li>\r\n \t<li>The pH of normal urine is 6.30, and the total phosphate concentration [latex]\\left(\\left[\\text{PO}_{4}^{3-}\\right]+\\left[{\\text{HPO}}_{4}^{2-}\\right]+\\left[{\\text{H}}_{2}{\\text{PO}}_{4}^{-}\\right]+\\left[\\text{H}_3\\text{PO}_4\\right]\\right)[\/latex]\u00a0is 0.020 <em>M<\/em>. What is the minimum concentration of Ca<sup>2+<\/sup> necessary to induce kidney stone formation? (See Question\u00a049 for additional information.)<\/li>\r\n \t<li>Magnesium metal (a component of alloys used in aircraft and a reducing agent used in the production of uranium, titanium, and other active metals) is isolated from sea water by the following sequence of reactions: [latex]{\\text{Mg}}^{\\text{2+}}\\left(aq\\right)+{\\text{Ca(OH)}}_{2}\\left(aq\\right)\\longrightarrow {\\text{Mg(OH)}}_{2}\\left(s\\right)+{\\text{Ca}}^{\\text{2+}}\\left(aq\\right)[\/latex][latex]{\\text{Mg(OH)}}_{2}\\left(s\\right)+\\text{2HCl}(aq)\\longrightarrow {\\text{MgCl}}_{2}\\left(s\\right)+{\\text{2H}}_{2}\\text{O}\\left(l\\right)[\/latex]\r\n[latex]{\\text{MgCl}}_{2}\\left(l\\right)\\stackrel{\\text{electrolysis}}{\\longrightarrow }\\text{Mg}\\left(s\\right)+{\\text{Cl}}_{2}\\left(g\\right)[\/latex]Sea water has a density of 1.026 g\/cm<sup>3<\/sup> and contains 1272 parts per million of magnesium as Mg<sup>2+<\/sup>(<em>aq<\/em>) by mass. What mass, in kilograms, of Ca(OH)<sub>2<\/sub> is required to precipitate 99.9% of the magnesium in 1.00 \u00d7 10<sup>3<\/sup> L of sea water?<\/li>\r\n \t<li>Hydrogen sulfide is bubbled into a solution that is 0.10 <em>M<\/em> in both Pb<sup>2+<\/sup> and Fe<sup>2+<\/sup> and 0.30 <em>M<\/em> in HCl. After the solution has come to equilibrium it is saturated with H<sub>2<\/sub>S ([H<sub>2<\/sub>S] = 0.10 <em>M<\/em>). What concentrations of Pb<sup>2+<\/sup> and Fe<sup>2+<\/sup> remain in the solution? For a saturated solution of H<sub>2<\/sub>S we can use the equilibrium:[latex]{\\text{H}}_{2}\\text{S}\\left(aq\\right)+{\\text{2H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{2H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{S}}^{2-}\\left(aq\\right)K=1.0\\times 1{0}^{-26}[\/latex](Hint: The [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] changes as metal sulfides precipitate.)<\/li>\r\n \t<li>Perform the following calculations involving concentrations of iodate ions:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>The iodate ion concentration of a saturated solution of La(IO<sub>3<\/sub>)<sub>3<\/sub> was found to be 3.1 \u00d7 10<sup>\u20133<\/sup> mol\/L. Find the <em>K<\/em><sub>sp<\/sub>.<\/li>\r\n \t<li>Find the concentration of iodate ions in a saturated solution of Cu(IO<sub>3<\/sub>)<sub>2<\/sub> (<em>K<\/em><sub>sp<\/sub> = 7.4 \u00d7 10<sup>\u20138<\/sup>).<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Calculate the molar solubility of AgBr in 0.035 <em>M<\/em> NaBr (<em>K<\/em><sub>sp<\/sub> = 5 \u00d7 10<sup>\u201313<\/sup>).<\/li>\r\n \t<li>How many grams of Pb(OH)<sub>2<\/sub> will dissolve in 500 mL of a 0.050-<em>M<\/em> PbCl<sub>2<\/sub> solution (<em>K<\/em><sub>sp<\/sub> = 1.2 \u00d7 10<sup>\u201315<\/sup>)?<\/li>\r\n \t<li>Use\u00a0<a href=\"http:\/\/employees.oneonta.edu\/viningwj\/sims\/common_ion_effect_s1.html\" target=\"_blank\">this simulation to study the process of salts dissolving and forming saturated solutions<\/a>\u00a0to complete the following exercise. Using 0.01 g CaF<sub>2<\/sub>, give the K<sub>sp<\/sub> values found in a 0.2-<em>M<\/em> solution of each of the salts. Discuss why the values change as you change soluble salts.<\/li>\r\n \t<li>How many grams of Milk of Magnesia, Mg(OH)<sub>2<\/sub> (<em>s<\/em>) (58.3 g\/mol), would be soluble in 200 mL of water. K<sub>sp<\/sub> = 7.1 \u00d7 10<sup>\u201312<\/sup>. Include the ionic reaction and the expression for K<sub>sp<\/sub> in your answer. What is the pH? (<em>K<\/em><sub>w<\/sub> = 1 \u00d7 10<sup>\u201314<\/sup> = [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] [OH<sup>\u2013<\/sup>])<\/li>\r\n \t<li>Two hypothetical salts, LM<sub>2<\/sub> and LQ, have the same molar solubility in H<sub>2<\/sub>O. If K<sub>sp<\/sub> for LM<sub>2<\/sub> is 3.20 \u00d7 10<sup>\u20135<\/sup>, what is the K<sub>sp<\/sub> value for LQ?<\/li>\r\n \t<li>Which of the following carbonates will form first? Which of the following will form last?\u00a0Explain.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]{\\text{MgCO}}_{3}{K}_{\\text{sp}}=3.5\\times 1{0}^{-8}[\/latex]<\/li>\r\n \t<li>[latex]{\\text{CaCO}}_{3}{K}_{\\text{sp}}=4.2\\times 1{0}^{-7}[\/latex]<\/li>\r\n \t<li>[latex]{\\text{SrCO}}_{3}{K}_{\\text{sp}}=3.9\\times 1{0}^{-9}[\/latex]<\/li>\r\n \t<li>[latex]{\\text{BaCO}}_{3}{K}_{\\text{sp}}=4.4\\times 1{0}^{-5}[\/latex]<\/li>\r\n \t<li>[latex]{\\text{MnCO}}_{3}{K}_{\\text{sp}}=5.1\\times 1{0}^{-9}[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>How many grams of Zn(CN)<sub>2<\/sub>(<em>s<\/em>) (117.44 g\/mol) would be soluble in 100 mL of H<sub>2<\/sub>O? Include the balanced reaction and the expression for <em>K<\/em><sub>sp<\/sub> in your answer. The <em>K<\/em><sub>sp<\/sub> value for Zn(CN)<sub>2<\/sub>(<em>s<\/em>) is 3.0 \u00d7 10<sup>\u201316<\/sup>.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"71248\"]Show Selected Answers[\/reveal-answer]\r\n[hidden-answer a=\"71248\"]\r\n\r\n1.\u00a0In dissolution, one unit of substance produces a quantity of discrete ions or polyatomic ions that equals the number of times that the subunit appears in the formula.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\begin{array}{ccc}\\text{AgI}\\left(s\\right)\\rightleftharpoons &amp; {\\text{Ag}}^{\\text{+}}\\left(aq\\right)+&amp; {\\text{I}}^{-}\\left(aq\\right)\\\\ &amp; x&amp; {x}\\end{array}[\/latex]\r\nDissolving AgI(<em>s<\/em>) must produce the same amount of I<sup>\u2013<\/sup> ion as it does Ag<sup>+<\/sup> ion.<\/li>\r\n \t<li>[latex]\\begin{array}{ccc}{\\text{CaCO}}_{3}\\left(s\\right)\\rightleftharpoons &amp; {\\text{Ca}}^{\\text{2+}}\\left(aq\\right)+&amp; {\\text{CO}}_{3}^{2-}\\left(aq\\right)\\\\ &amp; {x}&amp; x\\end{array}[\/latex]\r\nDissolving CaCO<sub>3<\/sub>(<em>s<\/em>) must produce the same amount of Ca<sup>2+<\/sup> ion as it does [latex]{\\text{CO}}_{3}^{2-}[\/latex] ion.<\/li>\r\n \t<li>[latex]\\begin{array}{lll}\\text{Mg}{\\left(\\text{OH}\\right)}_{2}\\left(s\\right)\\longrightarrow \\hfill &amp; {\\text{Mg}}^{\\text{2+}}\\left(aq\\right)\\hfill &amp; +2{\\text{OH}}^{-}\\left(aq\\right)\\hfill \\\\ \\hfill &amp; x\\hfill &amp; {\\text{2}x}\\hfill \\end{array}[\/latex]\r\nWhen one unit of Mg(OH)<sub>2<\/sub> dissolves, two ions of OH<sup>\u2013<\/sup> are formed for each Mg<sup>2+<\/sup> ion.<\/li>\r\n \t<li>[latex]\\begin{array}{ccc}{\\text{Mg}}_{3}{\\left({\\text{PO}}_{4}\\right)}_{2}\\left(s\\right)\\rightleftharpoons &amp; {\\text{3Mg}}^{\\text{2+}}\\left(aq\\right)+&amp; {\\text{2PO}}_{4}^{3-}\\left(aq\\right)\\\\ &amp; {3x}&amp; 2x\\end{array}[\/latex]\r\nOne unit of Mg<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub> provides two units of [latex]{\\text{PO}}_{4}^{3-}[\/latex] ion and three units of Mg<sup>2+<\/sup> ion.<\/li>\r\n \t<li>[latex]\\begin{array}{cccc}{\\text{Ca}}_{5}{\\left({\\text{PO}}_{4}\\right)}_{3}\\text{OH}\\left(s\\right)\\rightleftharpoons &amp; {\\text{5Ca}}^{\\text{2+}}\\left(aq\\right)+&amp; {\\text{3PO}}_{4}^{3-}\\left(aq\\right)+&amp; {\\text{OH}}^{-}\\left(aq\\right)\\\\ &amp; {5x}&amp; {3x}&amp; x\\end{array}[\/latex]\r\nOne unit of Ca<sub>5<\/sub>(PO<sub>4<\/sub>)<sub>3<\/sub>OH dissolves into five units of Ca<sup>2+<\/sup> ion, three units of [latex]{\\text{PO}}_{4}^{3-}[\/latex] ion, and one unit of OH<sup>\u2013<\/sup> ion.<\/li>\r\n<\/ol>\r\n3.\u00a0There is no change. A solid has an activity of 1 whether there is a little or a lot.\r\n\r\n5.\u00a0The solubility of silver bromide at the new temperature must be known. Normally the solubility increases and some of the solid silver bromide will dissolve.\r\n\r\n7.\u00a0CaF<sub>2<\/sub>, MnCO<sub>3<\/sub>, and ZnS; each is a salt of a weak acid and the hydronium ion from water reacts with the anion, causing more solid to dissolve to maintain the equilibrium concentration of the anion\r\n\r\n9.\u00a0The answers for each compound are as follows:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\text{LaF}_{3}\\left(s\\right)\\rightleftharpoons\\text{La}^{3+}\\left(aq\\right)+{3F}^{-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,{;}\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{sp}}=\\left[\\text{La}^{3+}\\right]\\left[\\text{F}^{-}\\right]^{3}[\/latex]<\/li>\r\n \t<li>[latex]\\text{CaCO}_{3}\\left(s\\right)\\rightleftharpoons\\text{Ca}^{2+}\\left(aq\\right)+\\text{CO}_{3}^{2-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,{;}\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{sp}}=\\left[\\text{Ca}^{2+}\\right]\\left[\\text{CO}_{3}^{2-}\\right][\/latex]<\/li>\r\n \t<li>[latex]\\text{Ag}_{2}\\text{SO}_{4}\\left(s\\right)\\rightleftharpoons2\\text{Ag}^{+}\\left(aq\\right)+\\text{SO}_{4}^{2-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,{;}\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{sp}}=\\left[\\text{Ag}^{+}\\right]^{2}\\left[\\text{SO}_{4}^{2-}\\right][\/latex]<\/li>\r\n \t<li>[latex]\\text{Pb}\\left(\\text{OH}\\right)_{2}\\left(s\\right)\\rightleftharpoons\\text{Pb}^{2+}\\left(aq\\right)+2\\text{OH}^{-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,{;}\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{sp}}=\\left[\\text{Pb}^{2+}\\right]\\left[\\text{OH}^{-}\\right]^{2}[\/latex]<\/li>\r\n<\/ol>\r\n11.\u00a0Convert each concentration into molar units. Multiply each concentration by 10 to determine the mass in 1 L, and then divide the molar mass.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>BaSeO<sub>4<\/sub>\r\n<ul>\r\n \t<li>[latex]\\frac{0.{\\text{118 g L}}^{-1}}{280.{\\text{28 g mol}}^{-1}}=\\text{4.21}\\times {10}^{-4}\\text{}M[\/latex]<\/li>\r\n \t<li><em>K<\/em> = [Ba<sup>2+<\/sup>] [latex]\\left[{\\text{SeO}}_{4}^{2-}\\right][\/latex] = (4.21 \u00d7 10<sup>\u20134<\/sup>)(4.21 \u00d7 10<sup>\u20134<\/sup>) = 1.77 \u00d7 10<sup>\u20137<\/sup><\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Ba(BrO<sub>3<\/sub>)<sub>2<\/sub>\u00b7H<sub>2<\/sub>O:\r\n<ul>\r\n \t<li>[latex]\\frac{3.0{\\text{g L}}^{-1}}{{\\text{411.147 g mol}}^{-1}}=\\text{7.3}\\times {10}^{-3}\\text{}M[\/latex]<\/li>\r\n \t<li><em>K<\/em> = [Ba<sup>2+<\/sup>] [latex]{\\left[{\\text{BrO}}_{3}{}^{-}\\right]}^{2}[\/latex] = (7.3 \u00d7 10<sup>\u20133<\/sup>)(2 \u00d7 7.3 \u00d7 10<sup>\u20133<\/sup>)<sup>2<\/sup> = 1.6 \u00d7 10<sup>\u20136<\/sup><\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>NH<sub>4<\/sub>MgAsO<sub>4<\/sub>\u00b76H<sub>2<\/sub>O:\r\n<ul>\r\n \t<li>[latex]\\frac{0.{\\text{38 g L}}^{-1}}{{\\text{289.3544 g mol}}^{-1}}=\\text{1.3}\\times {10}^{-3}\\text{}M[\/latex]<\/li>\r\n \t<li><em>K<\/em> = [latex]\\left[{\\text{NH}}_{4}{}^{\\text{+}}\\right][\/latex] [Mg<sup>2+<\/sup>] [latex]\\left[{\\text{AsO}}_{4}{}^{3-}\\right][\/latex] = (1.3 \u00d7 10<sup>\u20133<\/sup>)<sup>3<\/sup> = 2.2 \u00d7 10<sup>\u20139<\/sup><\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>La<sub>2<\/sub>(MoO<sub>4<\/sub>)<sub>3<\/sub>:\r\n<ul>\r\n \t<li>[latex]\\frac{0.0{\\text{179 g L}}^{-1}}{{\\text{757.62 g mol}}^{-1}}=\\text{2.36}\\times {10}^{-5}\\text{}M[\/latex]<\/li>\r\n \t<li><em>K<\/em> = [La<sup>3+<\/sup>]<sup>2<\/sup> [latex]{\\left[{\\text{MoO}}_{4}{}^{2-}\\right]}^{3}[\/latex] = (2 \u00d7 2.36 \u00d7 10<sup>\u20135<\/sup>)<sup>2<\/sup>(3 \u00d7 2.36 \u00d7 10<sup>\u20135<\/sup>)<sup>3<\/sup> = 2.228 \u00d7 10<sup>\u20139<\/sup> \u00d7 3.549 \u00d7 10<sup>\u201313<\/sup> =\u00a07.91\u00a0 \u00d7 \u00a010<sup>\u201322<\/sup><\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ol>\r\n13.\u00a0Let <em>x<\/em> be the molar solubility.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li><em>K<\/em><sub>sp<\/sub> = [latex]\\left[{\\text{K}}^{\\text{+}}\\right]\\left[{\\text{HC}}_{4}{\\text{H}}_{4}{\\text{O}}_{6}^{-}\\right][\/latex] = 3 \u00d7 10<sup>\u20134<\/sup> = <em>x<\/em><sup>2<\/sup>, <em>x<\/em> = 2 \u00d7 10<sup>\u20132<\/sup><em>M<\/em>;<\/li>\r\n \t<li><em>K<\/em><sub>sp<\/sub> = [Pb<sup>2+<\/sup>][I<sup>\u2013<\/sup>]<sup>2<\/sup> = 8.7 \u00d7 10<sup>9<\/sup> = <em>x<\/em>(2<em>x<\/em>)<sup>3<\/sup> = 4<em>x<\/em><sup>3<\/sup>, <em>x<\/em> = 1.3 \u00d7 10<sup>\u20133<\/sup><em>M<\/em>;<\/li>\r\n \t<li><em>K<\/em><sub>sp<\/sub> = [latex]{\\left[{\\text{Ag}}^{\\text{+}}\\right]}^{4}\\left[{\\text{Fe(CN)}}_{6}{}^{\\text{4-}}\\right][\/latex] = 1.55 \u00d7 10<sup>\u201341<\/sup> = (4<em>x<\/em>)<sup>4<\/sup><em>x<\/em> = 256<em>x<\/em><sup>5<\/sup>, <em>x<\/em> = 2.27 \u00d7 10<sup>\u20139<\/sup><em>M<\/em>;<\/li>\r\n \t<li><em>K<\/em><sub>sp<\/sub> = [latex]\\left[{\\text{Hg}}_{2}{}^{\\text{2+}}\\right]{\\left[{\\text{I}}^{-}\\right]}^{2}[\/latex] = 4.5 \u00d7 10<sup>\u201329<\/sup> = [<em>x<\/em>][2<em>x<\/em>]<sup>2<\/sup> = 4<em>x<\/em><sup>3<\/sup>, <em>x<\/em> = 2.2 \u00d7 10<sup>\u201310<\/sup><em>M<\/em><\/li>\r\n<\/ol>\r\n15.\u00a0The correct concentrations are as follows:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li><em>K<\/em><sub>sp<\/sub> = 1.8 \u00d7 10<sup>\u201310<\/sup> = [Ag<sup>+<\/sup>][Cl<sup>\u2013<\/sup>] = <em>x<\/em>(<em>x<\/em> + 0.025), where <em>x<\/em> = [Ag<sup>+<\/sup>]. Assume that <em>x<\/em> is small when compared with 0.025 and therefore ignore it:\r\n<ul>\r\n \t<li>[latex]x=\\frac{1.8\\times {10}^{-10}}{0.025}=7.2\\times {10}^{-9}\\text{}M=\\left[{\\text{Ag}}^{\\text{+}}\\right],\\text{}\\left[{\\text{Cl}}^{-}\\right]=0.02\\text{5}M[\/latex]<\/li>\r\n \t<li>Check: [latex]\\frac{7.2\\times {10}^{-9}\\text{}M}{0.025\\text{}M}\\times \\text{100%}=\\text{2.9}\\times {10}^{-5}%[\/latex], an insignificant change;<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li><em>K<\/em><sub>sp<\/sub> = 3.9 \u00d7 10<sup>\u201311<\/sup> = [Ca<sup>2+<\/sup>][F<sup>\u2013<\/sup>]<sup>2<\/sup> = <em>x<\/em>(2<em>x<\/em> + 0.00133 <em>M<\/em>)<sup>2<\/sup>, where <em>x<\/em> = [Ca<sup>2+<\/sup>]. Assume that <em>x<\/em> is small when compared with 0.0013 <em>M<\/em> and disregard it:\r\n<ul>\r\n \t<li>[latex]x=\\frac{\\text{3.9}\\times {10}^{-11}}{{\\left(0.00133\\right)}^{2}}=2.2\\times {10}^{-5}\\text{}M=\\left[{\\text{Ca}}^{\\text{2+}}\\right],\\text{}\\left[{\\text{F}}^{-}\\right]=0.0013\\text{}M[\/latex]<\/li>\r\n \t<li>Check: [latex]\\frac{\\text{2.25}\\times {10}^{-5}\\text{}M}{0.00133\\text{}M}\\times \\text{100%}=\\text{1.69%}[\/latex]. This value is less than 5% and can be ignored.<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Find the concentration of K<sub>2<\/sub>SO<sub>4<\/sub>:\r\n<ul>\r\n \t<li>[latex]\\frac{\\text{19.50 g}}{174.2{\\text{60 g mol}}^{-1}}=\\text{0.1119 mol}[\/latex]\r\n[latex]\\frac{0.1119\\text{mol}}{0.5\\text{L}}=\\text{0.2238}M={\\text{[SO}}_{4}{}^{2-}][\/latex]\r\n<em>K<\/em><sub>sp<\/sub> = 1.18 \u00d7 10<sup>\u201318<\/sup> = [Ag<sup>+<\/sup>]<sup>2<\/sup> [latex]{\\text{[SO}}_{4}{}^{2-}][\/latex] = 4<em>x<\/em><sup>2<\/sup>(<em>x<\/em> + 0.2238)\r\n[latex]{x}^{2}=\\frac{1.18\\times {10}^{-18}}{4\\left(0.2238\\right)}=\\text{1.32}\\times {10}^{-18}[\/latex]\r\n<em>x<\/em> = 1.15 \u00d7 10<sup>\u20139<\/sup>[Ag<sup>+<\/sup>] = 2<em>x<\/em> = 2.30 \u00d7 10<sup>\u20139<\/sup><em>M<\/em><\/li>\r\n \t<li>Check: [latex]\\frac{1.15\\times {10}^{-9}}{0.2238}\\times 100%=\\text{5.14}\\times {10}^{-7}[\/latex]; the condition is satisfied.<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Find the concentration of OH<sup>\u2013<\/sup> from the pH:\r\n<ul>\r\n \t<li>pOH = 14.00 \u2013 11.45 = 2.55\r\n[OH<sup>\u2013<\/sup>] = 2.8 \u00d7 10<sup>\u20133<\/sup><em>M\r\n<\/em><em>K<\/em><sub>sp<\/sub> = 4.5 \u00d7 10<sup>\u201317<\/sup> = [Zn<sup>2+<\/sup>][OH<sup>\u2013<\/sup>]<sup>2<\/sup> = <em>x<\/em>(2<em>x<\/em> + 2.8 \u00d7 10<sup>\u20133<\/sup>)<sup>2\r\n<\/sup>Assume that <em>x<\/em> is small when compared with 2.8 \u00d7 10<sup>\u20133<\/sup>:\r\n[latex]x=\\frac{4.5\\times {10}^{-17}}{{\\left(2.8\\times {10}^{-3}\\right)}^{2}}=\\text{5.7}\\times 10 - 12\\text{}M={\\text{[Zn}}^{\\text{2+}}\\text{]}[\/latex]<\/li>\r\n \t<li>Check: [latex]\\frac{5.7\\times {10}^{-12}}{2.8\\times {10}^{-3}}\\times \\text{100%}=\\text{2.0}\\times {10}^{-7}%[\/latex]; <em>x<\/em> is less than 5% of [OH<sup>\u2013<\/sup>] and is, therefore, negligible. In each case the change in initial concentration of the common ion is less than 5%.<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ol>\r\n17. \u00a0(a) <em>K<\/em><sub>sp<\/sub> = 1.9 \u00d7 10<sup>\u20134<\/sup> = [Ti<sup>+<\/sup>][Cl<sup>\u2013<\/sup>]; Let <em>x<\/em> = [Cl<sup>\u2013<\/sup>]:\r\n\r\n1.9 \u00d7 10<sup>\u20134<\/sup> = (<em>x<\/em> = 0.025)<em>x<\/em>\r\n\r\nAssume that <em>x<\/em> is small when compared with 0.025:\r\n\r\n[latex]x=\\frac{1.9\\times {10}^{-4}}{0.025}=7.6\\times {10}^{-3}\\text{}M[\/latex]\r\n\r\nCheck: [latex]\\frac{7.6\\times {10}^{-3}}{0.025}\\times \\text{100%}=\\text{30%}[\/latex]\r\n\r\nThis value is too large to drop <em>x<\/em>. Therefore solve by using the quadratic equation:\r\n\r\n[latex]\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}[\/latex]\r\n\r\n<em>x<\/em><sup>2<\/sup> + 0.025<em>x<\/em> \u2013 1.9 \u00d7 10<sup>\u20134<\/sup> = 0\r\n\r\n[latex]\\begin{array}{l}{ }x=\\frac{-0.025\\pm \\sqrt{6.25\\times {10}^{-4}+7.6\\times {10}^{-4}}}{2}=\\frac{-0.025\\pm \\sqrt{1.385\\times {10}^{-3}}}{2}\\\\ =\\frac{-0.025\\pm 0.0372}{2}=0.00\\text{61}M\\end{array}[\/latex]\r\n\r\n(Use only the positive answer for physical sense.)\r\n\r\n[Ti<sup>+<\/sup>] = 0.025 + 0.0061 = 3.1 \u00d7 10<sup>\u20132<\/sup><em>M<\/em>\r\n\r\n[Cl<sup>\u2013<\/sup>] = 6.1 \u00d7 10<sup>\u20133<\/sup>\r\n\r\n(b) <em>K<\/em><sub>sp<\/sub> = 1.7 \u00d7 10<sup>\u20136<\/sup> = [Ba<sup>2+<\/sup>][F<sup>\u2013<\/sup>]<sup>2<\/sup>; Let <em>x<\/em> = [Ba<sup>2+<\/sup>]\r\n\r\n1.7 \u00d7 10<sup>\u20136<\/sup> = <em>x<\/em>(<em>x<\/em> + 0.0313)<sup>2<\/sup> = 1.7 \u00d7 10<sup>\u20133<\/sup><em>M<\/em>\r\n\r\nCheck: [latex]\\frac{1.7\\times {10}^{-3}}{0.0313}\\times \\text{100%}=\\text{5.5%}[\/latex]\r\n\r\nThis value is too large to drop <em>x<\/em>, and the entire equation must be solved. One method to find the answer is to solve by successive approximations. Begin by choosing the value of <em>x<\/em> that has just been calculated:\r\n\r\n<em>x<\/em>\u2032(5.4 \u00d7 10<sup>\u20135<\/sup> + 0.0313)<sup>2<\/sup> = 1.7 \u00d7 10<sup>\u20133<\/sup> or\r\n\r\n[latex]x\\prime =\\frac{1.7\\times {10}^{-6}}{1.089\\times {10}^{-3}}=\\text{1.6}\\times {10}^{-3}[\/latex]\r\n\r\nA third approximation using this last calculation is as follows:\r\n\r\n<em>x<\/em>\u2032\u2032(1.7 \u00d7 10<sup>\u20133<\/sup> + 0.0313)<sup>2<\/sup> = 1.7 \u00d7 10<sup>\u20136<\/sup> or\r\n\r\n[latex]x\\prime \\prime =\\frac{1.7\\times {10}^{-6}}{1.089\\times {10}^{-3}}=\\text{1.6}\\times {10}^{-3}[\/latex]\r\n\r\nThis value is well within 5% and is acceptable.\r\n\r\n[Ba<sup>2+<\/sup>] = 1.6 \u00d7 10<sup>\u20133<\/sup><em>M<\/em>\r\n\r\n[F<sup>\u2013<\/sup>] = (1.6 \u00d7 10<sup>\u20133<\/sup> + 0.0313) = 0.0329 <em>M<\/em>;\r\n\r\n(c) Find the molar concentration of the Mg(NO<sub>3<\/sub>)<sub>2<\/sub>. The molar mass of Mg(NO<sub>3<\/sub>)<sub>2<\/sub> is 148.3149 g\/mol. The number of moles is [latex]\\frac{\\text{8.156 g}}{148.314{\\text{9 g mol}}^{-1}}=\\text{0.05499 mol}[\/latex]\r\n\r\n[latex]M=\\frac{\\text{0.05499 mol}}{\\text{2.250 L}}=\\text{0.02444 M}[\/latex]\r\n\r\nLet <em>x<\/em> = [latex]\\left[{\\text{C}}_{2}{\\text{O}}_{4}^{2-}\\right][\/latex] and assume that <em>x<\/em> is small when compared with 0.02444 <em>M<\/em>.\r\n\r\n<em>K<\/em><sub>sp<\/sub> = 8.6 \u00d7 10<sup>\u20135<\/sup> = [Mg<sup>2+<\/sup>] [latex]\\left[{\\text{C}}_{2}{\\text{O}}_{4}^{2-}\\right][\/latex] = (<em>x<\/em>)(<em>x<\/em> + 0.02444)\r\n\r\n0.02444<em>x<\/em> = 8.6 \u00d7 10<sup>\u20135<\/sup>\r\n\r\n<em>x<\/em> = [latex]\\left[{\\text{C}}_{2}{\\text{O}}_{4}^{2-}\\right][\/latex] = 3.5 \u00d7 10<sup>\u20133<\/sup>\r\n\r\nCheck: [latex]\\frac{3.5\\times {10}^{-3}}{0.02444}\\times \\text{100%}=\\text{14%}[\/latex]\r\n\r\nThis value is greater than 5%, so the quadratic equation must be used to solve for <em>x<\/em>:\r\n\r\n[latex]\\begin{array}{l}{}\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}\\\\{x}^{2}+0.02444x - 8.6\\times {10}^{-5}=0\\\\x=\\frac{-0.02444\\pm \\sqrt{5.973\\times {10}^{-4}+3.44\\times {10}^{-4}}}{2}=\\text{}\\frac{-0.02444\\pm 0.03068}{2}\\end{array}[\/latex]\r\n\r\n(Use only the positive answer for physical sense.)\r\n\r\n<em>x<\/em>\u2032 = [latex]\\left[{\\text{C}}_{2}{\\text{O}}_{4}{}^{2-}\\right][\/latex] = 3.5 \u00d7 10<sup>\u20133<\/sup><em>M<\/em>\r\n\r\n[Mg<sup>2+<\/sup>] = 3.1 \u00d7 10<sup>\u20133<\/sup> + 0.02444 = 0.0275 <em>M<\/em>\r\n\r\n(d) pH = 12.700; pOH = 1.300\r\n\r\n[OH<sup>\u2013<\/sup>] = 0.0501 <em>M<\/em>; Let <em>x<\/em> = [Ca<sup>2+<\/sup>]\r\n\r\n<em>K<\/em><sub>sp<\/sub> = 7.9 \u00d7 10<sup>\u20136<\/sup> = [Ca<sup>2+<\/sup>][OH<sup>\u2013<\/sup>]<sup>2<\/sup> = (<em>x<\/em>)(<em>x<\/em> + 0.050)<sup>2<\/sup>\r\n\r\nAssume that <em>x<\/em> is small when compared with 0.050 <em>M<\/em>:\r\n\r\n<em>x<\/em> = [Ca<sup>2+<\/sup>] = 3.15 \u00d7 10<sup>\u20133<\/sup> (one additional significant figure is carried)\r\n\r\nCheck: [latex]\\frac{3.15\\times {10}^{-3}}{0.050}\\times \\text{100%}=\\text{6.28%}[\/latex]\r\n\r\nThis value is greater than 5%, so a more exact method, such as successive approximations, must be used. Begin by choosing the value of <em>x<\/em> that has just been calculated:\r\n\r\n<em>x<\/em>\u2032(3.15 \u00d7 10<sup>\u20133<\/sup> + 0.0501)<sup>2<\/sup> = 7.9 \u00d7 10<sup>\u20136<\/sup> or\r\n\r\n[latex]x\\prime =\\frac{7.9\\times {10}^{-6}}{2.836\\times {10}^{-3}}=\\text{2.8}\\times {10}^{-3}[\/latex]\r\n\r\n[Ca<sup>2+<\/sup>] = 2.8 \u00d7 10<sup>\u20133<\/sup><em>M<\/em>\r\n\r\n[OH<sup>\u2013<\/sup>] = (2.8 \u00d7 10<sup>\u20133<\/sup> + 0.0501) = 0.053 \u00d7 10<sup>\u20132<\/sup><em>M<\/em>\r\n\r\nIn each case, the initial concentration of the common ion changes by more than 5%.\r\n\r\n19.\u00a0The changes in concentration are greater than 5% and thus exceed the maximum value for disregarding the change.\r\n\r\n21.\u00a0Ca(OH)<sub>2<\/sub>: [Ca<sup>2+<\/sup>][OH<sup>\u2013<\/sup>]<sup>2<\/sup> = 7.9 \u00d7 10<sup>\u20136<\/sup>\r\n\r\nLet <em>x<\/em> be [Ca<sup>2+<\/sup>] = molar solubility; then [OH<sup>\u2013<\/sup>] = 2<em>x<\/em>\r\n\r\n<em>K<\/em><sub>sp<\/sub> = <em>x<\/em>(2<em>x<\/em>)<sup>2<\/sup> = 4<em>x<\/em><sup>3<\/sup> = 7.9 \u00d7 10<sup>\u20136<\/sup>\r\n\r\n<em>x<\/em><sup>3<\/sup> = 1.975 \u00d7 10<sup>\u20136<\/sup>\r\n\r\n<em>x<\/em> = 0.013 <em>M<\/em>\r\n\r\nCaCO<sub>3<\/sub>: [Ca<sup>2+<\/sup>] [latex]{\\text{[CO}}_{3}^{2-}][\/latex] = 4.8 \u00d7 10<sup>\u20139<\/sup>\r\n\r\nLet <em>x<\/em> be [Ca<sup>2+<\/sup>]; then [latex]{\\text{[CO}}_{3}^{2-}][\/latex] = [Ca<sup>2+<\/sup>] = molar solubility\r\n\r\n<em>K<\/em><sub>sp<\/sub> = <em>x<\/em><sup>2<\/sup> = 4.8 \u00d7 10<sup>\u20139<\/sup>\r\n\r\n<em>x<\/em> = 6.9 \u00d7 10<sup>\u20135<\/sup><em>M<\/em>\r\n\r\nCaSO<sub>4<\/sub>\u00b72H<sub>2<\/sub>O: [Ca<sup>2+<\/sup>] [latex]{\\text{[SO}}_{4}^{2-}]\\text{\\cdot }[\/latex] [H<sub>2<\/sub>O]<sup>2<\/sup> = 2.4 10<sup>\u20135<\/sup>\r\n\r\nLet <em>x<\/em> be [Ca<sup>2+<\/sup>] = molar solubility = [latex]{\\text{[SO}}_{4}^{2-}][\/latex]; then [H<sub>2<\/sub>O] = 2<em>x<\/em>\r\n\r\n<em>K<\/em><sub>sp<\/sub> = (<em>x<\/em>)(<em>x<\/em>)(2<em>x<\/em>)<sup>2<\/sup> = 2.4 \u00d7 10<sup>\u20135<\/sup>\r\n\r\n<em>x<\/em><sup>4<\/sup> = 6.0 \u00d7 10<sup>\u20136<\/sup>\r\n\r\n<em>x<\/em> = 0.049 <em>M<\/em>\r\n\r\nThis value is more than three times the value given by the <a href=\"http:\/\/hbcponline.com\/faces\/contents\/ContentsSearch.xhtml\"><em>Handbook of Chemistry and Physics<\/em><\/a> of (0.014 <em>M<\/em>) and reflects the complex interaction of water within the precipitate:\r\n\r\nCaC<sub>2<\/sub>O<sub>4<\/sub>\u00b7H<sub>2<\/sub>O: [Ca<sup>2+<\/sup>] [latex]\\left[{\\text{C}}_{2}{\\text{O}}_{4}^{2-}\\right][\/latex] [H<sub>2<\/sub>O] = 2.27 \u00d7 10<sup>\u20139<\/sup>\r\n\r\nLet <em>x<\/em> be [Ca<sup>2+<\/sup>] = molar solubility = [latex]\\left[{\\text{C}}_{2}{\\text{O}}_{4}^{2-}\\right][\/latex] = [H<sub>2<\/sub>O]\r\n\r\n<em>x<\/em><sup>3<\/sup> = 2.27 \u00d7 10<sup>\u20139<\/sup>\r\n\r\n<em>x<\/em> = 1.3 \u00d7 10<sup>\u20133<\/sup><em>M<\/em>\r\n\r\nIn this case, the interaction of water is also complex and the solubility is considerably less than that calculated.\r\n\r\nCa<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub>: [Ca<sup>2+<\/sup>]<sup>3<\/sup> [latex]{\\left[{\\text{PO}}_{4}^{3-}\\right]}^{2}[\/latex] = 1 \u00d7 10<sup>\u201325<\/sup>\r\n\r\nUpon solution there are three Ca<sup>2+<\/sup> and two [latex]{\\text{PO}}_{4}^{3-}[\/latex] ions. Let the concentration of Ca<sup>2+<\/sup> formed upon solution be <em>x<\/em>. Then [latex]\\frac{2}{3}x[\/latex] is the concentration of [latex]{\\text{PO}}_{4}{}^{\\text{3-}}:[\/latex]\r\n\r\n[latex]{x}^{3}{\\left(\\frac{2}{3}x\\right)}^{2}{x}^{3}=1\\times {10}^{-25}=\\text{0.4444}{x}^{5}[\/latex]\r\n\r\n<em>x<\/em> = 1 \u00d7 10<sup>\u20135<\/sup><em>M<\/em> = [Ca<sup>2+<\/sup>]\r\n\r\nThe solubility is then one-third the concentration of Ca<sup>2+<\/sup>, or 4 \u00d7 10<sup>\u20136<\/sup>.\r\n\r\n23. First, find the concentration in a saturated solution of CaSO<sub>4<\/sub>. Before placing the CaSO<sub>4<\/sub> in water, the concentrations of Ca<sup>2+<\/sup> and [latex]{\\text{SO}}_{4}^{2-}[\/latex] are 0. Let <em>x<\/em> be the change in concentration of Ca<sup>2+<\/sup>, which is equal to the concentration of [latex]{\\text{SO}}_{4}^{2-}:[\/latex]\r\n\r\n<em>K<\/em><sub>sp<\/sub> = [Ca<sup>2+<\/sup>] [latex]\\left[{\\text{SO}}_{4}^{2-}\\right][\/latex] = 2.4 \u00d7 10<sup>\u20135<\/sup>\r\n\r\n<em>x<\/em> \u00d7 <em>x<\/em> = <em>x<\/em><sup>2<\/sup> = 2.4 \u00d7 10<sup>\u20135<\/sup>\r\n\r\n[latex]x=\\sqrt{2.4\\times {10}^{-5}}[\/latex]\r\n\r\n<em>x<\/em> = 4.9 \u00d7 10<sup>\u20133<\/sup><em>M<\/em> = [latex]\\left[{\\text{SO}}_{4}^{2-}\\right][\/latex] = [Ca<sup>2+<\/sup>]\r\n\r\nSince this concentration is higher than 2.60 \u00d7 10<sup>\u20133<\/sup><em>M<\/em>, \u201cgyp\u201d water does not meet the standards.\r\n\r\n25.\u00a0The amount of CaSO<sub>4<\/sub>\u00b72H<sub>2<\/sub>O that dissolves is limited by the presence of a substantial amount of [latex]{\\text{SO}}_{4}^{2-}[\/latex] already in solution from the 0.010 <em>M<\/em> [latex]{\\text{SO}}_{4}{}^{-}[\/latex]. This is a common-ion problem. Let <em>x<\/em> be the change in concentration of Ca<sup>2+<\/sup> and of [latex]{\\text{SO}}_{4}^{2-}[\/latex] that dissociates from CaSO<sub>4<\/sub>:\r\n\r\n[latex]{\\text{CaSO}}_{4}\\left(s\\right)\\longrightarrow {\\text{Ca}}^{\\text{2+}}\\left(aq\\right)+{\\text{SO}}_{4}{}^{\\text{2-}}\\left(aq\\right)[\/latex]\r\n\r\n<em>K<\/em><sub>sp<\/sub> = [Ca<sup>2+<\/sup>] [latex]{\\text{[SO}}_{4}^{2-}][\/latex] = 2.4 \u00d7 10<sup>\u20135<\/sup>\r\n\r\nAddition of 0.010 <em>M<\/em> [latex]{\\text{SO}}_{4}^{2-}[\/latex] generated from the complete dissociation of 0.010 <em>M<\/em> SO<sub>4<\/sub> gives\r\n\r\n[<em>x<\/em>][<em>x<\/em> + 0.010] = 2.4 \u00d7 10<sup>\u20135<\/sup>. Here, <em>x<\/em> cannot be neglected in comparison with 0.010 <em>M<\/em>; the quadratic equation must be used. In standard form:\r\n\r\n<em>x<\/em><sup>2<\/sup> + 0.010<em>x<\/em> \u2013 2.4 \u00d7 10<sup>\u20135<\/sup> = 0\r\n\r\n[latex]x=\\frac{-0.01\\pm \\sqrt{1\\times {10}^{-4}+9.6\\times {10}^{-5}}}{2}=\\frac{-0.01\\pm 1.4\\times {10}^{-2}}{2}[\/latex]\r\n\r\nOnly the positive value will give a meaningful answer:\r\n\r\n<em>x<\/em> = 2.0 \u00d7 10<sup>\u20133<\/sup> = [Ca<sup>2+<\/sup>]\r\n\r\nThis is also the concentration of CaSO<sub>4<\/sub>\u00b72H<sub>2<\/sub>O that has dissolved. The mass of the salt in 1 L is\r\n\r\nMass (CaSO<sub>4<\/sub>\u00b72H<sub>2<\/sub>O) = 2.0 \u00d7 10<sup>\u20133<\/sup> mol\/L \u00d7 172.16 g\/mol = 0.34 g\/L\r\n\r\nNote that the presence of the common ion, [latex]{\\text{SO}}_{4}^{2-}[\/latex], has caused a decrease in the concentration of Ca<sup>2+<\/sup> that otherwise would be in solution:\r\n\r\n[latex]\\sqrt{2.4\\times {10}^{-5}}=\\text{4.9}\\times {10}^{-5}\\text{}M[\/latex]\r\n\r\n27.\u00a0In each of the following, allow <em>x<\/em> to be the molar concentration of the ion occurring only once in the formula.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li><em>K<\/em><sub>sp<\/sub> = [Ag<sup>+<\/sup>][Cl<sup>\u2013<\/sup>] = 1.5 \u2013 10<sup>\u201316<\/sup> = [<em>x<\/em><sup>2<\/sup>], [<em>x<\/em>] = 1.2 \u2013 10<sup>\u20138<\/sup><em>M<\/em>, [Ag<sup>+<\/sup>] = [I<sup>\u2013<\/sup>] = 1.2 \u00d7 10<sup>\u20138<\/sup><em>M<\/em><\/li>\r\n \t<li><em>K<\/em><sub>sp<\/sub> = [latex]{{\\text{[Ag}}^{\\text{+}}]}^{2}\\left[{\\text{SO}}_{4}^{2-}\\right][\/latex] = 1.18 \u00d7 10<sup>\u20135<\/sup> = [2<em>x<\/em>]<sup>2<\/sup>[<em>x<\/em>], 4<em>x<\/em><sup>3<\/sup> = 1.18 \u00d7 10<sup>\u20135<\/sup>, <em>x<\/em> = 1.43 \u00d7 10<sup>\u20132<\/sup><em>M\r\n<\/em>As there are 2 Ag<sup>+<\/sup> ions for each [latex]{\\text{SO}}_{4}^{2-}[\/latex] ion, [Ag<sup>+<\/sup>] = 2.86 \u00d7 10<sup>\u20132<\/sup><em>M<\/em>, [latex]{\\text{[SO}}_{4}^{2-}][\/latex] = 1.43 \u00d7 10<sup>\u20132<\/sup><em>M<\/em>; (c) <em>K<\/em>sp = [Mn<sup>2+<\/sup>]<sup>2<\/sup>[OH<sup>\u2013<\/sup>]<sup>2<\/sup> = 4.5 \u00d7 10<sup>\u201314<\/sup> = [<em>x<\/em>][2<em>x<\/em>]<sup>2<\/sup>, 4<em>x<\/em><sup>3<\/sup> = 4.5 \u00d7 10<sup>\u201314<\/sup>, <em>x<\/em> = 2.24 \u00d7 10<sup>\u20135<\/sup><em>M<\/em>.<\/li>\r\n \t<li>Since there are two OH<sup>\u2013<\/sup> ions for each Mn<sup>2+<\/sup> ion, multiplication of <em>x<\/em> by 2 gives 4.48 \u00d7 10<sup>\u20135<\/sup><em>M<\/em>. If the value of <em>x<\/em> is rounded to the correct number of significant figures, [Mn<sup>2+<\/sup>] = 2.2 \u00d7 10<sup>\u20135<\/sup><em>M<\/em>. [OH<sup>\u2013<\/sup>] = 4.5 \u00d7 10<sup>\u20135<\/sup><em>M<\/em>. If the value of <em>x<\/em> is rounded before determining the value of [OH<sup>\u2013<\/sup>], the resulting value of [OH<sup>\u2013<\/sup>] is 4.4 \u00d7 10<sup>\u20135<\/sup><em>M<\/em>. We normally maintain one additional figure in the calculator throughout all calculations before rounding.<\/li>\r\n \t<li><em>K<\/em><sub>sp<\/sub> = [Sr<sup>2+<\/sup>][OH<sup>\u2013<\/sup>]<sup>2<\/sup> = 3.2 \u00d7 10<sup>\u20134<\/sup> = [<em>x<\/em>][2<em>x<\/em>]<sup>2<\/sup>, 4<em>x<\/em><sup>3<\/sup> = 3.2 \u00d7 10<sup>\u20134<\/sup>, <em>x<\/em> = 4.3 \u00d7 10<sup>\u20132<\/sup><em>M<\/em>.\r\nSubstitution gives [Sr<sup>2+<\/sup>] = 4.3 \u00d7 10<sup>\u20132<\/sup> M, [OH<sup>\u2013<\/sup>] = 8.6 \u00d7 10<sup>\u20132<\/sup><em>M<\/em><\/li>\r\n \t<li><em>K<\/em><sub>sp<\/sub> = [Mg<sup>2+<\/sup>]<sup>2<\/sup>[OH<sup>\u2013<\/sup>]<sup>2<\/sup> = 1.5 \u00d7 10<sup>\u201311<\/sup> = [<em>x<\/em>][2<em>x<\/em>]<sup>2<\/sup>, 4<em>x<\/em><sup>3<\/sup> = 1.5 \u00d7 10<sup>\u201311<\/sup>, <em>x<\/em> = 1.55 \u00d7 10<sup>\u20134<\/sup><em>M<\/em>, 2<em>x<\/em> = 3.1 \u00d7 10<sup>\u20134<\/sup>.\r\nSubstitution and taking the correct number of significant figures gives [Mg<sup>2+<\/sup>] = 1.6 \u00d7 10<sup>\u20134<\/sup><em>M<\/em>, [OH\u2013] = 3.1 \u00d7 10<sup>\u20134<\/sup><em>M<\/em>. If the number is rounded first, the first value is still [Mg<sup>2+<\/sup>] = 1.6 \u00d7 10<sup>\u20134<\/sup><em>M<\/em>, but the second is [OH<sup>\u2013<\/sup>] = 3.2 \u00d7 10<sup>\u20134<\/sup><em>M<\/em>.<\/li>\r\n<\/ol>\r\n29.\u00a0In each case the value of <em>K<\/em><sub>sp<\/sub> is found by multiplication of the concentrations raised to the ion\u2019s stoichiometric power. Molar units are not normally shown in the value of <em>K<\/em>.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>TlCl: <em>K<\/em><sub>sp<\/sub> = (1.4 \u00d7 10<sup>\u20132<\/sup>)(1.2 \u00d7 10<sup>\u20132<\/sup>) = 2.0 \u00d7 10<sup>\u20134<\/sup><\/li>\r\n \t<li>Ce(IO<sub>3<\/sub>)<sub>4<\/sub>: <em>K<\/em><sub>sp<\/sub> = (1.8 \u00d7 10<sup>\u20134<\/sup>)(2.6 \u00d7 10<sup>\u201313<\/sup>)<sup>4<\/sup> = 5.1 \u00d7 10<sup>\u201317<\/sup><\/li>\r\n \t<li>Gd<sub>2<\/sub>(SO<sub>4<\/sub>)<sub>3<\/sub>: <em>K<\/em><sub>sp<\/sub> = (0.132)<sup>2<\/sup>(0.198)<sup>3<\/sup> = 1.35 \u00d7 10<sup>\u20134<\/sup><\/li>\r\n \t<li>Ag<sub>2<\/sub>SO<sub>4<\/sub>: <em>K<\/em><sub>sp<\/sub> = (2.40 \u00d7 10<sup>\u20132<\/sup>)<sup>2<\/sup>(2.05 \u00d7 10<sup>\u20132<\/sup>) = 1.18 \u00d7 10<sup>\u20135<\/sup><\/li>\r\n \t<li>BaSO<sub>4<\/sub>: <em>K<\/em><sub>sp<\/sub> = (0.500)(2.16 \u00d7 10<sup>\u201310<\/sup>) = 1.08 \u00d7 10<sup>\u201310<\/sup><\/li>\r\n<\/ol>\r\n31.\u00a0(a) [latex]{\\text{CaCO}}_{3}:{\\text{CaCO}}_{3}\\left(s\\right)\\longrightarrow {\\text{Ca}}^{\\text{2+}}\\left(aq\\right)+{\\text{CO}}_{3}{}^{2-}\\left(aq\\right)[\/latex]\r\n\r\n<em>K<\/em><sub>sp<\/sub> = [Ca<sup>2+<\/sup>] [latex]{\\text{[CO}}_{3}^{2-}][\/latex] = 4.8 \u00d7 10<sup>\u20139<\/sup>\r\n\r\ntest <em>K<\/em><sub>sp<\/sub> against <em>Q<\/em> = [Ca<sup>2+<\/sup>] [latex]{\\text{[CO}}_{3}^{2-}][\/latex]\r\n\r\n<em>Q<\/em> = [Ca<sup>2+<\/sup>] [latex]{\\text{[CO}}_{3}^{2-}][\/latex] = (0.003)(0.003) = 9 \u00d7 10<sup>\u20136<\/sup>\r\n\r\n<em>K<\/em><sub>sp<\/sub> = 4.8 \u00d7 10<sup>\u20139<\/sup> &lt; 9 \u00d7 10<sup>\u20136<\/sup>\r\n\r\nThe ion product does exceed <em>K<\/em><sub>sp<\/sub>, so CaCO<sub>3<\/sub> does precipitate.\r\n\r\n(b) [latex]{\\text{Co(OH)}}_{2}:{\\text{Co(OH)}}_{2}\\left(s\\right)\\longrightarrow {\\text{Co}}^{\\text{2+}}\\left(aq\\right)+{\\text{2OH}}^{-}\\left(aq\\right)[\/latex]\r\n\r\n<em>K<\/em><sub>sp<\/sub> = [Co<sup>2+<\/sup>][OH<sup>\u2013<\/sup>]<sup>2<\/sup> = 2 \u00d7 10<sup>\u201316<\/sup>\r\n\r\ntest <em>K<\/em><sub>sp<\/sub> against <em>Q<\/em> = [Co<sup>2+<\/sup>][OH<sup>\u2013<\/sup>]<sup>2<\/sup>\r\n\r\n<em>Q<\/em> = [Co<sup>2+<\/sup>][OH<sup>\u2013<\/sup>]<sup>2<\/sup> = (0.01)(1 \u00d7 10<sup>\u20137<\/sup>)<sup>2<\/sup> = 1 \u00d7 10<sup>\u201316<\/sup>\r\n\r\n<em>K<\/em><sub>sp<\/sub> = 2 \u00d7 10<sup>\u201316<\/sup> &gt; 1 \u00d7 10<sup>\u201316<\/sup>\r\n\r\nThe ion product does not exceed <em>K<\/em><sub>sp<\/sub>, so the compound does not precipitate.\r\n\r\n(c) CaHPO<sub>4<\/sub>: (<em>K<\/em><sub>sp<\/sub> = 5 \u00d7 10<sup>\u20136<\/sup>):\r\n\r\n<em>Q<\/em> = [Ca<sup>2+<\/sup>] [latex]{\\text{[HPO}}_{4}^{2-}][\/latex] = (0.01)(2 \u00d7 10<sup>\u20136<\/sup>) = 2 \u00d7 10<sup>\u20138<\/sup> &lt; <em>K<\/em><sub>sp<\/sub>\r\n\r\nThe ion product does not exceed <em>K<\/em><sub>sp<\/sub>, so compound does not precipitate.\r\n\r\n(d) Pb<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub>: (<em>K<\/em><sub>sp<\/sub> = 3 \u00d7 10<sup>\u201344<\/sup>):\r\n\r\n<em>Q<\/em> = [Pb<sup>2+<\/sup>]<sup>3<\/sup> [latex]{{\\text{[PO}}_{4}^{3-}]}^{2}[\/latex] = (0.01)<sup>3<\/sup>(1 \u00d7 10<sup>\u201313<\/sup>)<sup>2<\/sup> = 1 \u00d7 10<sup>\u201332<\/sup> &gt; <em>K<\/em><sub>sp<\/sub>\r\n\r\nThe ion product exceeds <em>K<\/em><sub>sp<\/sub>, so the compound precipitates.\r\n\r\n33.\u00a0Precipitation of [latex]{\\text{SO}}_{4}^{2-}[\/latex] will begin when the ion product of the concentration of the [latex]{\\text{SO}}_{4}^{2-}[\/latex] and Ba<sup>2+<\/sup> ions exceeds the <em>K<\/em><sub>sp<\/sub> of BaSO<sub>4<\/sub>.\r\n\r\n<em>K<\/em><sub>sp<\/sub> = 1.08 \u00d7 10<sup>\u201310<\/sup> = [Ba<sup>2+<\/sup>] [latex]\\left[{\\text{SO}}_{4}^{2-}\\right][\/latex] = (0.0758) [latex]{\\text{[SO}}_{4}^{2-}][\/latex]\r\n\r\n[latex]\\left[\\text{SO}_{4}^{2-}\\right]=\\frac{1.08\\times{10}^{-10}}{0.0758}=\\text{1.42}\\times{10}^{-9}M[\/latex]\r\n\r\n35.\u00a0Precipitation of Ag<sub>3<\/sub>PO<sub>4<\/sub> will begin when the ion product of the concentrations of the Ag<sup>+<\/sup> and [latex]{\\text{PO}}_{4}{}^{\\text{3-}}[\/latex] ions exceeds <em>K<\/em><sub>sp<\/sub>:\r\n\r\n[latex]{\\text{Ag}}_{3}{\\text{PO}}_{4}\\left(s\\right)\\longrightarrow {\\text{3Ag}}^{\\text{+}}\\left(aq\\right)+{\\text{PO}}_{4}{}^{\\text{3-}}\\left(aq\\right)[\/latex]\r\n\r\n<em>K<\/em><sub>sp<\/sub> = 1.8 \u00d7 10<sup>\u201318<\/sup> = [Ag<sup>+<\/sup>]<sup>3<\/sup> [latex]{\\text{[PO}}_{4}{}^{\\text{3-}}][\/latex] = (0.0125)<sup>3<\/sup> [latex]{\\text{[PO}}_{4}{}^{3-}][\/latex]\r\n\r\n[latex]\\left[\\text{PO}_{4}^{3-}\\right]=\\frac{1.08\\times{10}^{-18}}{\\left(0.0125\\right)^{3}}=\\text{9.2}\\times {10}^{-13}M[\/latex]\r\n\r\n37.\u00a0[latex]{\\text{Ag}}_{2}{\\text{CO}}_{3}\\left(s\\right)\\longrightarrow {\\text{2Ag}}^{\\text{+}}\\left(aq\\right)+{\\text{CO}}_{3}{}^{\\text{2-}}\\left(aq\\right)[\/latex]\r\n\r\n[Ag<sup>+<\/sup>]<sup>2<\/sup> [latex]\\left[{\\text{CO}}_{3}{}^{\\text{2-}}\\right][\/latex] = <em>K<\/em><sub>sp<\/sub> = 8.2 \u00d7 10<sup>\u201312<\/sup>\r\n\r\n[Ag<sup>+<\/sup>]<sup>2<\/sup>(2.5 \u00d7 10<sup>\u20136<\/sup>) = 8.2 \u00d7 10<sup>\u201312<\/sup>\r\n\r\n[latex]\\left[\\text{Ag}^{+}\\right]^{2}=\\frac{8.2\\times{10}^{-12}}{2.50\\times{10}^{-6}}=\\text{3.28}\\times{10}^{-6}[\/latex]\r\n\r\n[Ag<sup>+<\/sup>] = 1.8 \u00d7 10<sup>\u20133<\/sup><em>M<\/em>\r\n\r\n39.\u00a0In the <em>K<\/em><sub>sp<\/sub> expression, substitute the concentration of Ca<sup>2+<\/sup> and solve for [F<sup>\u2013<\/sup>].\r\n\r\n<em>K<\/em><sub>sp<\/sub> = 3.9 \u00d7 10<sup>\u201311<\/sup> = [Ca<sup>2+<\/sup>][F<sup>\u2013<\/sup>]<sup>2<\/sup> = (1.0 \u00d7 10<sup>\u20134<\/sup>)[F<sup>\u2013<\/sup>]<sup>2<\/sup>\r\n\r\n[latex]{\\left[{\\text{F}}^{-}\\right]}^{2}=\\frac{3.9\\times {10}^{-11}}{1.0\\times {10}^{-4}}=\\text{3.9}\\times {10}^{-7}[\/latex]\r\n\r\n[F<sup>\u2013<\/sup>] = 6.2 \u00d7 10<sup>\u20134<\/sup>\r\n\r\n41.\u00a0(a) 2.28 L; (b) 7.3 \u00d7 10<sup>\u20137<\/sup> g\r\n\r\n43.\u00a0When 99.9% of Cu<sup>2+<\/sup> has precipitated as CuS, then 0.1% remains in solution.\r\n\r\n[latex]\\frac{0.1}{100}[\/latex] \u00d7 0.010 mol\/L = 1 \u00d7 10<sup>\u20135<\/sup><em>M<\/em> = [Cu<sup>2+<\/sup>]\r\n\r\n[Cu<sup>2+<\/sup>][S<sup>2\u2013<\/sup>] = <em>K<\/em><sub>sp<\/sub> = 6.7 \u00d7 10<sup>\u201342<\/sup>\r\n\r\n(1 \u00d7 10<sup>\u20135<\/sup>)[S<sup>2\u2013<\/sup>] = 6.7 \u00d7 10<sup>\u201342<\/sup>\r\n\r\n[S<sup>2\u2013<\/sup>] = 7 \u00d7 10<sup>\u201337<\/sup><em>M<\/em>\r\n\r\n[Cd<sup>2+<\/sup>][S<sup>2\u2013<\/sup>]<em>K<\/em><sub>sp<\/sub> = 2.8 \u00d7 10<sup>\u201335<\/sup>\r\n\r\n[Cd<sup>2+<\/sup>](7 \u00d7 10<sup>\u201337<\/sup>) = 2.8 \u00d7 10<sup>\u201335<\/sup>\r\n\r\n[Cd<sup>2+<\/sup>] = 4 \u00d7 10<sup>1<\/sup><em>M<\/em>\r\n\r\nThus [Cd<sup>2+<\/sup>] can increase to about 40 <em>M<\/em> before precipitation begins. [Cd<sup>2+<\/sup>] is only 0.010 <em>M<\/em>, so 100% of it is dissolved.\r\n\r\n45.\u00a0To compare ions of the same oxidation state, look for compounds with a common counter ion that have very different <em>K<\/em><sub>sp<\/sub> values, one of which has a relatively large <em>K<\/em><sub>sp<\/sub>\u2014that is, a compound that is somewhat soluble.\r\n\r\n(a) [latex]{\\text{Hg}}_{2}{}^{\\text{2+}}[\/latex] and Cu<sup>2+<\/sup>: Add [latex]{\\text{SO}}_{4}{}^{\\text{2-}}[\/latex]. CuSO<sub>4<\/sub> is soluble (see\u00a0<a class=\"target-chapter\" href=\".\/chapter\/solubility-products\/\" target=\"_blank\">Solubility Products<\/a>), but <em>K<\/em><sub>sp<\/sub> for Hg<sub>2<\/sub>SO<sub>4<\/sub> is 6.2 \u00d7 10<sup>\u20137<\/sup>. When only 0.1% [latex]{\\text{Hg}}_{2}{}^{\\text{2+}}[\/latex] remains in solution:\r\n\r\n[latex]\\left[{\\text{Hg}}_{2}{}^{\\text{2+}}\\right]=\\frac{0.1%}{100%}\\times \\text{0.10}=1\\times {10}^{-4}\\text{}M[\/latex]\r\n\r\nand\r\n\r\n[latex]\\left[{\\text{SO}}_{4}^{2-}\\right]=\\frac{{K}_{\\text{sp}}}{\\left[{\\text{Hg}}_{2}{}^{\\text{2+}}\\right]}=\\frac{6.2\\times {10}^{-7}}{1\\times {10}^{-4}}=\\text{6.2}\\times {10}^{-3}\\text{}M[\/latex];\r\n\r\n(b) [latex]{\\text{SO}}_{4}^{2-}[\/latex] and Cl<sup>\u2013<\/sup>: Add Ba<sup>2+<\/sup>. BaCl<sub>2<\/sub> is soluble (see the section on catalysis), but <em>K<\/em><sub>sp<\/sub> for BaSO<sub>4<\/sub> is 1.08 \u00d7 10<sup>\u201310<\/sup>. When only 0.1% [latex]{\\text{SO}}_{4}^{2-}[\/latex] remains in solution, [latex]{\\text{[SO}}_{4}^{2-}][\/latex] = 1 \u00d7 10<sup>\u20134<\/sup><em>M<\/em> and\r\n\r\n[latex]\\left[{\\text{Ba}}^{\\text{2+}}\\right]=\\frac{1.08\\times {10}^{-10}}{1\\times {10}^{-4}}=1\\times {10}^{-6}\\text{}M[\/latex];\r\n\r\n(c) Hg<sup>2+<\/sup> and Co<sup>2+<\/sup>: Add S<sup>2\u2013<\/sup>: For the least soluble form of CoS, <em>K<\/em><sub>sp<\/sub> = 6.7 \u00d7 10<sup>\u201329<\/sup> and for HgS, <em>K<\/em><sub>sp<\/sub> = 2 \u00d7 10<sup>\u201359<\/sup>. CoS will not begin to precipitate until:\r\n\r\n[Co<sup>2+<\/sup>][S<sup>2\u2013<\/sup>] = <em>K<\/em><sub>sp<\/sub> = 6.7 \u00d7 10<sup>\u201329<\/sup>\r\n\r\n(0.10)[S<sup>2\u2013<\/sup>] = 6.7 \u00d7 10<sup>\u201329<\/sup>\r\n\r\n[S<sup>2\u2013<\/sup>] = 6.7 \u00d7 10<sup>\u201328<\/sup>\r\n\r\nAt that concentration:\r\n\r\n[Hg<sup>2+<\/sup>](6.7 \u00d7 10<sup>\u201328<\/sup>) = 2 \u00d7 10<sup>\u201359<\/sup>\r\n\r\n[Hg<sup>2+<\/sup>] = 3 \u00d7 10<sup>\u201332<\/sup><em>M<\/em>\r\n\r\nThat is, it is virtually 100% precipitated. For a saturated (0.10 <em>M<\/em>) H<sub>2<\/sub>S solution, the corresponding [latex]\\left[\\text{H}_{3}\\text{O}^{+}\\right][\/latex] is:\r\n\r\n[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]=\\frac{\\left[{\\text{H}}_{2}\\text{S}\\right]}{\\left[{\\text{S}}^{2-}\\right]}{K}_{\\text{a}}=\\frac{\\left(0.10\\right)}{\\left(6.7\\times {10}^{-28}\\right)}\\times \\text{1.0}\\times {10}^{-26}[\/latex]\r\n\r\n[latex]\\left[\\text{H}_{3}\\text{O}^{+}\\right]=1.5M[\/latex]\r\n\r\nA solution more basic than this would supply enough S<sup>2\u2013<\/sup> for CoS to precipitate.\r\n\r\n(d) Zn<sup>2+<\/sup> and Sr<sup>2+<\/sup>: Add OH<sup>\u2013<\/sup> until [OH<sup>\u2013<\/sup>] = 0.050 <em>M<\/em>. For Sr(OH)<sub>2<\/sub>\u00b78H<sub>2<\/sub>O, <em>K<\/em><sub>sp<\/sub> = 3.2 \u00d7 10<sup>\u20134<\/sup>. For Zn(OH)<sub>2<\/sub>, <em>K<\/em><sub>sp<\/sub> = 4.5 \u00d7 10<sup>\u201311<\/sup>. When Zn<sup>2+<\/sup> is 99.9% precipitated, then [Zn<sup>2+<\/sup>] = 1 \u00d7 10<sup>\u20134<\/sup><em>M<\/em> and\r\n\r\n[latex]{\\left[{\\text{OH}}^{-}\\right]}^{2}=\\frac{{K}_{\\text{sp}}}{\\left[{\\text{Zn}}^{\\text{2+}}\\right]}\\text{=}\\frac{4.5\\times {10}^{-11}}{1\\times {10}^{-4}}=\\text{4.5}\\times {10}^{-7}[\/latex]\r\n\r\n[OH<sup>\u2013<\/sup>] = 7 \u00d7 10<sup>\u20134<\/sup><em>M<\/em>\r\n\r\nWhen Sr(OH)<sub>2<\/sub>\u00b78H<sub>2<\/sub>O just begins to precipitate:\r\n\r\n[latex]{\\left[{\\text{OH}}^{-}\\right]}^{2}=\\frac{{K}_{\\text{sp}}}{\\left[{\\text{Sr}}^{\\text{2+}}\\right]}=\\frac{3.2\\times {10}^{-4}}{0.10}=\\text{3.2}\\times {10}^{-3}[\/latex]\r\n\r\n[OH<sup>\u2013<\/sup>] = 0.057 <em>M<\/em>\r\n\r\nIf [OH<sup>\u2013<\/sup>] is maintained less than 0.056 <em>M<\/em>, then Zn<sup>2+<\/sup> will precipitate and Sr<sup>2+<\/sup> will not.\r\n\r\n(e) Ba<sup>2+<\/sup> and Mg<sup>2+<\/sup>: Add [latex]{\\text{SO}}_{4}^{2-}[\/latex]. MgSO<sub>4<\/sub> is soluble and BaSO<sub>4<\/sub> is not (<em>K<\/em><sub>sp<\/sub> = 2 \u00d7 10<sup>\u201310<\/sup>).\r\n\r\n(f) [latex]{\\text{CO}}_{3}^{2-}[\/latex] and OH<sup>\u2013<\/sup>: Add Ba<sup>2+<\/sup>. For Ba(OH)<sub>2<\/sub>, 8H<sub>2<\/sub>O, <em>K<\/em><sub>sp<\/sub> = 5.0 \u00d7 10<sup>\u20133<\/sup>; for BaCO<sub>3<\/sub>, <em>K<\/em><sub>sp<\/sub> = 8.1 \u00d7 10<sup>\u20139<\/sup>. When 99.9% of [latex]{\\text{CO}}_{3}^{2-}[\/latex] has been precipitated [latex]{\\text{[CO}}_{3}^{2-}][\/latex] = 1 \u00d7 10<sup>\u20134<\/sup><em>M<\/em> and\r\n\r\n[latex]\\left[\\text{Ba}^{2+}\\right]=\\frac{K_{\\text{sp}}}{\\left[\\text{CO}_{3}^{2-}\\right]}=\\frac{8.1\\times{10}^{-9}}{1\\times{10}^{-4}}=\\text{8.1}\\times{10}^{-5}M[\/latex]\r\n\r\nBa(OH)<sub>2<\/sub>\u00b78H<sub>2<\/sub>O begins to precipitate when:\r\n\r\n[latex]{\\text{[Ba}}^{\\text{2+}}\\text{]}=\\frac{{K}_{\\text{sp}}}{{\\left[{\\text{OH}}^{-}\\right]}^{2}}=\\frac{5.0\\times {10}^{-3}}{{\\left(0.10\\right)}^{2}}=\\text{0.50}M[\/latex]\r\n\r\nAs long as [Ba<sup>2+<\/sup>] is maintained at less than 0.50 <em>M<\/em>, BaCO<sub>3<\/sub> precipitates and Ba(OH)<sub>2<\/sub>\u00b78H<sub>2<\/sub>O does not.\r\n\r\n47.\u00a0Compare the concentration of Ag<sup>+<\/sup> as determined from the two solubility product expressions. The one requiring the smaller [Ag<sup>+<\/sup>] will precipitate first.\r\n\r\nFor AgCl: <em>K<\/em><sub>sp<\/sub> = 1.8 \u00d7 10<sup>\u201310<\/sup> = [Ag<sup>+<\/sup>][Cl<sup>\u2013<\/sup>]\r\n\r\n[latex]\\left[{\\text{Ag}}^{\\text{+}}\\right]=\\frac{1.8\\times {10}^{-10}}{\\left[0.10\\right]}\\text{=1.8}\\times {10}^{-9}\\text{}M[\/latex]\r\n\r\nFor AgI: <em>K<\/em><sub>sp<\/sub> = 1.5 \u00d7 10<sup>\u201316<\/sup> = [Ag<sup>+<\/sup>][I<sup>\u2013<\/sup>]\r\n\r\n[latex]\\left[{\\text{Ag}}^{\\text{+}}\\right]=\\frac{1.5\\times {10}^{-16}}{1.0\\times {10}^{-2}}=\\text{1.5}\\times {10}^{-9}\\text{}M[\/latex]\r\n\r\nAs the value of [Ag<sup>+<\/sup>] is smaller for AgI, AgI will precipitate first.\r\n\r\n49.\u00a0The dissolution of Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub> yields:\r\n\r\n[latex]{\\text{Ca}}_{3}{{\\text{(PO}}_{4})}_{2}\\left(s\\right)\\rightleftharpoons {\\text{3Ca}}^{\\text{2+}}\\left(aq\\right)+{\\text{2PO}}_{4}^{3-}\\left(aq\\right)[\/latex]\r\n\r\nGiven the concentration of Ca<sup>2+<\/sup> in solution, the maximum [latex]\\left[\\text{PO}_{4}^{3-}\\right][\/latex] can be calculated by using the <em>K<\/em><sub>sp<\/sub> expression for Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub>:\r\n<p style=\"text-align: center;\"><em>K<\/em><sub>sp<\/sub> = 1 \u00d7 10<sup>\u201325<\/sup> = [Ca<sup>2+<\/sup>]<sup>3<\/sup> [latex]{{\\text{[PO}}_{4}{}^{3-]}}^{2}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\left[{\\text{Ca}}^{\\text{2+}}\\right]}_{\\text{urine}}=\\frac{0.10\\text{}g\\left(\\frac{1\\text{}mol}{40.08\\text{}g}\\right)}{1.4\\text{L}}=\\text{1.8}\\times {10}^{-3}\\text{}M[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]{{\\text{[PO}}_{4}{}^{3-}]}^{2}=\\frac{1\\times {10}^{-25}}{{\\left(1.8\\times {10}^{-3}\\right)}^{3}}=\\text{1.7}\\times {10}^{-17}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]{{\\text{[PO}}_{4}{}^{3-}]}^{2}=4\\times {10}^{-9}\\text{}M[\/latex]<\/p>\r\n51.\u00a0Calculate the amount of Mg<sup>2+<\/sup> present in sea water; then use <em>K<\/em><sub>sp<\/sub> to calculate the amount of Ca(OH)<sub>2<\/sub> required to precipitate the magnesium.\r\n<p style=\"text-align: center;\">mass Mg = 1.00 \u00d7 10<sup>3<\/sup> L \u00d7 1000 cm<sup>3<\/sup>\/L \u00d7 1.026 g\/cm<sup>3<\/sup> \u00d7 1272 ppm \u00d7 10<sup>\u20136<\/sup> ppm<sup>\u20131<\/sup> = 1.305 \u00d7 10<sup>3<\/sup> g<\/p>\r\nThe concentration is 1.305 g\/L. If 99.9% is to be recovered 0.999 \u00d7 1.305 g\/L = 1.304 g\/L will be obtained. The molar concentration is:\r\n<p style=\"text-align: center;\">[latex]\\frac{1.304{\\text{g L}}^{-1}}{24.305{\\text{g mol}}^{-1}}=\\text{0.05365}M[\/latex]<\/p>\r\nAs the Ca(OH)<sub>2<\/sub> reacts with Mg<sup>2+<\/sup> on a 1:1 mol basis, the amount of Ca(OH)<sub>2<\/sub> required to precipitate 99.9% of the Mg<sup>2+<\/sup> in 1 L is:\r\n\r\n0.05365 M \u00d7 74.09 g\/mol Ca(OH)<sub>2<\/sub> = 3.97 g\/L\r\n\r\nFor treatment of 1000 L, 1000 L \u00d7 3.97 g\/L = 3.97 \u00d7 10<sup>3<\/sup> g = 3.97 kg. However, additional [OH<sup>\u2013<\/sup>] must be added to maintain the equilibrium:\r\n<p style=\"text-align: center;\">[latex]{\\text{Mg(OH)}}_{2}\\left(s\\right)\\rightleftharpoons {\\text{Mg}}^{\\text{2+}}\\left(aq\\right)+{\\text{2OH}}^{-}\\left(aq\\right)\\left({K}_{\\text{sp}}=\\text{1.5}\\times {10}^{-11}\\right)[\/latex]<\/p>\r\nWhen the initial 1.035 g\/L is reduced to 0.1% of the original, the molarity is calculated as:\r\n<p style=\"text-align: center;\">[latex]\\frac{0.001\\times {\\text{1.305 g L}}^{-1}}{24.305{\\text{g mol}}^{-1}}=\\text{5.369}\\times {10}^{-5}\\text{}M[\/latex]<\/p>\r\nThe added amount of OH<sup>\u2013<\/sup> required is found from the solubility product:\r\n<p style=\"text-align: center;\">[Mg<sup>2+<\/sup>][OH<sup>\u2013<\/sup>]<sup>2<\/sup> = (5.369 \u00d7 10<sup>\u20135<\/sup>)[OH<sup>\u2013<\/sup>]<sup>2<\/sup> = 1.5 \u00d7 10<sup>\u201311<\/sup> = <em>K<\/em><sub>sp<\/sub><\/p>\r\n<p style=\"text-align: center;\">[OH<sup>\u2013<\/sup>] = 5.29 \u00d7 10<sup>\u20134<\/sup><\/p>\r\nThus an additional [latex]\\frac{1}{2}[\/latex] \u00d7 5.29 \u00d7 10<sup>\u20134<\/sup> mol (2.65 \u00d7 10<sup>\u20134<\/sup> mol) of Ca(OH)<sub>2<\/sub> per liter is required to supply the OH<sup>\u2013<\/sup>. For 1000 L:\r\n<p style=\"text-align: center;\">[latex]{\\text{mass Ca(OH)}}_{2}=2.65\\times {10}^{-4}{\\text{mol Ca(OH)}}_{2}{\\text{L}}^{-1}\\times \\text{1.00}\\times {10}^{3}\\text{L}\\times \\frac{74.0946\\text{g}}{{\\text{mol Ca(OH)}}_{2}}=\\text{20 g}[\/latex]<\/p>\r\nThe total Ca(OH)<sub>2<\/sub> required is 3.97 kg + 0.020 kg = 3.99 kg.\r\n\r\n53.\u00a0(a) <em>K<\/em><sub>sp<\/sub> = [La<sup>3+<\/sup>] [latex]{\\left[{\\text{IO}}_{3}{}^{-}\\right]}^{3}[\/latex] = [latex]\\left(\\frac{1}{3}\\times \\text{3.1}\\times {10}^{-3}\\right)[\/latex] (3.1 \u00d7 10<sup>\u20133<\/sup>)<sup>3<\/sup> = (0.0010)(3.0 \u00d7 10<sup>\u20138<\/sup>) = 3.0 \u00d7 10<sup>\u201311<\/sup>;\r\n\r\n(b) <em>K<\/em><sub>sp<\/sub> = [Cu<sup>2+<\/sup>] [latex]{\\left[{\\text{IO}}_{3}{}^{-}\\right]}^{2}[\/latex] = <em>x<\/em>(2<em>x<\/em>)<sup>2<\/sup> = 7.4 \u00d7 10<sup>\u20138<\/sup>\r\n\r\n4<em>x<\/em><sup>3<\/sup> = 7.4 \u00d7 10<sup>\u20138<\/sup>\r\n\r\n<em>x<\/em><sup>3<\/sup> = 1.85 \u00d7 10<sup>\u20138<\/sup>\r\n\r\n<em>x<\/em> = 2.64 \u00d7 10<sup>\u20133<\/sup>\r\n\r\n[Cu<sup>2+<\/sup>] = 2.6 \u00d7 10<sup>\u20133<\/sup>\r\n\r\n[latex]\\left[{\\text{IO}}_{3}{}^{-}\\right][\/latex] = 2<em>x<\/em> = 5.3 \u00d7 10<sup>\u20133<\/sup>\r\n\r\n55.\u00a01.8 \u00d7 10<sup>\u20135<\/sup> g Pb(OH)<sub>2<\/sub>\r\n\r\n57.\u00a0[latex]{\\text{Mg(OH)}}_{2}\\text{(s)}\\rightleftharpoons {\\text{Mg}}^{\\text{2+}}+{\\text{2OH}}^{-}{K}_{\\text{sp}}=\\left[{\\text{Mg}}^{\\text{2+}}\\right]{\\left[{\\text{OH}}^{-}\\right]}^{2}[\/latex]\r\n\r\n7.1 \u00d7 10<sup>\u201312<\/sup> = (<em>x<\/em>)(2<em>x<\/em>)<sup>2<\/sup> = 4<em>x<\/em><sup>3<\/sup>\r\n\r\n<em>x<\/em> = 1.21 \u00d7 10<sup>\u20134<\/sup><em>M<\/em> = [Mg<sup>2+<\/sup>]\r\n\r\n[latex]\\frac{1.21\\times {10}^{-4}\\text{mol}}{\\text{L}}\\times 0.20\\text{0 L}\\times \\frac{{\\text{1 mol Mg(OH)}}_{2}}{{\\text{1 mol Mg}}^{\\text{2+}}}\\times \\frac{{\\text{58.3 g Mg(OH)}}_{2}}{{\\text{1 mol Mg(OH)}}_{2}}=\\text{1.14}\\times {10}^{-3}{\\text{Mg(OH)}}_{2}[\/latex]\r\n\r\n59.\u00a0SrCO<sub>3<\/sub> will form first, since it has the smallest <em>K<\/em><sub>sp<\/sub> value it is the least soluble. BaCO<sub>3<\/sub> will be the last to precipitate, it has the largest <em>K<\/em><sub>sp<\/sub> value.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Glossary<\/h2>\r\n<strong>common ion effect:<\/strong> effect on equilibrium when a substance with an ion in common with the dissolved species is added to the solution; causes a decrease in the solubility of an ionic species, or a decrease in the ionization of a weak acid or base\r\n\r\n<strong>molar solubility: <\/strong>solubility of a compound expressed in units of moles per liter (mol\/L)\r\n\r\n<strong>selective precipitation: <\/strong>process in which ions are separated using differences in their solubility with a given precipitating reagent\r\n\r\n<strong>solubility product (<em>K<\/em><sub>sp<\/sub>): <\/strong>equilibrium constant for the dissolution of a slightly soluble electrolyte","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<p>By the end of this module, you will be able to:<\/p>\n<ul>\n<li>Write chemical equations and equilibrium expressions representing solubility equilibria<\/li>\n<li>Carry out equilibrium computations involving solubility, equilibrium expressions, and solute concentrations<\/li>\n<\/ul>\n<\/div>\n<p>The preservation of medical laboratory blood samples, mining of sea water for magnesium, formulation of over-the-counter medicines such as Milk of Magnesia and antacids, and treating the presence of hard water in your home\u2019s water supply are just a few of the many tasks that involve controlling the equilibrium between a slightly soluble ionic solid and an aqueous solution of its ions.<\/p>\n<p>In some cases, we want to prevent\u00a0dissolution from occurring. Tooth decay, for example, occurs when the calcium hydroxylapatite, which has the formula Ca<sub>5<\/sub>(PO<sub>4<\/sub>)<sub>3<\/sub>(OH), in our teeth dissolves. The dissolution process is aided when bacteria in our mouths feast on the sugars in our diets to produce lactic acid, which reacts with the hydroxide ions in the calcium hydroxylapatite. Preventing the dissolution prevents the decay. On the other hand, sometimes we want a substance to dissolve. We want the calcium carbonate in a chewable antacid to dissolve because the CO<sub>3<\/sub><sup>2\u2212<\/sup> ions produced in this process help soothe an upset stomach.<\/p>\n<p>In this section, we will find out how we can control the dissolution of a slightly soluble ionic solid by the application of Le Ch\u00e2telier\u2019s principle. We will also learn how to use the equilibrium constant of the reaction to determine the concentration of ions present in a solution.<\/p>\n<h2>The Solubility Product Constant<\/h2>\n<p>Silver chloride is what\u2019s known as a sparingly soluble ionic solid (Figure\u00a01). Recall from the solubility rules in an earlier chapter that halides of Ag<sup>+<\/sup> are not normally soluble. However, when we add an excess of solid AgCl to water, it dissolves to a small extent and produces a mixture consisting of a very dilute solution of Ag<sup>+<\/sup> and Cl<sup>\u2013<\/sup> ions in equilibrium with undissolved silver chloride:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{AgCl}\\left(s\\right){\\underset{\\text{precipitation}}{\\overset{\\text{dissolution}}{\\longrightleftharpoons}}}{\\text{Ag}}^{+}\\left(aq\\right)+{\\text{Cl}}^{-}\\left(aq\\right)[\/latex]<\/p>\n<p>This equilibrium, like other equilibria, is dynamic; some of the solid AgCl continues to dissolve, but at the same time, Ag<sup>+<\/sup> and Cl<sup>\u2013<\/sup> ions in the solution combine to produce an equal amount of the solid. At equilibrium, the opposing processes have equal rates.<\/p>\n<div style=\"width: 890px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213936\/CNX_Chem_15_01_AgCl.jpg\" alt=\"Two beakers are shown with a bidirectional arrow between them. Both beakers are just over half filled with a clear, colorless liquid. The beaker on the left shows a cubic structure composed of alternating green and slightly larger grey spheres. Evenly distributed in the region outside, 11 space filling models are shown. These are each composed of a central red sphere with two smaller white spheres attached in a bent arrangement. In the beaker on the right, the green and grey spheres are no longer connected in a cubic structure. Nine green spheres, 10 grey spheres, and 11 red and white molecules are evenly mixed and distributed throughout the liquid in the beaker.\" width=\"880\" height=\"354\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure\u00a01. Silver chloride is a sparingly soluble ionic solid. When it is added to water, it dissolves slightly and produces a mixture consisting of a very dilute solution of Ag+ and Cl\u2013 ions in equilibrium with undissolved silver chloride.<\/p>\n<\/div>\n<p>The equilibrium constant for the equilibrium between a slightly soluble ionic solid and a solution of its ions is called the <b>solubility product (<em>K<\/em><sub>sp<\/sub>)<\/b> of the solid. Recall from the chapter on solutions and colloids that we use an ion\u2019s concentration as an approximation of its activity in a dilute solution. For silver chloride, at equilibrium:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{AgCl}\\left(s\\right)\\rightleftharpoons {\\text{Ag}}^{\\text{+}}\\left(aq\\right)+{\\text{Cl}}^{-}\\left(aq\\right){K}_{\\text{sp}}=\\left[{\\text{Ag}}^{\\text{+}}\\left(aq\\right)\\right]\\left[{\\text{Cl}}^{-}\\left(aq\\right)\\right][\/latex]<\/p>\n<p>When looking at dissolution reactions such as this, the solid is listed as a reactant, whereas the ions are listed as products. The solubility product constant, as with every equilibrium constant expression, is written as the product of the concentrations of each of the ions, raised to the power of their stoichiometric coefficients. Here, the solubility product constant is equal to Ag<sup>+<\/sup> and Cl<sup>\u2013<\/sup> when a solution of silver chloride is in equilibrium with undissolved AgCl. There is no denominator representing the reactants in this equilibrium expression since the reactant is a pure solid; therefore [AgCl] does not appear in the expression for <em>K<\/em><sub>sp<\/sub>.<\/p>\n<p>Some common solubility products are listed in Table\u00a01\u00a0according to their <em>K<\/em><sub>sp<\/sub> values, whereas a more extensive compilation of products appears in <a class=\"target-chapter\" href=\".\/chapter\/solubility-products\/\" target=\"_blank\">Solubility Products<\/a>. Each of these equilibrium constants is much smaller than 1 because the compounds listed are only slightly soluble. A small <em>K<\/em><sub>sp<\/sub> represents a system in which the equilibrium lies to the left, so that relatively few hydrated ions would be present in a saturated solution.<\/p>\n<table id=\"fs-idp15839744\" class=\"span-all\" summary=\"No Summary Text\">\n<thead>\n<tr valign=\"middle\">\n<th colspan=\"2\">Table\u00a01. Common Solubility Products by Decreasing Equilibrium Constants<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"middle\">\n<td>Substance<\/td>\n<td><em>K<\/em><sub>sp<\/sub> at 25 \u00b0C<\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td>CuCl<\/td>\n<td>1.2 \u00d7 10<sup>\u20136<\/sup><\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td>CuBr<\/td>\n<td>6.27 \u00d7 10<sup>\u20139<\/sup><\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td>AgI<\/td>\n<td>1.5 \u00d7 10<sup>\u201316<\/sup><\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td>PbS<\/td>\n<td>7 \u00d7 10<sup>\u201329<\/sup><\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td>Al(OH)<sub>3<\/sub><\/td>\n<td>2 \u00d7 10<sup>\u201332<\/sup><\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td>Fe(OH)<sub>3<\/sub><\/td>\n<td>4 \u00d7 10<sup>\u201338<\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"textbox examples\">\n<h3>Example 1:\u00a0Writing Equations and Solubility Products<\/h3>\n<p>Write the ionic equation for the dissolution and the solubility product expression for each of the following slightly soluble ionic compounds:<\/p>\n<ol>\n<li>AgI, silver iodide, a solid with antiseptic properties<\/li>\n<li>CaCO<sub>3<\/sub>, calcium carbonate, the active ingredient in many over-the-counter chewable antacids<\/li>\n<li>Mg(OH)<sub>2<\/sub>, magnesium hydroxide, the active ingredient in Milk of Magnesia<\/li>\n<li>Mg(NH<sub>4<\/sub>)PO<sub>4<\/sub>, magnesium ammonium phosphate, an essentially insoluble substance used in tests for magnesium<\/li>\n<li>Ca<sub>5<\/sub>(PO<sub>4<\/sub>)<sub>3<\/sub>OH, the mineral apatite, a source of phosphate for fertilizers<\/li>\n<\/ol>\n<p>(Hint: When determining how to break 4 and 5 up into ions, refer to the list of polyatomic ions in the section on chemical nomenclature.)<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q50078\">Show Answer<\/span><\/p>\n<div id=\"q50078\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]\\text{AgI}\\left(s\\right)\\rightleftharpoons\\text{Ag}^{+}\\left(aq\\right)+\\text{I}^{-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,{;}\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{sp}}=\\left[\\text{Ag}^{+}\\right]\\left[\\text{I}^{-}\\right][\/latex]<\/li>\n<li>[latex]\\text{CaCO}_{3}\\left(s\\right)\\rightleftharpoons\\text{Ca}^{2+}\\left(aq\\right)+\\text{CO}_{3}^{2-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,{;}\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{sp}}=\\left[\\text{Ca}^{2+}\\right]\\left[\\text{CO}_{3}^{2-}\\right][\/latex]<\/li>\n<li>[latex]\\text{Mg}\\left(\\text{OH}\\right)_{2}\\left(s\\right)\\rightleftharpoons\\text{Mg}^{2+}\\left(aq\\right)+2\\text{OH}^{-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,{;}\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{sp}}=\\left[\\text{Mg}^{2+}\\right]\\left[\\text{OH}^{-}\\right]^{2}[\/latex]<\/li>\n<li>[latex]\\text{Mg}\\left(\\text{NH}_{4}\\right)\\text{PO}_{4}\\left(s\\right)\\rightleftharpoons\\text{Mg}^{2+}\\left(aq\\right)+\\text{NH}_{4}^{+}\\left(aq\\right)+\\text{PO}_{4}^{3-}\\left(aq\\right)  \\,\\,\\,\\,\\,\\,\\,{;}\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{sp}}=\\left[\\text{Mg}^{2+}\\right]\\left[\\text{NH}_{4}^\\text{+}\\right]\\left[\\text{PO}_{4}^{3-}\\right][\/latex]<\/li>\n<li>[latex]\\text{Ca}_{5}\\left(\\text{PO}_{4}\\right)3\\text{OH}\\left(\\text{s}\\right)\\rightleftharpoons5\\text{Ca}^{2+}\\left(aq\\right)+3\\text{PO}_{4}^{3-}\\left(aq\\right)+\\text{OH}-\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,{;}\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{sp}}=\\left[\\text{Ca}^{2+}\\right]^{5}\\left[\\text{PO}_{4}^{3-}\\right]^{3}\\left[\\text{OH}^{-}\\right][\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<h4>Check Your Learning<\/h4>\n<p>Write the ionic equation for the dissolution and the solubility product for each of the following slightly soluble compounds:<\/p>\n<ol>\n<li>BaSO<sub>4<\/sub><\/li>\n<li>Ag<sub>2<\/sub>SO<sub>4<\/sub><\/li>\n<li>Al(OH)<sub>3<\/sub><\/li>\n<li>Pb(OH)Cl<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q636225\">Show Answer<\/span><\/p>\n<div id=\"q636225\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]\\text{BaSO}_{4}\\left(s\\right)\\rightleftharpoons\\text{Ba}^{2+}\\left(aq\\right)+\\text{SO}_{4}^{2-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,{;}\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{sp}}=\\left[\\text{Ba}^{2+}\\right]\\left[\\text{SO}_{4}^{2-}\\right][\/latex]<\/li>\n<li>[latex]{\\text{Ag}}_{2}{\\text{SO}}_{4}\\left(s\\right)\\rightleftharpoons {\\text{2Ag}}^{\\text{+}}\\left(aq\\right)+{\\text{SO}}_{4}{}^{2-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,{;}\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{sp}}={\\left[{\\text{Ag}}^{\\text{+}}\\right]}^{2}\\left[{\\text{SO}}_{4}{}^{2-}\\right][\/latex]<\/li>\n<li>[latex]\\text{Al}\\left(\\text{OH}\\right)_{3}\\left(s\\right)\\rightleftharpoons\\text{Al}^{2+}\\left(aq\\right)+3\\text{OH}^{-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,{;}\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{sp}}=\\left[\\text{Al}^{3+}\\right]\\left[\\text{OH}^{-}\\right]^{3}[\/latex]<\/li>\n<li>[latex]\\text{Pb}\\left(\\text{OH}\\right)\\text{Cl}\\left(s\\right)\\rightleftharpoons\\text{Pb}^{2+}\\left(aq\\right)+\\text{OH}^{-}\\left(aq\\right)+\\text{Cl}^{-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,{;}\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{sp}}=\\left[\\text{Pb}^{2+}\\right]\\left[\\text{OH}^{-}\\right]\\left[\\text{Cl}^{-}\\right][\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<p>Now we will extend the discussion of <em>K<\/em><sub>sp<\/sub> and show how the solubility product constant is determined from the solubility of its ions, as well as how <em>K<\/em><sub>sp<\/sub> can be used to determine the molar solubility of a substance.<\/p>\n<h2><em>K<\/em><sub>sp<\/sub> and Solubility<\/h2>\n<p>Recall that the definition of <em>solubility<\/em> is the maximum possible concentration of a solute in a solution at a given temperature and pressure. We can determine the solubility product of a slightly soluble solid from that measure of its solubility at a given temperature and pressure, provided that the only significant reaction that occurs when the solid dissolves is its dissociation into solvated ions, that is, the only equilibrium involved is:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{M}}_{p}{\\text{X}}_{q}\\left(s\\right)\\rightleftharpoons p{\\text{M}}^{\\text{m+}}\\left(aq\\right)+q{\\text{X}}^{\\text{n}-}\\left(aq\\right)[\/latex]<\/p>\n<p>In this case, we calculate the solubility product by taking the solid\u2019s solubility expressed in units of moles per liter (mol\/L), known as its <strong>molar solubility<\/strong>.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 2:\u00a0Calculation of <em>K<\/em><sub>sp<\/sub> from Equilibrium Concentrations<\/h3>\n<p>We began the chapter with an informal discussion of how the mineral fluorite\u00a0is formed. Fluorite, CaF<sub>2<\/sub>, is a slightly soluble solid that dissolves according to the equation:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{CaF}}_{2}\\left(s\\right)\\rightleftharpoons {\\text{Ca}}^{\\text{2+}}\\left(aq\\right)+{\\text{2F}}^{-}\\left(aq\\right)[\/latex]<\/p>\n<p>The concentration of Ca<sup>2+<\/sup> in a saturated solution of CaF<sub>2<\/sub> is 2.1 \u00d7 10<sup>\u20134<\/sup><em>M<\/em>; therefore, that of F<sup>\u2013<\/sup> is 4.2 \u00d7 10<sup>\u20134<\/sup><em>M<\/em>, that is, twice the concentration of Ca<sup>2+<\/sup>. What is the solubility product of fluorite?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q15268\">Show Answer<\/span><\/p>\n<div id=\"q15268\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, write out the <em>K<\/em><sub>sp<\/sub> expression, then substitute in concentrations and solve for <em>K<\/em><sub>sp<\/sub>:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{CaF}}_{2}\\left(s\\right)\\rightleftharpoons {\\text{Ca}}^{\\text{2+}}\\left(aq\\right)+{\\text{2F}}^{-}\\left(aq\\right)[\/latex]<\/p>\n<p>A saturated solution is a solution at equilibrium with the solid. Thus:<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{\\text{sp}}={\\text{[Ca}}^{\\text{2+}}\\text{]}\\text{[}{\\text{F}}^{-}{]}^{2}=\\text{(2.1}\\times {10}^{-4}\\big){\\left(4.2\\times {10}^{-4}\\right)}^{2}=\\text{3.7}\\times {10}^{-11}[\/latex]<\/p>\n<p>As with other equilibrium constants, we do not include units with <em>K<\/em><sub>sp<\/sub>.<\/p>\n<\/div>\n<\/div>\n<h4>Check Your Learning<\/h4>\n<p>In a saturated solution that is in contact with solid Mg(OH)<sub>2<\/sub>, the concentration of Mg<sup>2+<\/sup> is 3.7 \u00d7 10<sup>\u20135<\/sup><em>M<\/em>. What is the solubility product for Mg(OH)<sub>2<\/sub>?<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{Mg(OH)}}_{2}\\left(s\\right)\\rightleftharpoons {\\text{Mg}}^{\\text{2+}}\\left(aq\\right)+{\\text{2OH}}^{-}\\left(aq\\right)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q918992\">Show Answer<\/span><\/p>\n<div id=\"q918992\" class=\"hidden-answer\" style=\"display: none\">2.0 \u00d7 10<sup>\u201313<\/sup><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 3:\u00a0Determination of Molar Solubility from <em>K<\/em><sub>sp<\/sub><\/h3>\n<p>The <em>K<\/em><sub>sp<\/sub> of copper(I) bromide, CuBr, is 6.3 \u00d7 10<sup>\u20139<\/sup>. Calculate the molar solubility of copper bromide.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q315393\">Show Answer<\/span><\/p>\n<div id=\"q315393\" class=\"hidden-answer\" style=\"display: none\">\n<p>The solubility product constant of copper(I) bromide is 6.3 \u00d7 10<sup>\u20139<\/sup>.\u00a0The reaction is:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{CuBr}\\left(s\\right)\\rightleftharpoons {\\text{Cu}}^{\\text{+}}\\left(aq\\right)+{\\text{Br}}^{-}\\left(aq\\right)[\/latex]<\/p>\n<p>First, write out the solubility product equilibrium constant expression:<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{\\text{sp}}=\\text{[}{\\text{Cu}}^{\\text{+}}{]\\text{[}\\text{Br}}^{-}\\text{]}[\/latex]<\/p>\n<p>Create an ICE table (as introduced in the chapter on fundamental equilibrium concepts), leaving the CuBr column empty as it is a solid and does not contribute to the <em>K<\/em><sub>sp<\/sub>:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213937\/CNX_Chem_15_01_ICETable1_img.jpg\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), and Equilibrium concentration ( M ). The second column has the header of, \u201cC u B r equilibrium arrow C u superscript positive sign plus B r superscript negative sign.\u201d Under the second column is a subgroup of three rows and three columns. The first column is blank. The second column has the following: 0, x, 0 plus x equals x. The third column has the following 0, x, 0 plus x equals x.\" width=\"879\" height=\"238\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p>At equilibrium:<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{\\text{sp}}=\\left[\\text{Cu}^{+}\\right]\\left[\\text{Br}^{-}\\right][\/latex]<br \/>\n[latex]6.3\\times {10}^{-9}=\\left(x\\right)\\left(x\\right)={x}^{2}[\/latex]<br \/>\n[latex]x=\\sqrt{\\left(6.3\\times {10}^{-9}\\right)}=\\text{7.9}\\times {10}^{-5}[\/latex]<\/p>\n<p>Therefore, the molar solubility of CuBr is 7.9 \u00d7 10<sup>\u20135<\/sup><em>M<\/em>.<\/p>\n<\/div>\n<\/div>\n<h4>Check Your Learning<\/h4>\n<p>The <em>K<\/em><sub>sp<\/sub> of AgI is 1.5 \u00d7 10<sup>\u201316<\/sup>. Calculate the molar solubility of silver iodide.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q528472\">Show Answer<\/span><\/p>\n<div id=\"q528472\" class=\"hidden-answer\" style=\"display: none\">1.2 \u00d7 10<sup>\u20138<\/sup><em>M<\/em><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 4:\u00a0Determination of Molar Solubility from <em>K<\/em><sub>sp<\/sub>, Part II<\/h3>\n<p>The <em>K<\/em><sub>sp<\/sub> of calcium hydroxide, Ca(OH)<sub>2<\/sub>, is 8.0 \u00d7 10<sup>\u20136<\/sup>. Calculate the molar solubility of calcium hydroxide.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q17713\">Show Answer<\/span><\/p>\n<div id=\"q17713\" class=\"hidden-answer\" style=\"display: none\">\n<p>The solubility product constant of calcium hydroxide is 8.0 \u00d7 10<sup>\u20136<\/sup>.\u00a0The reaction is:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{Ca(OH)}}_{2}\\left(s\\right)\\rightleftharpoons {\\text{Ca}}^{\\text{2+}}\\left(aq\\right)+{\\text{2OH}}^{-}\\left(aq\\right)[\/latex]<\/p>\n<p>First, write out the solubility product equilibrium constant expression:<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{\\text{sp}}=\\left[\\text{Ca}^{2+}\\right]\\left[\\text{OH}^{-}\\right]^{2}[\/latex]<\/p>\n<p>Create an ICE table, leaving the Ca(OH)<sub>2<\/sub> column empty as it is a solid and does not contribute to the <em>K<\/em><sub>sp<\/sub>:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-4942 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/08\/23214454\/CNX_Chem_15_01_ICETable7_img.jpg\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), and Equilibrium concentration ( M ). The second column has the header of, \u201cC a ( O H ) subscript 2 equilibrium arrow C a superscript 2 positive sign plus 2 O H superscript negative sign.\u201d Under the second column is a subgroup of three rows and three columns. The first column is blank. The second column has the following: 0, x, and 0 plus x equals x. The third column has the following 0, 2 x, and 0 plus 2 x equals 2 x.\" width=\"880\" height=\"238\" \/><\/p>\n<p>At equilibrium:<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{\\text{sp}}=\\left[\\text{Ca}^{2+}\\right]\\left[\\text{OH}^{-}\\right]^{2}[\/latex]<br \/>\n[latex]8.0\\times{10}^{-6}=\\left(x\\right)\\left(2x\\right)^{2}=\\left(x\\right)\\left(4x^{2}\\right)=4{x}^{3}[\/latex]<br \/>\n[latex]x=\\sqrt[3]{\\frac{8.0\\times {10}^{-6}}{4}}=\\text{1.3}\\times {10}^{-2}[\/latex]<\/p>\n<p>Therefore, the molar solubility of Ca(OH)<sub>2<\/sub> is 1.3 \u00d7 10<sup>\u20132<\/sup><em>M<\/em>.<\/p>\n<\/div>\n<\/div>\n<h4>Check Your Learning<\/h4>\n<p>The <em>K<\/em><sub>sp<\/sub> of PbI<sub>2<\/sub> is 1.4 \u00d7 10<sup>\u20138<\/sup>. Calculate the molar solubility of lead(II) iodide.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q194233\">Show Answer<\/span><\/p>\n<div id=\"q194233\" class=\"hidden-answer\" style=\"display: none\">1.5 \u00d7 10<sup>\u20133<\/sup><em>M<\/em><\/div>\n<\/div>\n<\/div>\n<p>Note that solubility is not always given as a molar value. When the solubility of a compound is given in some unit other than moles per liter, we must convert the solubility into moles per liter (i.e., molarity) in order to use it in the solubility product constant expression. Example 5\u00a0shows how to perform those unit conversions before determining the solubility product equilibrium.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 5:\u00a0Determination of <em>K<\/em><sub>sp<\/sub> from Gram Solubility<\/h3>\n<p>Many of the pigments used by artists in oil-based paints (Figure\u00a02) are sparingly soluble in water. For example, the solubility of the artist\u2019s pigment chrome yellow, PbCrO<sub>4<\/sub>, is 4.3 \u00d7 10<sup>\u20135<\/sup> g\/L. Determine the solubility product equilibrium constant for PbCrO<sub>4<\/sub>.<\/p>\n<div style=\"width: 511px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213940\/CNX_Chem_15_01_OilPaints.jpg\" alt=\"A photograph is shown of a portion of an oil painting which reveals colors of orange, brown, yellow, green, blue, and purple colors in its strokes. A few water droplets rest on the surface.\" width=\"501\" height=\"334\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. Oil paints contain pigments that are very slightly soluble in water. In addition to chrome yellow (PbCrO<sub>4<\/sub>), examples include Prussian blue (Fe<sub>7<\/sub>(CN)<sub>18<\/sub>), the reddish-orange color vermilion (HgS), and green color veridian (Cr<sub>2<\/sub>O<sub>3<\/sub>). (credit: Sonny Abesamis)<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q586402\">Show Answer<\/span><\/p>\n<div id=\"q586402\" class=\"hidden-answer\" style=\"display: none\">\n<p>We are given the solubility of PbCrO<sub>4<\/sub> in grams per liter. If we convert this solubility into moles per liter, we can find the equilibrium concentrations of Pb<sup>2+<\/sup> and [latex]{\\text{CrO}}_{4}^{2-}[\/latex], then <em>K<\/em><sub>sp<\/sub>:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4943\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213942\/CNX_Chem_15_01_PbCrO4_img.jpg\" alt=\"This figure shows four horizontally oriented rectangles. The first three from the left are shaded green and the last one at the right is shaded white. Right pointing arrows between the rectangles are labeled \u201c1,\u201d \u201c2,\u201d and \u201c3\u201d moving left to right across the diagram. The first rectangle is labeled \u201cSolubility of P b C r O subscript 4, in g divdided by L.\u201d The second rectangle is labeled \u201c&#091; P b C r O subscript 4 &#093;, in m o l divided by L.\u201d The third is labeled \u201c&#091; P b superscript 2 plus&#093; and &#091; C r O subscript 4 superscript 2 negative &#093;.\u201d The fourth rectangle is labeled \u201cK subscript s p.\u201d\" width=\"880\" height=\"156\" \/><\/p>\n<ol>\n<li><em>Use the molar mass of PbCrO<sub>4<\/sub><\/em> [latex]\\left(\\frac{323.2\\text{g}}{1\\text{mol}}\\right)[\/latex] <em>to convert the solubility of PbCrO<sub>4<\/sub> in grams per liter into moles per liter:<br \/>\n<\/em>[latex]\\begin{array}{l}{ }\\left[{\\text{PbCrO}}_{4}\\right]&=&\\frac{4.3\\times {10}^{-5}{\\text{g PbCrO}}_{4}}{1\\text{L}}\\times \\frac{1{\\text{mol PbCrO}}_{4}}{323.2{\\text{g PbCrO}}_{4}}\\\\& =&\\frac{1.3\\times {10}^{-7}{\\text{mol PbCrO}}_{4}}{1\\text{L}}\\\\& =&\\text{}1.3\\times {10}^{-7}M\\end{array}[\/latex]<\/li>\n<li><em>The chemical equation for the dissolution indicates that 1 mol of PbCrO<sub>4<\/sub> gives 1 mol of Pb<sup>2+<\/sup>(aq) and 1 mol of<\/em> [latex]{\\text{CrO}}_{4}{}^{\\text{2-}}\\text{(}aq\\text{)}:[\/latex][latex]{\\text{PbCrO}}_{4}\\left(s\\right)\\rightleftharpoons {\\text{Pb}}^{2+}\\left(aq\\right)+{\\text{CrO}}_{4}{}^{2-}\\left(aq\\right)[\/latex]Thus, both [Pb<sup>2+<\/sup>] and [latex]\\left[{\\text{CrO}}_{4}{}^{\\text{2-}}\\right][\/latex] are equal to the molar solubility of PbCrO<sub>4<\/sub>:[latex]\\left[\\text{Pb}^{2+}\\right]=\\left[\\text{CrO}_{4}^{2-}\\right]=1.3\\times{10}^{-7}M[\/latex]<\/li>\n<li><em>Solve. K<\/em><sub>sp<\/sub> = [Pb<sup>2+<\/sup>] [latex]\\left[{\\text{CrO}}_{4}{}^{2-}\\right][\/latex] = (1.3 \u00d7 10<sup>\u20137<\/sup>)(1.3 \u00d7 10<sup>\u20137<\/sup>) = 1.7 \u00d7 10<sup>\u201314<\/sup><\/li>\n<\/ol>\n<\/div>\n<\/div>\n<h4>Check Your Learning<\/h4>\n<p>The solubility of TlCl [thallium(I) chloride], an intermediate formed when thallium is being isolated from ores, is 3.2 grams per liter at 20 \u00b0C. What is its solubility product?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q95775\">Show Answer<\/span><\/p>\n<div id=\"q95775\" class=\"hidden-answer\" style=\"display: none\">1.4 \u00d7 10<sup>\u20134<\/sup> (1.5 \u00d7 10<sup>\u20134<\/sup> if we round the solubility to two digits before calculating <em>K<\/em><sub>sp<\/sub>)<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 6:\u00a0Calculating the Solubility of Hg<sub>2<\/sub>Cl<sub>2<\/sub><\/h3>\n<p>Calomel, Hg<sub>2<\/sub>Cl<sub>2<\/sub>, is a compound composed of the diatomic ion of mercury(I), [latex]{\\text{Hg}}_{2}{}^{\\text{2+}}[\/latex], and chloride ions, Cl<sup>\u2013<\/sup>. Although most mercury compounds are now known to be poisonous, eighteenth-century physicians used calomel as a medication. Their patients rarely suffered any mercury poisoning from the treatments because calomel is quite insoluble:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{Hg}}_{2}{\\text{Cl}}_{2}\\left(s\\right)\\rightleftharpoons {\\text{Hg}}_{2}{}^{\\text{2+}}\\left(aq\\right)+{\\text{2Cl}}^{-}\\left(aq\\right){K}_{\\text{sp}}=\\text{1.1}\\times {10}^{-18}[\/latex]<\/p>\n<p>Calculate the molar solubility of Hg<sub>2<\/sub>Cl<sub>2<\/sub>.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q561036\">Show Answer<\/span><\/p>\n<div id=\"q561036\" class=\"hidden-answer\" style=\"display: none\">\n<p>The molar solubility of Hg<sub>2<\/sub>Cl<sub>2<\/sub> is equal to the concentration of [latex]{\\text{Hg}}_{2}{}^{\\text{2+}}[\/latex] ions because for each 1 mol of Hg<sub>2<\/sub>Cl<sub>2<\/sub> that dissolves, 1 mol of [latex]{\\text{Hg}}_{2}{}^{\\text{2+}}[\/latex] forms:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4944\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213944\/CNX_Chem_15_01_Format_img.jpg\" alt=\"This figure shows four horizontally oriented light green rectangles. Right pointing arrows are placed between them. The first rectangle is labeled \u201cDetermine the direction of change.\u201d The second rectangle is labeled \u201cDetermine x and the equilibrium concentrations.\u201d The third is labeled \u201cSolve for x and the equilibrium concentrations.\u201d The fourth rectangle is labeled \u201cCheck the math.\u201d\" width=\"880\" height=\"156\" \/><\/p>\n<ol>\n<li><em>Determine the direction of change.<\/em> Before any Hg<sub>2<\/sub>Cl<sub>2<\/sub> dissolves, <em>Q<\/em> is zero, and the reaction will shift to the right to reach equilibrium.<\/li>\n<li><em>Determine<\/em> x<em> and equilibrium concentrations.<\/em> Concentrations and changes are given in the following ICE table:<img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213945\/CNX_Chem_15_01_ICETable2_img.jpg\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header of, \u201cH g subscript 2 C l subscript 2 equilibrium arrow H g subscript 2 superscript 2 positive sign plus 2 C l superscript negative sign.\u201d Under the second column is a subgroup of three rows and three columns. The first column is blank. The second column has the following: 0, x, 0 plus x equals x. The third column has the following: 0, 2 x, 0 plus 2 x equals 2 x.\" width=\"838\" height=\"227\" data-media-type=\"image\/jpeg\" \/>Note that the change in the concentration of Cl<sup>\u2013<\/sup> (2<em>x<\/em>) is twice as large as the change in the concentration of [latex]{\\text{Hg}}_{2}{}^{\\text{2+}}[\/latex] (<em>x<\/em>) because 2 mol of Cl<sup>\u2013<\/sup> forms for each 1 mol of [latex]{\\text{Hg}}_{2}{}^{\\text{2+}}[\/latex] that forms. Hg<sub>2<\/sub>Cl<sub>2<\/sub> is a pure solid, so it does not appear in the calculation.<\/li>\n<li><em>Solve for<\/em> x<em> and the equilibrium concentrations.<\/em> We substitute the equilibrium concentrations into the expression for <em>K<\/em><sub>sp<\/sub> and calculate the value of <em>x<\/em>: [latex]{K}_{\\text{sp}}=\\left[\\text{Hg}_{2}^{2+}\\right]\\left[\\text{Cl}^{-}\\right]^{2}[\/latex][latex]1.1\\times{10}^{-18}=\\left(x\\right)\\left(2x\\right)^{2}[\/latex][latex]4{x}^{3}=\\text{1.1}\\times {10}^{-18}[\/latex]\n<p>[latex]x=\\sqrt[3]{\\left(\\frac{1.1\\times {10}^{-18}}{4}\\right)}=\\text{6.5}\\times {10}^{-7}\\text{}M[\/latex]<\/p>\n<p>[latex]\\left[{\\text{Hg}}_{2}{}^{\\text{2+}}\\right]=\\text{6.5}\\times {10}^{-7}\\text{}M=\\text{6.5}\\times {10}^{-7}\\text{}M[\/latex]<\/p>\n<p>[latex]\\left[\\text{Cl}^{-}\\right]=2x=2\\left(6.5\\times{10}^{-7}\\right)=1.3\\times {10}^{-6}M[\/latex]<\/p>\n<p>The molar solubility of Hg<sub>2<\/sub>Cl<sub>2<\/sub> is equal to [latex]{\\text{[Hg}}_{2}{}^{\\text{2+}}][\/latex], or 6.5 \u00d7 10<sup>\u20137<\/sup><em>M<\/em>.<\/li>\n<li><em>Check the work.<\/em> At equilibrium, <em>Q<\/em> = <em>K<\/em><sub>sp<\/sub>: [latex]Q=\\left[\\text{Hg}_{2}^{2+}\\right]\\left[\\text{Cl}^{-}\\right]^{2}=\\left(6.5\\times{10}^{-7}\\right)\\left(1.3\\times{10}^{-6}\\right)^{2}=1.1\\times{10}^{-18}[\/latex]<br \/>\nThe calculations check.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<h4>Check Your Learning<\/h4>\n<p>Determine the molar solubility of MgF<sub>2<\/sub> from its solubility product: <em>K<\/em><sub>sp<\/sub> = 6.4 \u00d7 10<sup>\u20139<\/sup>.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q24056\">Show Answer<\/span><\/p>\n<div id=\"q24056\" class=\"hidden-answer\" style=\"display: none\">1.2 \u00d7 10<sup>\u20133<\/sup><em>M<\/em><\/div>\n<\/div>\n<\/div>\n<p>Tabulated <em>K<\/em><sub>sp<\/sub> values can also be compared to reaction quotients calculated from experimental data to tell whether a solid will precipitate in a reaction under specific conditions: <em>Q<\/em> equals <em>K<\/em><sub>sp<\/sub> at equilibrium; if <em>Q<\/em> is less than <em>K<\/em><sub>sp<\/sub>, the solid will dissolve until <em>Q<\/em> equals <em>K<\/em><sub>sp<\/sub>; if <em>Q<\/em> is greater than <em>K<\/em><sub>sp<\/sub>, precipitation will occur at a given temperature until <em>Q<\/em> equals <em>K<\/em><sub>sp<\/sub>.<\/p>\n<div class=\"textbox shaded\">\n<h3 data-type=\"title\">Using Barium Sulfate for Medical Imaging<\/h3>\n<p>Various types of medical imaging techniques are used to aid diagnoses of illnesses in a noninvasive manner. One such technique utilizes the ingestion of a barium compound before taking an X-ray image. A suspension of barium sulfate, a chalky powder, is ingested by the patient. Since the <em>K<\/em><sub>sp<\/sub> of barium sulfate is 1.1 \u00d7 10<sup>\u201310<\/sup>, very little of it dissolves as it coats the lining of the patient\u2019s intestinal tract. Barium-coated areas of the digestive tract then appear on an X-ray as white, allowing for greater visual detail than a traditional X-ray (Figure\u00a03).<\/p>\n<div id=\"attachment_4945\" style=\"width: 611px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-4945\" class=\"wp-image-4945\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213946\/CNX_Chem_15_01_BariumXray.jpg\" alt=\"This figure contains one image. A black and white abdominal x-ray image is shown in which the intestinal tract of a person is clearly visible in white.\" width=\"601\" height=\"404\" \/><\/p>\n<p id=\"caption-attachment-4945\" class=\"wp-caption-text\">Figure\u00a03. The suspension of barium sulfate coats the intestinal tract, which allows for greater visual detail than a traditional X-ray. (credit modification of work by \u201cglitzy queen00\u201d\/Wikimedia Commons)<\/p>\n<\/div>\n<p>Further diagnostic testing can be done using barium sulfate and fluoroscopy. In fluoroscopy, a continuous X-ray is passed through the body so the doctor can monitor, on a TV or computer screen, the barium sulfate\u2019s movement as it passes through the digestive tract. Medical imaging using barium sulfate can be used to diagnose acid reflux disease, Crohn\u2019s disease, and ulcers in addition to other conditions.<\/p>\n<p>Visit <a href=\"http:\/\/www.hopkinsmedicine.org\/healthlibrary\/conditions\/adult\/digestive_disorders\/barium_x-rays_upper_and_lower_gi_85,p01275\/\" target=\"_blank\">this website for more information on how barium is used in medical diagnoses<\/a> and which conditions it is used to diagnose.<\/p>\n<\/div>\n<h2>Predicting Precipitation<\/h2>\n<p>The equation that describes the equilibrium between solid calcium carbonate and its solvated ions is:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{CaCO}}_{3}\\left(s\\right)\\rightleftharpoons {\\text{Ca}}^{\\text{2+}}\\left(aq\\right)+{\\text{CO}}_{3}^{2-}\\left(aq\\right)[\/latex]<\/p>\n<p>We can establish this equilibrium either by adding solid calcium carbonate to water or by mixing a solution that contains calcium ions with a solution that contains carbonate ions. If we add calcium carbonate to water, the solid will dissolve until the concentrations are such that the value of the reaction quotient (<em>Q<\/em> = [Ca<sup>2+<\/sup>] [latex]\\left[{\\text{CO}}_{3}^{2-}\\right][\/latex] ) is equal to the solubility product (<em>K<\/em><sub>sp<\/sub> = 4.8 \u00d7 10<sup>\u20139<\/sup>). If we mix a solution of calcium nitrate, which contains Ca<sup>2+<\/sup> ions, with a solution of sodium carbonate, which contains [latex]{\\text{CO}}_{3}^{2-}[\/latex] ions, the slightly soluble ionic solid CaCO<sub>3<\/sub> will precipitate, provided that the concentrations of Ca<sup>2+<\/sup> and [latex]{\\text{CO}}_{3}^{2-}[\/latex] ions are such that <em>Q<\/em> is greater than <em>K<\/em><sub>sp<\/sub> for the mixture. The reaction shifts to the left and the concentrations of the ions are reduced by formation of the solid until the value of <em>Q<\/em> equals <em>K<\/em><sub>sp<\/sub>. A saturated solution in equilibrium with the undissolved solid will result. If the concentrations are such that <em>Q<\/em> is less than <em>K<\/em><sub>sp<\/sub>, then the solution is not saturated and no precipitate will form.<\/p>\n<p>We can compare numerical values of <em>Q<\/em> with <em>K<\/em><sub>sp<\/sub> to predict whether precipitation will occur, as Example 7\u00a0shows. (Note: Since all forms of equilibrium constants are temperature dependent, we will assume a room temperature environment going forward in this chapter unless a different temperature value is explicitly specified.)<\/p>\n<div class=\"textbox examples\">\n<h3>Example 7:\u00a0Precipitation of Mg(OH)<sub>2<\/sub><\/h3>\n<p>The first step in the preparation of magnesium metal is the precipitation of Mg(OH)<sub>2<\/sub> from sea water by the addition of lime, Ca(OH)<sub>2<\/sub>, a readily available inexpensive source of OH<sup>\u2013<\/sup> ion:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{Mg(OH)}}_{2}\\left(s\\right)\\rightleftharpoons {\\text{Mg}}^{\\text{2+}}\\left(aq\\right)+{\\text{2OH}}^{-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,{;}\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{sp}}=\\text{2.1}\\times {10}^{-13}[\/latex]<\/p>\n<p>The concentration of Mg<sup>2+<\/sup>(<em>aq<\/em>) in sea water is 0.0537 <em>M<\/em>. Will Mg(OH)<sub>2<\/sub> precipitate when enough Ca(OH)<sub>2<\/sub> is added to give a [OH<sup>\u2013<\/sup>] of 0.0010 <em>M<\/em>?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q543435\">Show Answer<\/span><\/p>\n<div id=\"q543435\" class=\"hidden-answer\" style=\"display: none\">\n<p>This problem asks whether the reaction:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{Mg(OH)}}_{2}\\left(s\\right)\\rightleftharpoons {\\text{Mg}}^{\\text{2+}}\\left(aq\\right)+{\\text{2OH}}^{-}\\left(aq\\right)[\/latex]<\/p>\n<p>shifts to the left and forms solid Mg(OH)<sub>2<\/sub> when [Mg<sup>2+<\/sup>] = 0.0537 <em>M<\/em> and [OH<sup>\u2013<\/sup>] = 0.0010 <em>M<\/em>. The reaction shifts to the left if <em>Q<\/em> is greater than <em>K<\/em><sub>sp<\/sub>. Calculation of the reaction quotient under these conditions is shown here:<\/p>\n<p style=\"text-align: center;\">[latex]Q=\\left[\\text{Mg}^{2+}\\right]\\left[\\text{OH}^{-}\\right]^{2}=\\left(0.0537\\right)\\left(0.0010\\right)^{2}=5.4\\times{10}^{-8}[\/latex]<\/p>\n<p>Because <em>Q<\/em> is greater than <em>K<\/em><sub>sp<\/sub> (<em>Q<\/em> = 5.4 \u00d7 10<sup>\u20138<\/sup> is larger than <em>K<\/em><sub>sp<\/sub> = 2.1 \u00d7 10<sup>\u201313<\/sup>), we can expect the reaction to shift to the left and form solid magnesium hydroxide. Mg(OH)<sub>2<\/sub>(<em>s<\/em>) forms until the concentrations of magnesium ion and hydroxide ion are reduced sufficiently so that the value of <em>Q<\/em> is equal to <em>K<\/em><sub>sp<\/sub>.<\/p>\n<\/div>\n<\/div>\n<h4>Check Your Learning<\/h4>\n<p>Use the solubility product in\u00a0<a class=\"target-chapter\" href=\".\/chapter\/solubility-products\/\" target=\"_blank\">Solubility Products<\/a>\u00a0to determine whether CaHPO<sub>4<\/sub> will precipitate from a solution with [Ca<sup>2+<\/sup>] = 0.0001 <em>M<\/em> and [latex]\\left[{\\text{HPO}}_{4}{}^{2-}\\right][\/latex] = 0.001 <em>M<\/em>.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q41133\">Show Answer<\/span><\/p>\n<div id=\"q41133\" class=\"hidden-answer\" style=\"display: none\">No precipitation of CaHPO<sub>4<\/sub>; <em>Q<\/em> = 1 \u00d7 10<sup>\u20137<\/sup>, which is less than <em>K<\/em><sub>sp<\/sub><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 8:\u00a0Precipitation of AgCl upon Mixing Solutions<\/h3>\n<p>Does silver chloride precipitate when equal volumes of a 2.0 \u00d7 10<sup>\u20134<\/sup>&#8211;<em>M<\/em> solution of AgNO<sub>3<\/sub> and a 2.0 \u00d7 10<sup>\u20134<\/sup>&#8211;<em>M<\/em> solution of NaCl are mixed?<\/p>\n<p>(Note: The solution also contains Na<sup>+<\/sup> and [latex]{\\text{NO}}_{3}^{-}[\/latex] ions, but when referring to solubility rules, one can see that sodium nitrate is very soluble and cannot form a precipitate.)<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q81096\">Show Answer<\/span><\/p>\n<div id=\"q81096\" class=\"hidden-answer\" style=\"display: none\">\n<p>The equation for the equilibrium between solid silver chloride, silver ion, and chloride ion is:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{AgCl}\\left(s\\right)\\rightleftharpoons {\\text{Ag}}^{\\text{+}}\\left(aq\\right)+{\\text{Cl}}^{-}\\left(aq\\right)[\/latex]<\/p>\n<p>The solubility product is 1.8 \u00d7 10<sup>\u201310<\/sup> (see\u00a0<a class=\"target-chapter\" href=\".\/chapter\/solubility-products\/\" target=\"_blank\">Solubility Products<\/a>).<\/p>\n<p>AgCl will precipitate if the reaction quotient calculated from the concentrations in the mixture of AgNO<sub>3<\/sub> and NaCl is greater than <em>K<\/em><sub>sp<\/sub>. The volume doubles when we mix equal volumes of AgNO<sub>3<\/sub> and NaCl solutions, so each concentration is reduced to half its initial value. Consequently, immediately upon mixing, [Ag<sup>+<\/sup>] and [Cl<sup>\u2013<\/sup>] are both equal to:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{1}{2}\\left(2.0\\times {10}^{-4}\\right)\\text{}M=\\text{1.0}\\times {10}^{-4}\\text{}M[\/latex].<\/p>\n<p>The reaction quotient, <em>Q<\/em>, is <em>momentarily<\/em> greater than <em>K<\/em><sub>sp<\/sub> for AgCl, so a supersaturated solution is formed:<\/p>\n<p style=\"text-align: center;\">[latex]Q=\\left[\\text{Ag}^{+}\\right]\\left[\\text{Cl}^{-}\\right]=\\left(1.0\\times{10}^{-4}\\right)\\left(1.0\\times {10}^{-4}\\right)=1.0\\times{10}^{-8}\\gt{K}_{\\text{sp}}[\/latex]<\/p>\n<p>Since supersaturated solutions are unstable, AgCl will precipitate from the mixture until the solution returns to equilibrium, with <em>Q<\/em> equal to <em>K<\/em><sub>sp<\/sub>.<\/p>\n<\/div>\n<\/div>\n<h4>Check Your Learning<\/h4>\n<p>Will KClO<sub>4<\/sub> precipitate when 20 mL of a 0.050-<em>M<\/em> solution of K<sup>+<\/sup> is added to 80 mL of a 0.50-<em>M<\/em> solution of [latex]{\\text{ClO}}_{4}{}^{-}[\/latex]? (Remember to calculate the new concentration of each ion after mixing the solutions before plugging into the reaction quotient expression.)<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q920440\">Show Answer<\/span><\/p>\n<div id=\"q920440\" class=\"hidden-answer\" style=\"display: none\">No, <em>Q<\/em> = 4.0 \u00d7 10<sup>\u20133<\/sup>, which is less than <em>K<\/em><sub>sp<\/sub> = 1.07 \u00d7 10<sup>\u20132<\/sup><\/div>\n<\/div>\n<\/div>\n<p>In the previous two examples, we have seen that Mg(OH)<sub>2<\/sub> or AgCl precipitate when <em>Q<\/em> is greater than <em>K<\/em><sub>sp<\/sub>. In general, when a solution of a soluble salt of the M<sup>m+<\/sup> ion is mixed with a solution of a soluble salt of the X<sup>n\u2013<\/sup> ion, the solid, M<sub>p<\/sub>X<sub>q<\/sub> precipitates if the value of <em>Q<\/em> for the mixture of M<sup>m+<\/sup> and X<sup>n\u2013<\/sup> is greater than <em>K<\/em><sub>sp<\/sub> for M<sub>p<\/sub>X<sub>q<\/sub>. Thus, if we know the concentration of one of the ions of a slightly soluble ionic solid and the value for the solubility product of the solid, then we can calculate the concentration that the other ion must exceed for precipitation to begin. To simplify the calculation, we will assume that precipitation begins when the reaction quotient becomes equal to the solubility product constant.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 9:\u00a0Precipitation of Calcium Oxalate<\/h3>\n<p>Blood will not clot if calcium ions are removed from its plasma. Some blood collection tubes contain salts of the oxalate ion, [latex]{\\text{C}}_{2}{\\text{O}}_{4}^{2-}[\/latex], for this purpose (Figure\u00a04). At sufficiently high concentrations, the calcium and oxalate ions form solid, CaC<sub>2<\/sub>O<sub>4<\/sub>\u00b7H<sub>2<\/sub>O (which also contains water bound in the solid). The concentration of Ca<sup>2+<\/sup> in a sample of blood serum is 2.2 \u00d7 10<sup>\u20133<\/sup><em>M<\/em>. What concentration of [latex]{\\text{C}}_{2}{\\text{O}}_{4}^{2-}[\/latex] ion must be established before CaC<sub>2<\/sub>O<sub>4<\/sub>\u00b7H<sub>2<\/sub>O begins to precipitate?<\/p>\n<div style=\"width: 510px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213948\/CNX_Chem_15_01_Blood.jpg\" alt=\"A photograph is shown of 6 vials of blood resting on and near a black and white document. Two of the vials have purple caps, three have tan caps, and one has a red cap. Each has a label and the vials with tan caps have a small amount of an off-white material present in a layer at the base of the vial.\" width=\"500\" height=\"487\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure\u00a04. Anticoagulants can be added to blood that will combine with the Ca<sup>2+<\/sup> ions in blood serum and prevent the blood from clotting. (credit: modification of work by Neeta Lind)<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q639745\">Show Answer<\/span><\/p>\n<div id=\"q639745\" class=\"hidden-answer\" style=\"display: none\">\n<p>The equilibrium expression is:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{CaC}}_{2}{\\text{O}}_{4}\\left(s\\right)\\rightleftharpoons {\\text{Ca}}^{\\text{2+}}\\left(aq\\right)+{\\text{C}}_{2}{\\text{O}}_{4}{}^{2-}\\left(aq\\right)[\/latex]<\/p>\n<p>For this reaction:<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{\\text{sp}}=\\left[{\\text{Ca}}^{\\text{2+}}\\right]\\left[{\\text{C}}_{2}{\\text{O}}_{4}^{2-}\\right]=2.27\\times {10}^{-9}[\/latex] (see\u00a0<a class=\"target-chapter\" href=\".\/chapter\/solubility-products\/\" target=\"_blank\">Solubility Products<\/a>)<\/p>\n<p>CaC<sub>2<\/sub>O<sub>4<\/sub> does not appear in this expression because it is a solid. Water does not appear because it is the solvent.<\/p>\n<p>Solid CaC<sub>2<\/sub>O<sub>4<\/sub> does not begin to form until <em>Q<\/em> equals <em>K<\/em><sub>sp<\/sub>. Because we know <em>K<\/em><sub>sp<\/sub> and [Ca<sup>2+<\/sup>], we can solve for the concentration of [latex]{\\text{C}}_{2}{\\text{O}}_{4}^{2-}[\/latex] that is necessary to produce the first trace of solid:<\/p>\n<p style=\"text-align: center;\">[latex]Q={K}_{\\text{sp}}=\\left[\\text{Ca}^{2+}\\right]\\left[\\text{C}_{2}\\text{O}_{4}^{2-}\\right]=2.27\\times{10}^{-9}[\/latex]<br \/>\n[latex]\\left(2.2\\times {10}^{-3}\\right){\\text{[C}}_{2}{\\text{O}}_{4}{}^{\\text{2-}}]=\\text{2.27}\\times {10}^{-9}[\/latex]<br \/>\n[latex]\\left[{\\text{C}}_{2}{\\text{O}}_{4}{}^{2-}\\right]=\\frac{2.27\\times {10}^{-9}}{2.2\\times {10}^{-3}}=1.0\\times {10}^{-6}[\/latex]<\/p>\n<p>A concentration of [latex]\\left[{\\text{C}}_{2}{\\text{O}}_{4}^{2-}\\right][\/latex] = 1.0 \u00d7 10<sup>\u20136<\/sup><em>M<\/em> is necessary to initiate the precipitation of CaC<sub>2<\/sub>O<sub>4<\/sub> under these conditions.<\/p>\n<\/div>\n<\/div>\n<h4>Check Your Learning<\/h4>\n<p>If a solution contains 0.0020 mol of [latex]{\\text{CrO}}_{4}^{2-}[\/latex] per liter, what concentration of Ag<sup>+<\/sup> ion must be reached by adding solid AgNO<sub>3<\/sub> before Ag<sub>2<\/sub>CrO<sub>4<\/sub> begins to precipitate? Neglect any increase in volume upon adding the solid silver nitrate.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q24336\">Show Answer<\/span><\/p>\n<div id=\"q24336\" class=\"hidden-answer\" style=\"display: none\">7.0 \u00d7 10<sup>\u20135<\/sup><em>M<\/em><\/div>\n<\/div>\n<\/div>\n<p>It is sometimes useful to know the concentration of an ion that remains in solution after precipitation. We can use the solubility product for this calculation too: If we know the value of <em>K<\/em><sub>sp<\/sub> and the concentration of one ion in solution, we can calculate the concentration of the second ion remaining in solution. The calculation is of the same type as that in Example 9\u2014calculation of the concentration of a species in an equilibrium mixture from the concentrations of the other species and the equilibrium constant. However, the concentrations are different; we are calculating concentrations after precipitation is complete, rather than at the start of precipitation.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 10:\u00a0Concentrations Following Precipitation<\/h3>\n<p>Clothing washed in water that has a manganese [Mn<sup>2+<\/sup>(<em>aq<\/em>)] concentration exceeding 0.1 mg\/L (1.8 \u00d7 10<sup>\u20136<\/sup><em>M<\/em>) may be stained by the manganese upon oxidation, but the amount of Mn<sup>2+<\/sup> in the water can be reduced by adding a base. If a person doing laundry wishes to add a buffer to keep the pH high enough to precipitate the manganese as the hydroxide, Mn(OH)<sub>2<\/sub>, what pH is required to keep [Mn<sup>2+<\/sup>] equal to 1.8 \u00d7 10<sup>\u20136<\/sup><em>M<\/em>?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q943057\">Show Answer<\/span><\/p>\n<div id=\"q943057\" class=\"hidden-answer\" style=\"display: none\">\n<p>The dissolution of Mn(OH)<sub>2<\/sub> is described by the equation:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{Mn(OH)}}_{2}\\left(s\\right)\\rightleftharpoons {\\text{Mn}}^{\\text{2+}}\\left(aq\\right)+{\\text{2OH}}^{-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,{;}\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{sp}}=\\text{4.5}\\times {10}^{-14}[\/latex]<\/p>\n<p>We need to calculate the concentration of OH<sup>\u2013<\/sup> when the concentration of Mn<sup>2+<\/sup> is 1.8 \u00d7 10<sup>\u20136<\/sup><em>M<\/em>. From that, we calculate the pH. At equilibrium:<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{\\text{sp}}=\\left[\\text{Mn}^{2+}\\right]\\left[\\text{OH}^{-}\\right]^{2}[\/latex]\u00a0or\u00a0[latex]\\left(1.8\\times {10}^{-6}\\right){\\left[{\\text{OH}}^{-}\\right]}^{2}=\\text{4.5}\\times {10}^{-14}[\/latex]<\/p>\n<p>so\u00a0[latex]\\left[{\\text{OH}}^{-}\\right]=\\text{1.6}\\times {10}^{-4}M[\/latex].<\/p>\n<p>Now we calculate the pH from the pOH:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\text{pOH}=-\\text{log}\\left[{\\text{OH}}^{-}\\right]=-\\text{log}\\left(1.6\\times 10 - 4\\right)=\\text{3.80}\\\\ \\text{pH}=\\text{14.00}-\\text{pOH}=\\text{14.00}-\\text{3.80}=\\text{10.20}\\end{array}[\/latex]<\/p>\n<p>If the person doing laundry adds a base, such as the sodium silicate (Na<sub>4<\/sub>SiO<sub>4<\/sub>) in some detergents, to the wash water until the pH is raised to 10.20, the manganese ion will be reduced to a concentration of 1.8 \u00d7 10<sup>\u20136<\/sup><em>M<\/em>; at that concentration or less, the ion will not stain clothing.<\/p>\n<\/div>\n<\/div>\n<h4>Check Your Learning<\/h4>\n<p>The first step in the preparation of magnesium metal is the precipitation of Mg(OH)<sub>2<\/sub> from sea water by the addition of Ca(OH)<sub>2<\/sub>. The concentration of Mg<sup>2+<\/sup>(<em>aq<\/em>) in sea water is 5.37 \u00d7 10<sup>\u20132<\/sup><em>M<\/em>. Calculate the pH at which [Mg<sup>2+<\/sup>] is diminished to 1.0 \u00d7 10<sup>\u20135<\/sup><em>M<\/em> by the addition of Ca(OH)<sub>2<\/sub>.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q138585\">Show Answer<\/span><\/p>\n<div id=\"q138585\" class=\"hidden-answer\" style=\"display: none\">11.09<\/div>\n<\/div>\n<\/div>\n<p>Due to their light sensitivity, mixtures of silver halides are used in fiber optics for medical lasers, in photochromic eyeglass lenses (glass lenses that automatically darken when exposed to sunlight), and\u2014before the advent of digital photography\u2014in photographic film. Even though AgCl (<em>K<\/em><sub>sp<\/sub> = 1.6 \u00d7 10<sup>\u201310<\/sup>), AgBr (<em>K<\/em><sub>sp<\/sub> = 7.7 \u00d7 10<sup>\u201313<\/sup>), and AgI (<em>K<\/em><sub>sp<\/sub> = 8.3 \u00d7 10<sup>\u201317<\/sup>) are each quite insoluble, we cannot prepare a homogeneous solid mixture of them by adding Ag<sup>+<\/sup> to a solution of Cl<sup>\u2013<\/sup>, Br<sup>\u2013<\/sup>, and I<sup>\u2013<\/sup>; essentially all of the AgI will precipitate before any of the other solid halides form because of its smaller value for <em>K<\/em><sub>sp<\/sub>. However, we can prepare a homogeneous mixture of the solids by slowly adding a solution of Cl<sup>\u2013<\/sup>, Br<sup>\u2013<\/sup>, and I<sup>\u2013<\/sup> to a solution of Ag<sup>+<\/sup>.<\/p>\n<p>When two anions form slightly soluble compounds with the same cation, or when two cations form slightly soluble compounds with the same anion, the less soluble compound (usually, the compound with the smaller <em>K<\/em><sub>sp<\/sub>) generally precipitates first when we add a precipitating agent to a solution containing both anions (or both cations). When the <em>K<\/em><sub>sp<\/sub> values of the two compounds differ by two orders of magnitude or more (e.g., 10<sup>\u20132<\/sup> vs. 10<sup>\u20134<\/sup>), almost all of the less soluble compound precipitates before any of the more soluble one does. This is an example of <b>selective precipitation<\/b>, where a reagent is added to a solution of dissolved ions causing one of the ions to precipitate out before the rest.<\/p>\n<div class=\"textbox shaded\">\n<h3 data-type=\"title\">The Role of Precipitation in Wastewater Treatment<\/h3>\n<p>Solubility equilibria are useful tools in the treatment of wastewater carried out in facilities that may treat the municipal water in your city or town (Figure\u00a05). Specifically, selective precipitation is used to remove contaminants from wastewater before it is released back into natural bodies of water. For example, phosphate ions [latex]{\\text{(PO}}_{4}^{2-})[\/latex] are often present in the water discharged from manufacturing facilities. An abundance of phosphate causes excess algae to grow, which impacts the amount of oxygen available for marine life as well as making water unsuitable for human consumption.<\/p>\n<div style=\"width: 510px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213949\/CNX_Chem_15_01_Wastewater.jpg\" alt=\"A color photograph is shown of a high volume wastewater treatment facility. Nineteen large circular pools of water undergoing treatment are visible across the center of the photograph. A building and parking lot are visible in the foreground.\" width=\"500\" height=\"338\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure\u00a05. Wastewater treatment facilities, such as this one, remove contaminants from wastewater before the water is released back into the natural environment. (credit: \u201ceutrophication&amp;hypoxia\u201d\/Wikimedia Commons)<\/p>\n<\/div>\n<p>One common way to remove phosphates from water is by the addition of calcium hydroxide, known as lime, Ca(OH)<sub>2<\/sub>. The lime is converted into calcium carbonate, a strong base, in the water. As the water is made more basic, the calcium ions react with phosphate ions to produce hydroxylapatite, Ca<sub>5<\/sub>(PO4)<sub>3<\/sub>(OH), which then precipitates out of the solution:<\/p>\n<p style=\"text-align: center;\">[latex]5{\\text{Ca}}^{\\text{2+}}+{\\text{3PO}}_{4}{}^{3-}+{\\text{OH}}^{-}\\rightleftharpoons {\\text{Ca}}_{10}{{\\text{(PO}}_{4})}_{6}\\cdot {\\text{(OH)}}_{2}\\left(s\\right)[\/latex]<\/p>\n<p>The precipitate is then removed by filtration and the water is brought back to a neutral pH by the addition of CO<sub>2<\/sub> in a recarbonation process. Other chemicals can also be used for the removal of phosphates by precipitation, including iron(III) chloride and aluminum sulfate.<\/p>\n<p><a href=\"http:\/\/science.jrank.org\/pages\/5146\/Phosphorus-Removal.html\" target=\"_blank\">View this site for more information on how phosphorus is removed from wastewater.<\/a><\/p>\n<\/div>\n<p>Selective precipitation can also be used in qualitative analysis. In this method, reagents are added to an unknown chemical mixture in order to induce precipitation. Certain reagents cause specific ions to precipitate out; therefore, the addition of the reagent can be used to determine whether the ion is present in the solution.<\/p>\n<div class=\"textbox\">View <a href=\"http:\/\/employees.oneonta.edu\/viningwj\/sims\/common_ion_effect_s1.html\" target=\"_blank\">this simulation to study the process of salts dissolving and forming saturated solutions<\/a> and precipitates for specific compounds, or compounds for which you select the charges on the ions and the <em>K<\/em><sub>sp<\/sub><\/div>\n<div class=\"textbox examples\">\n<h3>Example 11:\u00a0Precipitation of Silver Halides<\/h3>\n<p>A solution contains 0.0010 mol of KI and 0.10 mol of KCl per liter. AgNO<sub>3<\/sub> is gradually added to this solution. Which forms first, solid AgI or solid AgCl?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q657940\">Show Answer<\/span><\/p>\n<div id=\"q657940\" class=\"hidden-answer\" style=\"display: none\">\n<p>The two equilibria involved are:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{AgCl}\\left(s\\right)\\rightleftharpoons {\\text{Ag}}^{\\text{+}}\\left(aq\\right)+{\\text{Cl}}^{-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,{;}\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{sp}}=\\text{1.8}\\times {10}^{-10}[\/latex]<br \/>\n[latex]\\text{AgI}\\left(s\\right)\\rightleftharpoons {\\text{Ag}}^{\\text{+}}\\left(aq\\right)+{\\text{I}}^{-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,{;}\\,\\,\\,\\,\\,\\,\\,K_{\\text{sp}}=\\text{1.5}\\times {10}^{-16}[\/latex]<\/p>\n<p>If the solution contained about <em>equal<\/em> concentrations of Cl<sup>\u2013<\/sup> and I<sup>\u2013<\/sup>, then the silver salt with the smallest <em>K<\/em><sub>sp<\/sub> (AgI) would precipitate first. The concentrations are not equal, however, so we should find the [Ag<sup>+<\/sup>] at which AgCl begins to precipitate and the [Ag<sup>+<\/sup>] at which AgI begins to precipitate. The salt that forms at the lower [Ag<sup>+<\/sup>] precipitates first.<\/p>\n<p>For AgI: AgI precipitates when <em>Q<\/em> equals <em>K<\/em><sub>sp<\/sub> for AgI (1.5 \u00d7 10<sup>\u201316<\/sup>). When [I<sup>\u2013<\/sup>] = 0.0010 <em>M<\/em>:<\/p>\n<p style=\"text-align: center;\">[latex]Q=\\left[\\text{Ag}^{+}\\right]\\left[\\text{I}^{-}\\right]=\\left[\\text{Ag}^{+}\\right]\\left(0.0010\\right)=1.5\\times {10}^{-16}[\/latex]<br \/>\n[latex][{\\text{Ag}}^{\\text{+}}\\text{]}=\\frac{\\text{1.8}\\times {10}^{-10}}{0.10}=\\text{1.8}\\times {10}^{-9}[\/latex]<\/p>\n<p>AgI begins to precipitate when [Ag<sup>+<\/sup>] is 1.5 \u00d7 10<sup>\u201313<\/sup><em>M<\/em>.<\/p>\n<p>For AgCl: AgCl precipitates when <em>Q<\/em> equals <em>K<\/em><sub>sp<\/sub> for AgCl (1.8 \u00d7 10<sup>\u201310<\/sup>). When [Cl<sup>\u2013<\/sup>] = 0.10 <em>M<\/em>:<\/p>\n<p style=\"text-align: center;\">[latex]{Q}_{\\text{sp}}=\\left[\\text{Ag}^{+}\\right]\\left[\\text{Cl}^{-}\\right]=\\left[\\text{Ag}^{+}\\right]\\left(0.10\\right)=1.8\\times{10}^{-10}[\/latex]<br \/>\n[latex][{\\text{Ag}}^{\\text{+}}\\text{]}=\\frac{1.8\\times {10}^{-10}}{0.10}\\text{=1.8}\\times {10}^{-9}M[\/latex]<\/p>\n<p>AgCl begins to precipitate when [Ag<sup>+<\/sup>] is 1.8 \u00d7 10<sup>\u20139<\/sup><em>M<\/em>.<\/p>\n<p>AgI begins to precipitate at a lower [Ag<sup>+<\/sup>] than AgCl, so AgI begins to precipitate first.<\/p>\n<\/div>\n<\/div>\n<h4>Check Your Learning<\/h4>\n<p>If silver nitrate solution is added to a solution which is 0.050 <em>M<\/em> in both Cl<sup>\u2013<\/sup> and Br<sup>\u2013<\/sup> ions, at what [Ag<sup>+<\/sup>] would precipitation begin, and what would be the formula of the precipitate?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q671573\">Show Answer<\/span><\/p>\n<div id=\"q671573\" class=\"hidden-answer\" style=\"display: none\">[Ag<sup>+<\/sup>] = 1.5 \u00d7 10<sup>\u201311<\/sup><em>M<\/em>; AgBr precipitates first<\/div>\n<\/div>\n<\/div>\n<h2>Common Ion Effect<\/h2>\n<p>As we saw when we discussed buffer solutions, the hydronium ion concentration of an aqueous solution of acetic acid decreases when the strong electrolyte sodium acetate, NaCH<sub>3<\/sub>CO<sub>2<\/sub>, is added. We can explain this effect using Le Ch\u00e2telier\u2019s principle. The addition of acetate ions causes the equilibrium to shift to the left, decreasing the concentration of [latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}[\/latex] to compensate for the increased acetate ion concentration. This increases the concentration of CH<sub>3<\/sub>CO<sub>2<\/sub>H:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}+{\\text{H}}_{2}\\text{O}\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}+{\\text{CH}}_{3}{\\text{CO}}_{2}^{-}[\/latex]<\/p>\n<p>Because sodium acetate and acetic acid have the acetate ion in common, the influence on the equilibrium is called the <b>common ion effect<\/b>.<\/p>\n<p>The common ion effect can also have a direct effect on solubility equilibria. Suppose we are looking at the reaction where silver iodide is dissolved:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{AgI}\\left(s\\right)\\rightleftharpoons {\\text{Ag}}^{\\text{+}}\\left(aq\\right)+{\\text{I}}^{-}\\left(aq\\right)[\/latex]<\/p>\n<p>If we were to add potassium iodide (KI) to this solution, we would be adding a substance that shares a common ion with silver iodide. Le Ch\u00e2telier\u2019s principle tells us that when a change is made to a system at equilibrium, the reaction will shift to counteract that change. In this example, there would be an excess of iodide ions, so the reaction would shift toward the left, causing more silver iodide to precipitate out of solution.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 12:\u00a0Common Ion Effect<\/h3>\n<p>Calculate the molar solubility of cadmium sulfide (CdS) in a 0.010-<em>M<\/em> solution of cadmium bromide (CdBr<sub>2<\/sub>). The <em>K<\/em><sub>sp<\/sub> of CdS is 1.0 \u00d7 10<sup>\u201328<\/sup>.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q848161\">Show Answer<\/span><\/p>\n<div id=\"q848161\" class=\"hidden-answer\" style=\"display: none\">\n<p>The first thing you should notice is that the cadmium sulfide is dissolved in a solution that contains cadmium ions. We need to use an ICE table to set up this problem and include the CdBr<sub>2<\/sub> concentration as a contributor of cadmium ions:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{CdS}\\left(s\\right)\\rightleftharpoons {\\text{Cd}}^{2+}\\left(aq\\right)+{\\text{S}}^{2-}\\left(aq\\right)[\/latex]<\/p>\n<p style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213951\/CNX_Chem_15_01_ICETable3_img.jpg\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), and Equilibrium concentration ( M ). The second column has the header, \u201cC d S equilibrium arrow C d to the second power plus S to the second power superscript negative sign.\u201d Under the second column is a subgroup of three rows and three columns. The first column is blank. The second column has the following: 0.010, x, 0.010 plus x. The third column has the following: 0, x, 0 plus x equals x.\" width=\"879\" height=\"238\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p style=\"text-align: center;\">[latex]{K}_{\\text{sp}}=\\left[{\\text{Cd}}^{\\text{2+}}\\right]\\left[{\\text{S}}^{2-}\\right]=1.0\\times {10}^{-28}[\/latex]<br \/>\n[latex]\\left(0.010+x\\right)\\left(x\\right)=\\text{1.0}\\times {10}^{-28}[\/latex]<br \/>\n[latex]{x}^{2}+\\text{0.010}x-\\text{1.0}\\times {10}^{-28}=0[\/latex]<\/p>\n<p>We can solve this equation using the quadratic formula, but we can also make an assumption to make this calculation much simpler. Since the <em>K<\/em><sub>sp<\/sub> value is so small compared with the cadmium concentration, we can assume that the change between the initial concentration and the equilibrium concentration is negligible, so that 0.010 + <em>x<\/em> ~ 0.010. Going back to our <em>K<\/em><sub>sp<\/sub> expression, we would now get:<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{\\text{sp}}=\\left[\\text{Cd}^{2+}\\right]\\left[\\text{S}^{2-}\\right]=1.0\\times{10}^{-28}[\/latex]<br \/>\n[latex]\\left(0.010\\right)\\left(x\\right)=\\text{1.0}\\times {10}^{-28}[\/latex]<br \/>\n[latex]x=\\text{1.0}\\times {10}^{-26}[\/latex]<\/p>\n<p>Therefore, the molar solubility of CdS in this solution is 1.0 \u00d7 10<sup>\u201326<\/sup><em>M<\/em>.<\/p>\n<\/div>\n<\/div>\n<h4>Check Your Learning<\/h4>\n<p>Calculate the molar solubility of aluminum hydroxide, Al(OH)<sub>3<\/sub>, in a 0.015-<em>M<\/em> solution of aluminum nitrate, Al(NO<sub>3<\/sub>)<sub>3<\/sub>. The <em>K<\/em><sub>sp<\/sub> of Al(OH)<sub>3<\/sub> is 2 \u00d7 10<sup>\u201332<\/sup>.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q176865\">Show Answer<\/span><\/p>\n<div id=\"q176865\" class=\"hidden-answer\" style=\"display: none\">1 \u00d7 10<sup>\u201310<\/sup><em>M<\/em><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Key Concepts and Summary<\/h3>\n<p>The equilibrium constant for an equilibrium involving the precipitation or dissolution of a slightly soluble ionic solid is called the solubility product, <em>K<\/em><sub>sp<\/sub>, of the solid. When we have a heterogeneous equilibrium involving the slightly soluble solid M<sub>p<\/sub>X<sub>q<\/sub> and its ions M<sup>m+<\/sup> and X<sup>n\u2013<\/sup>:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{M}_{\\text{p}}{\\text{X}}_{\\text{q}}\\left(s\\right)\\rightleftharpoons p{\\text{M}}^{\\text{m+}}\\left(aq\\right)+q{\\text{X}}^{\\text{n}-}\\left(aq\\right)[\/latex]<\/p>\n<p>We write the solubility product expression as:<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{\\text{sp}}={{\\text{[M}}^{\\text{m+}}\\text{]}}^{\\text{p}}{{\\text{[X}}^{\\text{n}-}]}^{\\text{q}}[\/latex]<\/p>\n<p>The solubility product of a slightly soluble electrolyte can be calculated from its solubility; conversely, its solubility can be calculated from its <em>K<\/em><sub>sp<\/sub>, provided the only significant reaction that occurs when the solid dissolves is the formation of its ions.<\/p>\n<p>A slightly soluble electrolyte begins to precipitate when the magnitude of the reaction quotient for the dissolution reaction exceeds the magnitude of the solubility product. Precipitation continues until the reaction quotient equals the solubility product.<\/p>\n<p>A reagent can be added to a solution of ions to allow one ion to selectively precipitate out of solution. The common ion effect can also play a role in precipitation reactions. In the presence of an ion in common with one of the ions in the solution, Le Ch\u00e2telier\u2019s principle applies and more precipitate comes out of solution so that the molar solubility is reduced.<\/p>\n<h4>Key Equations<\/h4>\n<ul>\n<li>[latex]\\text{Mp}_{\\text{Xq}}\\left(s\\right)\\rightleftharpoons p{\\text{M}}^{\\text{m+}}\\left(aq\\right)+q{\\text{X}}^{\\text{n}-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,{;}\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{sp}}=\\left[\\text{M}^{\\text{m+}}\\right]^{\\text{p}}\\left[\\text{X}^{\\text{n}-}\\right]^\\text{q}[\/latex]<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Exercises<\/h3>\n<ol>\n<li>Complete the changes in concentrations for each of the following reactions:\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\begin{array}{ccc}\\text{AgI}\\left(s\\right)\\longrightarrow & {\\text{Ag}}^{\\text{+}}\\left(aq\\right)& +{\\text{I}}^{-}\\left(aq\\right)\\\\ & x&\\text{ _____}\\end{array}[\/latex]<\/li>\n<li>[latex]\\begin{array}{ccc}{\\text{CaCO}}_{3}\\left(s\\right)\\longrightarrow & {\\text{Ca}}^{\\text{2+}}\\left(aq\\right)+& {\\text{CO}}_{3}{}^{2-}\\left(aq\\right)\\\\ & \\text{____}& x\\end{array}[\/latex]<\/li>\n<li>[latex]\\begin{array}{ccc}\\text{Mg}{\\left(\\text{OH}\\right)}_{2}\\left(s\\right)\\longrightarrow & {\\text{Mg}}^{\\text{2+}}\\left(aq\\right)+& 2{\\text{OH}}^{-}\\left(aq\\right)\\\\ & x&\\text{ _____}\\end{array}[\/latex]<\/li>\n<li>[latex]\\begin{array}{ccc}{\\text{Mg}}_{3}{\\left({\\text{PO}}_{4}\\right)}_{2}\\left(s\\right)\\longrightarrow & 3{\\text{Mg}}^{\\text{2+}}\\left(aq\\right)+& 2{\\text{PO}}_{4}{}^{3-}\\left(aq\\right)\\\\ & & x\\text{_____}\\end{array}[\/latex]<\/li>\n<li>[latex]\\begin{array}{cccc}{\\text{Ca}}_{5}{\\left({\\text{PO}}_{4}\\right)}_{3}\\text{OH}\\left(s\\right)\\longrightarrow & 5{\\text{Ca}}^{\\text{2+}}\\left(aq\\right)+& 3{\\text{PO}}_{4}^{3-}\\left(aq\\right)+& {\\text{OH}}^{-}\\left(aq\\right)\\\\ & \\text{_____}& \\text{_____}& x\\end{array}[\/latex]<\/li>\n<\/ol>\n<\/li>\n<li>Complete the changes in concentrations for each of the following reactions:\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\begin{array}{ccc}{\\text{BaSO}}_{4}\\left(s\\right)\\longrightarrow & {\\text{Ba}}^{\\text{2+}}\\left(aq\\right)+& {\\text{SO}}_{4}{}^{2-}\\left(aq\\right)\\\\ & x& \\text{_____}\\end{array}[\/latex]<\/li>\n<li>[latex]\\begin{array}{ccc}{\\text{Ag}}_{2}{\\text{SO}}_{4}\\left(s\\right)\\longrightarrow & 2{\\text{Ag}}^{\\text{+}}\\left(aq\\right)+& {\\text{SO}}_{4}^{2-}\\left(aq\\right)\\\\ & \\text{_____}& x\\end{array}[\/latex]<\/li>\n<li>[latex]\\begin{array}{ccc}\\text{Al}{\\left(\\text{OH}\\right)}_{3}\\left(s\\right)\\longrightarrow & {\\text{Al}}^{\\text{3+}}\\left(aq\\right)+& 3{\\text{OH}}^{-}\\left(aq\\right)\\\\ & x& \\text{_____}\\end{array}[\/latex]<\/li>\n<li>[latex]\\begin{array}{cccc}\\text{Pb}\\left(\\text{OH}\\right)\\text{Cl}\\left(s\\right)\\longrightarrow & {\\text{Pb}}^{\\text{2+}}\\left(aq\\right)+& {\\text{OH}}^{-}\\left(aq\\right)+& {\\text{Cl}}^{-}\\left(aq\\right)\\\\ & \\text{_____}\\text{}& x& \\text{_____}\\end{array}[\/latex]<\/li>\n<li>[latex]\\begin{array}{ccc}{\\text{Ca}}_{3}{\\left({\\text{AsO}}_{4}\\right)}_{2}\\left(s\\right)\\longrightarrow & 3{\\text{Ca}}^{\\text{2+}}\\left(aq\\right)+& 2{\\text{AsO}}_{4}^{3-}\\left(aq\\right)\\\\ & 3x& \\text{_____}\\end{array}[\/latex]<\/li>\n<\/ol>\n<\/li>\n<li>How do the concentrations of Ag<sup>+<\/sup> and [latex]{\\text{CrO}}_{4}^{2-}[\/latex] in a saturated solution above 1.0 g of solid Ag<sub>2<\/sub>CrO<sub>4<\/sub> change when 100 g of solid Ag<sub>2<\/sub>CrO<sub>4<\/sub> is added to the system? Explain.<\/li>\n<li>How do the concentrations of Pb<sup>2+<\/sup> and S<sup>2\u2013<\/sup> change when K<sub>2<\/sub>S is added to a saturated solution of PbS?<\/li>\n<li>What additional information do we need to answer the following question: How is the equilibrium of solid silver bromide with a saturated solution of its ions affected when the temperature is raised?<\/li>\n<li>Which of the following slightly soluble compounds has a solubility greater than that calculated from its solubility product because of hydrolysis of the anion present: CoSO<sub>3<\/sub>, CuI, PbCO<sub>3<\/sub>, PbCl<sub>2<\/sub>, Tl<sub>2<\/sub>S, KClO<sub>4<\/sub>?<\/li>\n<li>Which of the following slightly soluble compounds has a solubility greater than that calculated from its solubility product because of hydrolysis of the anion present: AgCl, BaSO<sub>4<\/sub>, CaF<sub>2<\/sub>, Hg<sub>2<\/sub>I<sub>2<\/sub>, MnCO<sub>3<\/sub>, ZnS, PbS?<\/li>\n<li>Write the ionic equation for dissolution and the solubility product (<em>K<\/em><sub>sp<\/sub>) expression for each of the following slightly soluble ionic compounds:\n<ol style=\"list-style-type: lower-alpha;\">\n<li>PbCl<sub>2<\/sub><\/li>\n<li>Ag<sub>2<\/sub>S<\/li>\n<li>Sr<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub><\/li>\n<li>SrSO<sub>4<\/sub><\/li>\n<\/ol>\n<\/li>\n<li>Write the ionic equation for the dissolution and the <em>K<\/em><sub>sp<\/sub> expression for each of the following slightly soluble ionic compounds:\n<ol style=\"list-style-type: lower-alpha;\">\n<li>LaF<sub>3<\/sub><\/li>\n<li>CaCO<sub>3<\/sub><\/li>\n<li>Ag<sub>2<\/sub>SO<sub>4<\/sub><\/li>\n<li>Pb(OH)<sub>2<\/sub><\/li>\n<\/ol>\n<\/li>\n<li>The <a href=\"http:\/\/hbcponline.com\/faces\/contents\/ContentsSearch.xhtml\" target=\"_blank\"><em>Handbook of Chemistry and Physics<\/em><\/a> gives solubilities of the following compounds in grams per 100 mL of water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each.\n<ol style=\"list-style-type: lower-alpha;\">\n<li>BaSiF<sub>6<\/sub>, 0.026 g\/100 mL (contains [latex]{\\text{SiF}}_{6}^{2-}[\/latex] ions)<\/li>\n<li>Ce(IO<sub>3<\/sub>)<sub>4<\/sub>, 1.5 \u00d7 10<sup>\u20132<\/sup> g\/100 mL<\/li>\n<li>Gd<sub>2<\/sub>(SO<sub>4<\/sub>)<sub>3<\/sub>, 3.98 g\/100 mL<\/li>\n<li>(NH<sub>4<\/sub>)<sub>2<\/sub>PtBr<sub>6<\/sub>, 0.59 g\/100 mL (contains [latex]{\\text{PtBr}}_{6}^{2-}[\/latex] ions)<\/li>\n<\/ol>\n<\/li>\n<li>The <a href=\"http:\/\/hbcponline.com\/faces\/contents\/ContentsSearch.xhtml\" target=\"_blank\"><em>Handbook of Chemistry and Physics<\/em><\/a> gives solubilities of the following compounds in grams per 100 mL of water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each.\n<ol style=\"list-style-type: lower-alpha;\">\n<li>BaSeO<sub>4<\/sub>, 0.0118 g\/100 mL<\/li>\n<li>Ba(BrO<sub>3<\/sub>)<sub>2<\/sub>\u00b7H<sub>2<\/sub>O, 0.30 g\/100 mL<\/li>\n<li>NH<sub>4<\/sub>MgAsO<sub>4<\/sub>\u00b76H<sub>2<\/sub>O, 0.038 g\/100 mL<\/li>\n<li>La<sub>2<\/sub>(MoO<sub>4<\/sub>)<sub>3<\/sub>, 0.00179 g\/100 mL<\/li>\n<\/ol>\n<\/li>\n<li>Use solubility products and predict which of the following salts is the most soluble, in terms of moles per liter, in pure water: CaF<sub>2<\/sub>, Hg<sub>2<\/sub>Cl<sub>2<\/sub>, PbI<sub>2<\/sub>, or Sn(OH)<sub>2<\/sub>.<\/li>\n<li>Assuming that no equilibria other than dissolution are involved, calculate the molar solubility of each of the following from its solubility product:\n<ol style=\"list-style-type: lower-alpha;\">\n<li>KHC<sub>4<\/sub>H<sub>4<\/sub>O<sub>6<\/sub><\/li>\n<li>PbI<sub>2<\/sub><\/li>\n<li>Ag<sub>4<\/sub>[Fe(CN)<sub>6<\/sub>], a salt containing the [latex]{\\text{Fe(CN)}}_{4}^{-}[\/latex] ion.<\/li>\n<li>Hg<sub>2<\/sub>I<sub>2<\/sub><\/li>\n<\/ol>\n<\/li>\n<li>Assuming that no equilibria other than dissolution are involved, calculate the molar solubility of each of the following from its solubility product:\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Ag<sub>2<\/sub>SO<sub>4<\/sub><\/li>\n<li>PbBr<sub>2<\/sub><\/li>\n<li>AgI<\/li>\n<li>CaC<sub>2<\/sub>O<sub>4<\/sub>\u00b7H<sub>2<\/sub>O<\/li>\n<\/ol>\n<\/li>\n<li>Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that changes in the initial concentrations of the common ions can be neglected.\n<ol style=\"list-style-type: lower-alpha;\">\n<li>AgCl(<em>s<\/em>) in 0.025 <em>M<\/em> NaCl<\/li>\n<li>CaF<sub>2<\/sub>(<em>s<\/em>) in 0.00133 <em>M<\/em> KF<\/li>\n<li>Ag<sub>2<\/sub>SO<sub>4<\/sub>(<em>s<\/em>) in 0.500 L of a solution containing 19.50 g of K<sub>2<\/sub>SO<sub>4<\/sub><\/li>\n<li>Zn(OH)<sub>2<\/sub>(<em>s<\/em>) in a solution buffered at a pH of 11.45<\/li>\n<\/ol>\n<\/li>\n<li>Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that changes in the initial concentrations of the common ions can be neglected.\n<ol style=\"list-style-type: lower-alpha;\">\n<li>TlCl(<em>s<\/em>) in 1.250 <em>M<\/em> HCl<\/li>\n<li>PbI<sub>2<\/sub>(<em>s<\/em>) in 0.0355 <em>M<\/em> CaI<sub>2<\/sub><\/li>\n<li>Ag<sub>2<\/sub>CrO<sub>4<\/sub>(<em>s<\/em>) in 0.225 L of a solution containing 0.856 g of K<sub>2<\/sub>CrO<sub>4<\/sub><\/li>\n<li>Cd(OH)<sub>2<\/sub>(<em>s<\/em>) in a solution buffered at a pH of 10.995<\/li>\n<\/ol>\n<\/li>\n<li>Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that it is not appropriate to neglect the changes in the initial concentrations of the common ions.\n<ol style=\"list-style-type: lower-alpha;\">\n<li>TlCl(<em>s<\/em>) in 0.025 <em>M<\/em> TlNO<sub>3<\/sub><\/li>\n<li>BaF<sub>2<\/sub>(<em>s<\/em>) in 0.0313 <em>M<\/em> KF<\/li>\n<li>MgC<sub>2<\/sub>O<sub>4<\/sub> in 2.250 L of a solution containing 8.156 g of Mg(NO<sub>3<\/sub>)<sub>2<\/sub><\/li>\n<li>Ca(OH)<sub>2<\/sub>(<em>s<\/em>) in an unbuffered solution initially with a pH of 12.700<\/li>\n<\/ol>\n<\/li>\n<li>Explain why the changes in concentrations of the common ions in Question\u00a017 can be neglected.<\/li>\n<li>Explain why the changes in concentrations of the common ions in Question\u00a018 cannot be neglected.<\/li>\n<li>Calculate the solubility of aluminum hydroxide, Al(OH)<sub>3<\/sub>, in a solution buffered at pH 11.00.<\/li>\n<li>Refer to\u00a0<a class=\"target-chapter\" href=\".\/chapter\/solubility-products\/\" target=\"_blank\">Solubility Products<\/a>\u00a0for solubility products for calcium salts. Determine which of the calcium salts listed is most soluble in moles per liter and which is most soluble in grams per liter.<\/li>\n<li>Most barium compounds are very poisonous; however, barium sulfate is often administered internally as an aid in the X-ray examination of the lower intestinal tract (Figure\u00a03). This use of BaSO<sub>4<\/sub> is possible because of its low solubility. Calculate the molar solubility of BaSO<sub>4<\/sub> and the mass of barium present in 1.00 L of water saturated with BaSO<sub>4<\/sub>.<\/li>\n<li>Public Health Service standards for drinking water set a maximum of 250 mg\/L (2.60 \u00d7 10<sup>\u20133<\/sup> M) of [latex]{\\text{SO}}_{4}^{2-}[\/latex] because of its cathartic action (it is a laxative). Does natural water that is saturated with CaSO<sub>4<\/sub> (\u201cgyp\u201d water) as a result or passing through soil containing gypsum, CaSO<sub>4<\/sub>\\cdot 2H<sub>2<\/sub>O, meet these standards? What is [latex]{\\text{SO}}_{4}^{2-}[\/latex] in such water?<\/li>\n<li>Perform the following calculations:\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Calculate [Ag<sup>+<\/sup>] in a saturated aqueous solution of AgBr.<\/li>\n<li>What will [Ag<sup>+<\/sup>] be when enough KBr has been added to make [Br<sup>\u2013<\/sup>] = 0.050 <em>M<\/em>?<\/li>\n<li>What will [Br<sup>\u2013<\/sup>] be when enough AgNO<sub>3<\/sub> has been added to make [Ag<sup>+<\/sup>] = 0.020 <em>M<\/em>?<\/li>\n<\/ol>\n<\/li>\n<li>The solubility product of CaSO<sub>4<\/sub>\u00b72H<sub>2<\/sub>O is 2.4 \u00d7 10<sup>\u20135<\/sup>. What mass of this salt will dissolve in 1.0 L of 0.010 <em>M<\/em> [latex]{\\text{SO}}_{4}^{2-}?[\/latex]<\/li>\n<li>Assuming that no equilibria other than dissolution are involved, calculate the concentrations of ions in a saturated solution of each of the following (see\u00a0<a class=\"target-chapter\" href=\".\/chapter\/solubility-products\/\" target=\"_blank\">Solubility Products<\/a>).\n<ol style=\"list-style-type: lower-alpha;\">\n<li>TlCl<\/li>\n<li>BaF<sub>2<\/sub><\/li>\n<li>Ag<sub>2<\/sub>CrO<sub>4<\/sub><\/li>\n<li>CaC<sub>2<\/sub>O<sub>4<\/sub>\u00b7H<sub>2<\/sub>O<\/li>\n<li>the mineral anglesite, PbSO<sub>4<\/sub><\/li>\n<\/ol>\n<\/li>\n<li>Assuming that no equilibria other than dissolution are involved, calculate the concentrations of ions in a saturated solution of each of the following (see\u00a0<a class=\"target-chapter\" href=\".\/chapter\/solubility-products\/\" target=\"_blank\">Solubility Products<\/a>):\n<ol style=\"list-style-type: lower-alpha;\">\n<li>AgI<\/li>\n<li>Ag<sub>2<\/sub>SO<sub>4<\/sub><\/li>\n<li>Mn(OH)<sub>2<\/sub><\/li>\n<li>\u00a0Sr(OH)<sub>2<\/sub>\u00b78H<sub>2<\/sub>O<\/li>\n<li>the mineral brucite, Mg(OH)<sub>2<\/sub><\/li>\n<\/ol>\n<\/li>\n<li>The following concentrations are found in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, calculate <em>K<\/em><sub>sp<\/sub> for each of the slightly soluble solids indicated:\n<ol style=\"list-style-type: lower-alpha;\">\n<li>AgBr: [Ag<sup>+<\/sup>] = 5.7 \u00d7 10<sup>\u20137<\/sup><em>M<\/em>, [Br<sup>\u2013<\/sup>] = 5.7 \u00d7 10<sup>\u20137<\/sup><em>M<\/em><\/li>\n<li>CaCO<sub>3<\/sub>: [Ca<sup>2+<\/sup>] = 5.3 \u00d7 10<sup>\u20133<\/sup><em>M<\/em>, [latex]\\left[\\text{CO}_{3}^{2-}\\right][\/latex] = 9.0 \u00d7 10<sup>\u20137<\/sup><em>M<\/em><\/li>\n<li>PbF<sub>2<\/sub>: [Pb<sup>2+<\/sup>] = 2.1 \u00d7 10<sup>\u20133<\/sup><em>M<\/em>, [F<sup>\u2013<\/sup>] = 4.2 \u00d7 10<sup>\u20133<\/sup><em>M<\/em><\/li>\n<li>Ag<sub>2<\/sub>CrO<sub>4<\/sub>: [Ag<sup>+<\/sup>] = 5.3 \u00d7 10<sup>\u20135<\/sup><em>M<\/em>, 3.2 \u00d7 10<sup>\u20133<\/sup><em>M<\/em><\/li>\n<li>InF<sub>3<\/sub>: [In<sup>3+<\/sup>] = 2.3 \u00d7 10<sup>\u20133<\/sup><em>M<\/em>, [F<sup>\u2013<\/sup>] = 7.0 \u00d7 10<sup>\u20133<\/sup><em>M<\/em><\/li>\n<\/ol>\n<\/li>\n<li>The following concentrations are found in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, calculate <em>K<\/em><sub>sp<\/sub> for each of the slightly soluble solids indicated:\n<ol style=\"list-style-type: lower-alpha;\">\n<li>TlCl: [Tl<sup>+<\/sup>] = 1.21 \u00d7 10<sup>\u20132<\/sup><em>M<\/em>, [Cl<sup>\u2013<\/sup>] = 1.2 \u00d7 10<sup>\u20132<\/sup><em>M<\/em><\/li>\n<li>Ce(IO<sub>3<\/sub>)<sub>4<\/sub>: [Ce<sup>4+<\/sup>] = 1.8 \u00d7 10<sup>\u20134<\/sup><em>M<\/em>, [latex]\\left[\\text{IO}_{3}^{-}\\right][\/latex] = 2.6 \u00d7 10<sup>\u201313<\/sup><em>M<\/em><\/li>\n<li>Gd<sub>2<\/sub>(SO<sub>4<\/sub>)<sub>3<\/sub>: [Gd<sup>3+<\/sup>] = 0.132 <em>M<\/em>, [latex]\\left[\\text{SO}_{4}^{2-}\\right][\/latex] = 0.198 <em>M<\/em><\/li>\n<li>Ag<sub>2<\/sub>SO<sub>4<\/sub>: [Ag<sup>+<\/sup>] = 2.40 \u00d7 10<sup>\u20132<\/sup><em>M<\/em>, [latex]\\left[{\\text{SO}}_{4}^{2-}\\right][\/latex] = 2.05 \u00d7 10<sup>\u20132<\/sup><em>M<\/em><\/li>\n<li>BaSO<sub>4<\/sub>: [Ba<sup>2+<\/sup>] = 0.500 <em>M<\/em>, [latex]\\left[{\\text{SO}}_{4}^{2-}\\right][\/latex] = 2.16 \u00d7 10<sup>\u201310<\/sup><em>M<\/em><\/li>\n<\/ol>\n<\/li>\n<li>Which of the following compounds precipitates from a solution that has the concentrations indicated? (See\u00a0<a class=\"target-chapter\" href=\".\/chapter\/solubility-products\/\" target=\"_blank\">Solubility Products<\/a>\u00a0for <em>K<\/em><sub>sp<\/sub> values.)\n<ol style=\"list-style-type: lower-alpha;\">\n<li>KClO<sub>4<\/sub>: [K<sup>+<\/sup>] = 0.01 <em>M<\/em>, [latex]\\left[{\\text{ClO}}_{4}^{-}\\right][\/latex] = 0.01 <em>M<\/em><\/li>\n<li>K<sub>2<\/sub>PtCl<sub>6<\/sub>: [K<sup>+<\/sup>] = 0.01 <em>M<\/em>, [latex]{\\text{[PtCl}}_{6}^{2-}][\/latex] = 0.01 <em>M<\/em><\/li>\n<li>PbI<sub>2<\/sub>: [Pb<sup>2+<\/sup>] = 0.003 <em>M<\/em>, [I<sup>\u2013<\/sup>] = 1.3 \u00d7 10<sup>\u20133<\/sup><em>M<\/em><\/li>\n<li>Ag<sub>2<\/sub>S: [Ag<sup>+<\/sup>] = 1 \u00d7 10<sup>\u201310<\/sup><em>M<\/em>, [S<sup>2\u2013<\/sup>] = 1 \u00d7 10<sup>\u201313<\/sup><em>M<\/em><\/li>\n<\/ol>\n<\/li>\n<li>Which of the following compounds precipitates from a solution that has the concentrations indicated? (See\u00a0<a class=\"target-chapter\" href=\".\/chapter\/solubility-products\/\" target=\"_blank\">Solubility Products<\/a>\u00a0for <em>K<\/em><sub>sp<\/sub> values.)\n<ol style=\"list-style-type: lower-alpha;\">\n<li>CaCO<sub>3<\/sub>: [Ca<sup>2+<\/sup>] = 0.003 <em>M<\/em>, [latex]{\\text{[CO}}_{3}^{2-}][\/latex] = 0.003 <em>M<\/em><\/li>\n<li>Co(OH)<sub>2<\/sub>: [Co<sup>2+<\/sup>] = 0.01 <em>M<\/em>, [OH<sup>\u2013<\/sup>] = 1 \u00d7 10<sup>\u20137<\/sup><em>M<\/em><\/li>\n<li>CaHPO<sub>4<\/sub>: [Ca<sup>2+<\/sup>] = 0.01 <em>M<\/em>, [latex]{\\text{[HPO}}_{4}^{2-}][\/latex] = 2 \u00d7 10<sup>\u20136<\/sup><em>M<\/em><\/li>\n<li>Pb<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub>: [Pb<sup>2+<\/sup>] = 0.01 <em>M<\/em>, [latex]{\\text{[PO}}_{4}^{3-}][\/latex] 1 \u00d7 10<sup>\u201313<\/sup><em>M<\/em><\/li>\n<\/ol>\n<\/li>\n<li>Calculate the concentration of Tl<sup>+<\/sup> when TlCl just begins to precipitate from a solution that is 0.0250 <em>M<\/em> in Cl<sup>\u2013<\/sup>.<\/li>\n<li>Calculate the concentration of sulfate ion when BaSO<sub>4<\/sub> just begins to precipitate from a solution that is 0.0758 <em>M<\/em> in Ba<sup>2+<\/sup>.<\/li>\n<li>Calculate the concentration of Sr<sup>2+<\/sup> when SrF<sub>2<\/sub> starts to precipitate from a solution that is 0.0025 <em>M<\/em> in F<sup>\u2013<\/sup>.<\/li>\n<li>Calculate the concentration of [latex]{\\text{PO}}_{4}^{3-}[\/latex] when Ag<sub>3<\/sub>PO<sub>4<\/sub> starts to precipitate from a solution that is 0.0125 <em>M<\/em> in Ag<sup>+<\/sup>.<\/li>\n<li>Calculate the concentration of F<sup>\u2013<\/sup> required to begin precipitation of CaF<sub>2<\/sub> in a solution that is 0.010 <em>M<\/em> in Ca<sup>2+<\/sup>.<\/li>\n<li>Calculate the concentration of Ag<sup>+<\/sup> required to begin precipitation of Ag<sub>2<\/sub>CO<sub>3<\/sub> in a solution that is 2.50 \u00d7 10<sup>\u20136<\/sup><em>M<\/em> in [latex]{\\text{CO}}_{3}^{2-}[\/latex].<\/li>\n<li>What [Ag<sup>+<\/sup>] is required to reduce [latex]\\left[{\\text{CO}}_{3}^{2-}\\right][\/latex] to 8.2 \u00d7 10<sup>\u20134<\/sup><em>M<\/em> by precipitation of Ag<sub>2<\/sub>CO<sub>3<\/sub>?<\/li>\n<li>What [F<sup>\u2013<\/sup>] is required to reduce [Ca<sup>2+<\/sup>] to 1.0 \u00d7 10<sup>\u20134<\/sup><em>M<\/em> by precipitation of CaF<sub>2<\/sub>?<\/li>\n<li>A volume of 0.800 L of a 2 \u00d7 10<sup>\u20134<\/sup>&#8211;<em>M<\/em> Ba(NO<sub>3<\/sub>)<sub>2<\/sub> solution is added to 0.200 L of 5 \u00d7 10<sup>\u20134<\/sup><em>M<\/em> Li<sub>2<\/sub>SO<sub>4<\/sub>. Does BaSO<sub>4<\/sub> precipitate? Explain your answer.<\/li>\n<li>Perform these calculations for nickel(II) carbonate.\n<ol style=\"list-style-type: lower-alpha;\">\n<li>With what volume of water must a precipitate containing NiCO<sub>3<\/sub> be washed to dissolve 0.100 g of this compound? Assume that the wash water becomes saturated with NiCO<sub>3<\/sub> (<em>K<\/em><sub>sp<\/sub> = 1.36 \u00d7 10<sup>\u20137<\/sup>).<\/li>\n<li>If the NiCO<sub>3<\/sub> were a contaminant in a sample of CoCO<sub>3<\/sub> (<em>K<\/em><sub>sp<\/sub> = 1.0 \u00d7 10<sup>\u201312<\/sup>), what mass of CoCO<sub>3<\/sub> would have been lost? Keep in mind that both NiCO<sub>3<\/sub> and CoCO<sub>3<\/sub> dissolve in the same solution.<\/li>\n<\/ol>\n<\/li>\n<li>Iron concentrations greater than 5.4 \u00d7 10<sup>\u20136<\/sup><em>M<\/em> in water used for laundry purposes can cause staining. What [OH<sup>\u2013<\/sup>] is required to reduce [Fe<sup>2+<\/sup>] to this level by precipitation of Fe(OH)<sub>2<\/sub>?<\/li>\n<li>A solution is 0.010 <em>M<\/em> in both Cu<sup>2+<\/sup> and Cd<sup>2+<\/sup>. What percentage of Cd<sup>2+<\/sup> remains in the solution when 99.9% of the Cu<sup>2+<\/sup> has been precipitated as CuS by adding sulfide?<\/li>\n<li>A solution is 0.15 <em>M<\/em> in both Pb<sup>2+<\/sup> and Ag<sup>+<\/sup>. If Cl<sup>\u2013<\/sup> is added to this solution, what is [Ag<sup>+<\/sup>] when PbCl<sub>2<\/sub> begins to precipitate?<\/li>\n<li>What reagent might be used to separate the ions in each of the following mixtures, which are 0.1 <em>M<\/em> with respect to each ion? In some cases it may be necessary to control the pH. (Hint: Consider the <em>K<\/em><sub>sp<\/sub> values given in\u00a0<a class=\"target-chapter\" href=\".\/chapter\/solubility-products\/\" target=\"_blank\">Solubility Products<\/a>.)\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]{\\text{Hg}}_{2}^{2+}[\/latex] and Cu<sup>2+<\/sup><\/li>\n<li>[latex]{\\text{SO}}_{4}^{2-}[\/latex] and Cl<sup>\u2013<\/sup><\/li>\n<li>Hg<sup>2+<\/sup> and Co<sup>2+<\/sup><\/li>\n<li>Zn<sup>2+<\/sup> and Sr<sup>2+<\/sup><\/li>\n<li>Ba<sup>2+<\/sup> and Mg<sup>2+<\/sup><\/li>\n<li>[latex]{\\text{CO}}_{3}^{2-}[\/latex] and OH<sup>\u2013<\/sup><\/li>\n<\/ol>\n<\/li>\n<li>A solution contains 1.0 \u00d7 10<sup>\u20135<\/sup> mol of KBr and 0.10 mol of KCl per liter. AgNO<sub>3<\/sub> is gradually added to this solution. Which forms first, solid AgBr or solid AgCl?<\/li>\n<li>A solution contains 1.0 \u00d7 10<sup>\u20132<\/sup> mol of KI and 0.10 mol of KCl per liter. AgNO<sub>3<\/sub> is gradually added to this solution. Which forms first, solid AgI or solid AgCl?<\/li>\n<li>The calcium ions in human blood serum are necessary for coagulation (Figure\u00a04). Potassium oxalate, K<sub>2<\/sub>C<sub>2<\/sub>O<sub>4<\/sub>, is used as an anticoagulant when a blood sample is drawn for laboratory tests because it removes the calcium as a precipitate of CaC<sub>2<\/sub>O<sub>4<\/sub>\u00b7H<sub>2<\/sub>O. It is necessary to remove all but 1.0% of the Ca<sup>2+<\/sup> in serum in order to prevent coagulation. If normal blood serum with a buffered pH of 7.40 contains 9.5 mg of Ca<sup>2+<\/sup> per 100 mL of serum, what mass of K<sub>2<\/sub>C<sub>2<\/sub>O<sub>4<\/sub> is required to prevent the coagulation of a 10 mL blood sample that is 55% serum by volume? (All volumes are accurate to two significant figures. Note that the volume of serum in a 10-mL blood sample is 5.5 mL. Assume that the K<sub>sp<\/sub> value for CaC<sub>2<\/sub>O<sub>4<\/sub> in serum is the same as in water.)<\/li>\n<li>About 50% of urinary calculi (kidney stones) consist of calcium phosphate, Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub>. The normal mid range calcium content excreted in the urine is 0.10 g of Ca<sup>2+<\/sup> per day. The normal mid range amount of urine passed may be taken as 1.4 L per day. What is the maximum concentration of phosphate ion that urine can contain before a calculus begins to form?<\/li>\n<li>The pH of normal urine is 6.30, and the total phosphate concentration [latex]\\left(\\left[\\text{PO}_{4}^{3-}\\right]+\\left[{\\text{HPO}}_{4}^{2-}\\right]+\\left[{\\text{H}}_{2}{\\text{PO}}_{4}^{-}\\right]+\\left[\\text{H}_3\\text{PO}_4\\right]\\right)[\/latex]\u00a0is 0.020 <em>M<\/em>. What is the minimum concentration of Ca<sup>2+<\/sup> necessary to induce kidney stone formation? (See Question\u00a049 for additional information.)<\/li>\n<li>Magnesium metal (a component of alloys used in aircraft and a reducing agent used in the production of uranium, titanium, and other active metals) is isolated from sea water by the following sequence of reactions: [latex]{\\text{Mg}}^{\\text{2+}}\\left(aq\\right)+{\\text{Ca(OH)}}_{2}\\left(aq\\right)\\longrightarrow {\\text{Mg(OH)}}_{2}\\left(s\\right)+{\\text{Ca}}^{\\text{2+}}\\left(aq\\right)[\/latex][latex]{\\text{Mg(OH)}}_{2}\\left(s\\right)+\\text{2HCl}(aq)\\longrightarrow {\\text{MgCl}}_{2}\\left(s\\right)+{\\text{2H}}_{2}\\text{O}\\left(l\\right)[\/latex]<br \/>\n[latex]{\\text{MgCl}}_{2}\\left(l\\right)\\stackrel{\\text{electrolysis}}{\\longrightarrow }\\text{Mg}\\left(s\\right)+{\\text{Cl}}_{2}\\left(g\\right)[\/latex]Sea water has a density of 1.026 g\/cm<sup>3<\/sup> and contains 1272 parts per million of magnesium as Mg<sup>2+<\/sup>(<em>aq<\/em>) by mass. What mass, in kilograms, of Ca(OH)<sub>2<\/sub> is required to precipitate 99.9% of the magnesium in 1.00 \u00d7 10<sup>3<\/sup> L of sea water?<\/li>\n<li>Hydrogen sulfide is bubbled into a solution that is 0.10 <em>M<\/em> in both Pb<sup>2+<\/sup> and Fe<sup>2+<\/sup> and 0.30 <em>M<\/em> in HCl. After the solution has come to equilibrium it is saturated with H<sub>2<\/sub>S ([H<sub>2<\/sub>S] = 0.10 <em>M<\/em>). What concentrations of Pb<sup>2+<\/sup> and Fe<sup>2+<\/sup> remain in the solution? For a saturated solution of H<sub>2<\/sub>S we can use the equilibrium:[latex]{\\text{H}}_{2}\\text{S}\\left(aq\\right)+{\\text{2H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{2H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{S}}^{2-}\\left(aq\\right)K=1.0\\times 1{0}^{-26}[\/latex](Hint: The [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] changes as metal sulfides precipitate.)<\/li>\n<li>Perform the following calculations involving concentrations of iodate ions:\n<ol style=\"list-style-type: lower-alpha;\">\n<li>The iodate ion concentration of a saturated solution of La(IO<sub>3<\/sub>)<sub>3<\/sub> was found to be 3.1 \u00d7 10<sup>\u20133<\/sup> mol\/L. Find the <em>K<\/em><sub>sp<\/sub>.<\/li>\n<li>Find the concentration of iodate ions in a saturated solution of Cu(IO<sub>3<\/sub>)<sub>2<\/sub> (<em>K<\/em><sub>sp<\/sub> = 7.4 \u00d7 10<sup>\u20138<\/sup>).<\/li>\n<\/ol>\n<\/li>\n<li>Calculate the molar solubility of AgBr in 0.035 <em>M<\/em> NaBr (<em>K<\/em><sub>sp<\/sub> = 5 \u00d7 10<sup>\u201313<\/sup>).<\/li>\n<li>How many grams of Pb(OH)<sub>2<\/sub> will dissolve in 500 mL of a 0.050-<em>M<\/em> PbCl<sub>2<\/sub> solution (<em>K<\/em><sub>sp<\/sub> = 1.2 \u00d7 10<sup>\u201315<\/sup>)?<\/li>\n<li>Use\u00a0<a href=\"http:\/\/employees.oneonta.edu\/viningwj\/sims\/common_ion_effect_s1.html\" target=\"_blank\">this simulation to study the process of salts dissolving and forming saturated solutions<\/a>\u00a0to complete the following exercise. Using 0.01 g CaF<sub>2<\/sub>, give the K<sub>sp<\/sub> values found in a 0.2-<em>M<\/em> solution of each of the salts. Discuss why the values change as you change soluble salts.<\/li>\n<li>How many grams of Milk of Magnesia, Mg(OH)<sub>2<\/sub> (<em>s<\/em>) (58.3 g\/mol), would be soluble in 200 mL of water. K<sub>sp<\/sub> = 7.1 \u00d7 10<sup>\u201312<\/sup>. Include the ionic reaction and the expression for K<sub>sp<\/sub> in your answer. What is the pH? (<em>K<\/em><sub>w<\/sub> = 1 \u00d7 10<sup>\u201314<\/sup> = [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] [OH<sup>\u2013<\/sup>])<\/li>\n<li>Two hypothetical salts, LM<sub>2<\/sub> and LQ, have the same molar solubility in H<sub>2<\/sub>O. If K<sub>sp<\/sub> for LM<sub>2<\/sub> is 3.20 \u00d7 10<sup>\u20135<\/sup>, what is the K<sub>sp<\/sub> value for LQ?<\/li>\n<li>Which of the following carbonates will form first? Which of the following will form last?\u00a0Explain.\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]{\\text{MgCO}}_{3}{K}_{\\text{sp}}=3.5\\times 1{0}^{-8}[\/latex]<\/li>\n<li>[latex]{\\text{CaCO}}_{3}{K}_{\\text{sp}}=4.2\\times 1{0}^{-7}[\/latex]<\/li>\n<li>[latex]{\\text{SrCO}}_{3}{K}_{\\text{sp}}=3.9\\times 1{0}^{-9}[\/latex]<\/li>\n<li>[latex]{\\text{BaCO}}_{3}{K}_{\\text{sp}}=4.4\\times 1{0}^{-5}[\/latex]<\/li>\n<li>[latex]{\\text{MnCO}}_{3}{K}_{\\text{sp}}=5.1\\times 1{0}^{-9}[\/latex]<\/li>\n<\/ol>\n<\/li>\n<li>How many grams of Zn(CN)<sub>2<\/sub>(<em>s<\/em>) (117.44 g\/mol) would be soluble in 100 mL of H<sub>2<\/sub>O? Include the balanced reaction and the expression for <em>K<\/em><sub>sp<\/sub> in your answer. The <em>K<\/em><sub>sp<\/sub> value for Zn(CN)<sub>2<\/sub>(<em>s<\/em>) is 3.0 \u00d7 10<sup>\u201316<\/sup>.<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q71248\">Show Selected Answers<\/span><\/p>\n<div id=\"q71248\" class=\"hidden-answer\" style=\"display: none\">\n<p>1.\u00a0In dissolution, one unit of substance produces a quantity of discrete ions or polyatomic ions that equals the number of times that the subunit appears in the formula.<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\begin{array}{ccc}\\text{AgI}\\left(s\\right)\\rightleftharpoons & {\\text{Ag}}^{\\text{+}}\\left(aq\\right)+& {\\text{I}}^{-}\\left(aq\\right)\\\\ & x& {x}\\end{array}[\/latex]<br \/>\nDissolving AgI(<em>s<\/em>) must produce the same amount of I<sup>\u2013<\/sup> ion as it does Ag<sup>+<\/sup> ion.<\/li>\n<li>[latex]\\begin{array}{ccc}{\\text{CaCO}}_{3}\\left(s\\right)\\rightleftharpoons & {\\text{Ca}}^{\\text{2+}}\\left(aq\\right)+& {\\text{CO}}_{3}^{2-}\\left(aq\\right)\\\\ & {x}& x\\end{array}[\/latex]<br \/>\nDissolving CaCO<sub>3<\/sub>(<em>s<\/em>) must produce the same amount of Ca<sup>2+<\/sup> ion as it does [latex]{\\text{CO}}_{3}^{2-}[\/latex] ion.<\/li>\n<li>[latex]\\begin{array}{lll}\\text{Mg}{\\left(\\text{OH}\\right)}_{2}\\left(s\\right)\\longrightarrow \\hfill & {\\text{Mg}}^{\\text{2+}}\\left(aq\\right)\\hfill & +2{\\text{OH}}^{-}\\left(aq\\right)\\hfill \\\\ \\hfill & x\\hfill & {\\text{2}x}\\hfill \\end{array}[\/latex]<br \/>\nWhen one unit of Mg(OH)<sub>2<\/sub> dissolves, two ions of OH<sup>\u2013<\/sup> are formed for each Mg<sup>2+<\/sup> ion.<\/li>\n<li>[latex]\\begin{array}{ccc}{\\text{Mg}}_{3}{\\left({\\text{PO}}_{4}\\right)}_{2}\\left(s\\right)\\rightleftharpoons & {\\text{3Mg}}^{\\text{2+}}\\left(aq\\right)+& {\\text{2PO}}_{4}^{3-}\\left(aq\\right)\\\\ & {3x}& 2x\\end{array}[\/latex]<br \/>\nOne unit of Mg<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub> provides two units of [latex]{\\text{PO}}_{4}^{3-}[\/latex] ion and three units of Mg<sup>2+<\/sup> ion.<\/li>\n<li>[latex]\\begin{array}{cccc}{\\text{Ca}}_{5}{\\left({\\text{PO}}_{4}\\right)}_{3}\\text{OH}\\left(s\\right)\\rightleftharpoons & {\\text{5Ca}}^{\\text{2+}}\\left(aq\\right)+& {\\text{3PO}}_{4}^{3-}\\left(aq\\right)+& {\\text{OH}}^{-}\\left(aq\\right)\\\\ & {5x}& {3x}& x\\end{array}[\/latex]<br \/>\nOne unit of Ca<sub>5<\/sub>(PO<sub>4<\/sub>)<sub>3<\/sub>OH dissolves into five units of Ca<sup>2+<\/sup> ion, three units of [latex]{\\text{PO}}_{4}^{3-}[\/latex] ion, and one unit of OH<sup>\u2013<\/sup> ion.<\/li>\n<\/ol>\n<p>3.\u00a0There is no change. A solid has an activity of 1 whether there is a little or a lot.<\/p>\n<p>5.\u00a0The solubility of silver bromide at the new temperature must be known. Normally the solubility increases and some of the solid silver bromide will dissolve.<\/p>\n<p>7.\u00a0CaF<sub>2<\/sub>, MnCO<sub>3<\/sub>, and ZnS; each is a salt of a weak acid and the hydronium ion from water reacts with the anion, causing more solid to dissolve to maintain the equilibrium concentration of the anion<\/p>\n<p>9.\u00a0The answers for each compound are as follows:<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\text{LaF}_{3}\\left(s\\right)\\rightleftharpoons\\text{La}^{3+}\\left(aq\\right)+{3F}^{-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,{;}\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{sp}}=\\left[\\text{La}^{3+}\\right]\\left[\\text{F}^{-}\\right]^{3}[\/latex]<\/li>\n<li>[latex]\\text{CaCO}_{3}\\left(s\\right)\\rightleftharpoons\\text{Ca}^{2+}\\left(aq\\right)+\\text{CO}_{3}^{2-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,{;}\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{sp}}=\\left[\\text{Ca}^{2+}\\right]\\left[\\text{CO}_{3}^{2-}\\right][\/latex]<\/li>\n<li>[latex]\\text{Ag}_{2}\\text{SO}_{4}\\left(s\\right)\\rightleftharpoons2\\text{Ag}^{+}\\left(aq\\right)+\\text{SO}_{4}^{2-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,{;}\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{sp}}=\\left[\\text{Ag}^{+}\\right]^{2}\\left[\\text{SO}_{4}^{2-}\\right][\/latex]<\/li>\n<li>[latex]\\text{Pb}\\left(\\text{OH}\\right)_{2}\\left(s\\right)\\rightleftharpoons\\text{Pb}^{2+}\\left(aq\\right)+2\\text{OH}^{-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,{;}\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{sp}}=\\left[\\text{Pb}^{2+}\\right]\\left[\\text{OH}^{-}\\right]^{2}[\/latex]<\/li>\n<\/ol>\n<p>11.\u00a0Convert each concentration into molar units. Multiply each concentration by 10 to determine the mass in 1 L, and then divide the molar mass.<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>BaSeO<sub>4<\/sub>\n<ul>\n<li>[latex]\\frac{0.{\\text{118 g L}}^{-1}}{280.{\\text{28 g mol}}^{-1}}=\\text{4.21}\\times {10}^{-4}\\text{}M[\/latex]<\/li>\n<li><em>K<\/em> = [Ba<sup>2+<\/sup>] [latex]\\left[{\\text{SeO}}_{4}^{2-}\\right][\/latex] = (4.21 \u00d7 10<sup>\u20134<\/sup>)(4.21 \u00d7 10<sup>\u20134<\/sup>) = 1.77 \u00d7 10<sup>\u20137<\/sup><\/li>\n<\/ul>\n<\/li>\n<li>Ba(BrO<sub>3<\/sub>)<sub>2<\/sub>\u00b7H<sub>2<\/sub>O:\n<ul>\n<li>[latex]\\frac{3.0{\\text{g L}}^{-1}}{{\\text{411.147 g mol}}^{-1}}=\\text{7.3}\\times {10}^{-3}\\text{}M[\/latex]<\/li>\n<li><em>K<\/em> = [Ba<sup>2+<\/sup>] [latex]{\\left[{\\text{BrO}}_{3}{}^{-}\\right]}^{2}[\/latex] = (7.3 \u00d7 10<sup>\u20133<\/sup>)(2 \u00d7 7.3 \u00d7 10<sup>\u20133<\/sup>)<sup>2<\/sup> = 1.6 \u00d7 10<sup>\u20136<\/sup><\/li>\n<\/ul>\n<\/li>\n<li>NH<sub>4<\/sub>MgAsO<sub>4<\/sub>\u00b76H<sub>2<\/sub>O:\n<ul>\n<li>[latex]\\frac{0.{\\text{38 g L}}^{-1}}{{\\text{289.3544 g mol}}^{-1}}=\\text{1.3}\\times {10}^{-3}\\text{}M[\/latex]<\/li>\n<li><em>K<\/em> = [latex]\\left[{\\text{NH}}_{4}{}^{\\text{+}}\\right][\/latex] [Mg<sup>2+<\/sup>] [latex]\\left[{\\text{AsO}}_{4}{}^{3-}\\right][\/latex] = (1.3 \u00d7 10<sup>\u20133<\/sup>)<sup>3<\/sup> = 2.2 \u00d7 10<sup>\u20139<\/sup><\/li>\n<\/ul>\n<\/li>\n<li>La<sub>2<\/sub>(MoO<sub>4<\/sub>)<sub>3<\/sub>:\n<ul>\n<li>[latex]\\frac{0.0{\\text{179 g L}}^{-1}}{{\\text{757.62 g mol}}^{-1}}=\\text{2.36}\\times {10}^{-5}\\text{}M[\/latex]<\/li>\n<li><em>K<\/em> = [La<sup>3+<\/sup>]<sup>2<\/sup> [latex]{\\left[{\\text{MoO}}_{4}{}^{2-}\\right]}^{3}[\/latex] = (2 \u00d7 2.36 \u00d7 10<sup>\u20135<\/sup>)<sup>2<\/sup>(3 \u00d7 2.36 \u00d7 10<sup>\u20135<\/sup>)<sup>3<\/sup> = 2.228 \u00d7 10<sup>\u20139<\/sup> \u00d7 3.549 \u00d7 10<sup>\u201313<\/sup> =\u00a07.91\u00a0 \u00d7 \u00a010<sup>\u201322<\/sup><\/li>\n<\/ul>\n<\/li>\n<\/ol>\n<p>13.\u00a0Let <em>x<\/em> be the molar solubility.<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li><em>K<\/em><sub>sp<\/sub> = [latex]\\left[{\\text{K}}^{\\text{+}}\\right]\\left[{\\text{HC}}_{4}{\\text{H}}_{4}{\\text{O}}_{6}^{-}\\right][\/latex] = 3 \u00d7 10<sup>\u20134<\/sup> = <em>x<\/em><sup>2<\/sup>, <em>x<\/em> = 2 \u00d7 10<sup>\u20132<\/sup><em>M<\/em>;<\/li>\n<li><em>K<\/em><sub>sp<\/sub> = [Pb<sup>2+<\/sup>][I<sup>\u2013<\/sup>]<sup>2<\/sup> = 8.7 \u00d7 10<sup>9<\/sup> = <em>x<\/em>(2<em>x<\/em>)<sup>3<\/sup> = 4<em>x<\/em><sup>3<\/sup>, <em>x<\/em> = 1.3 \u00d7 10<sup>\u20133<\/sup><em>M<\/em>;<\/li>\n<li><em>K<\/em><sub>sp<\/sub> = [latex]{\\left[{\\text{Ag}}^{\\text{+}}\\right]}^{4}\\left[{\\text{Fe(CN)}}_{6}{}^{\\text{4-}}\\right][\/latex] = 1.55 \u00d7 10<sup>\u201341<\/sup> = (4<em>x<\/em>)<sup>4<\/sup><em>x<\/em> = 256<em>x<\/em><sup>5<\/sup>, <em>x<\/em> = 2.27 \u00d7 10<sup>\u20139<\/sup><em>M<\/em>;<\/li>\n<li><em>K<\/em><sub>sp<\/sub> = [latex]\\left[{\\text{Hg}}_{2}{}^{\\text{2+}}\\right]{\\left[{\\text{I}}^{-}\\right]}^{2}[\/latex] = 4.5 \u00d7 10<sup>\u201329<\/sup> = [<em>x<\/em>][2<em>x<\/em>]<sup>2<\/sup> = 4<em>x<\/em><sup>3<\/sup>, <em>x<\/em> = 2.2 \u00d7 10<sup>\u201310<\/sup><em>M<\/em><\/li>\n<\/ol>\n<p>15.\u00a0The correct concentrations are as follows:<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li><em>K<\/em><sub>sp<\/sub> = 1.8 \u00d7 10<sup>\u201310<\/sup> = [Ag<sup>+<\/sup>][Cl<sup>\u2013<\/sup>] = <em>x<\/em>(<em>x<\/em> + 0.025), where <em>x<\/em> = [Ag<sup>+<\/sup>]. Assume that <em>x<\/em> is small when compared with 0.025 and therefore ignore it:\n<ul>\n<li>[latex]x=\\frac{1.8\\times {10}^{-10}}{0.025}=7.2\\times {10}^{-9}\\text{}M=\\left[{\\text{Ag}}^{\\text{+}}\\right],\\text{}\\left[{\\text{Cl}}^{-}\\right]=0.02\\text{5}M[\/latex]<\/li>\n<li>Check: [latex]\\frac{7.2\\times {10}^{-9}\\text{}M}{0.025\\text{}M}\\times \\text{100%}=\\text{2.9}\\times {10}^{-5}%[\/latex], an insignificant change;<\/li>\n<\/ul>\n<\/li>\n<li><em>K<\/em><sub>sp<\/sub> = 3.9 \u00d7 10<sup>\u201311<\/sup> = [Ca<sup>2+<\/sup>][F<sup>\u2013<\/sup>]<sup>2<\/sup> = <em>x<\/em>(2<em>x<\/em> + 0.00133 <em>M<\/em>)<sup>2<\/sup>, where <em>x<\/em> = [Ca<sup>2+<\/sup>]. Assume that <em>x<\/em> is small when compared with 0.0013 <em>M<\/em> and disregard it:\n<ul>\n<li>[latex]x=\\frac{\\text{3.9}\\times {10}^{-11}}{{\\left(0.00133\\right)}^{2}}=2.2\\times {10}^{-5}\\text{}M=\\left[{\\text{Ca}}^{\\text{2+}}\\right],\\text{}\\left[{\\text{F}}^{-}\\right]=0.0013\\text{}M[\/latex]<\/li>\n<li>Check: [latex]\\frac{\\text{2.25}\\times {10}^{-5}\\text{}M}{0.00133\\text{}M}\\times \\text{100%}=\\text{1.69%}[\/latex]. This value is less than 5% and can be ignored.<\/li>\n<\/ul>\n<\/li>\n<li>Find the concentration of K<sub>2<\/sub>SO<sub>4<\/sub>:\n<ul>\n<li>[latex]\\frac{\\text{19.50 g}}{174.2{\\text{60 g mol}}^{-1}}=\\text{0.1119 mol}[\/latex]<br \/>\n[latex]\\frac{0.1119\\text{mol}}{0.5\\text{L}}=\\text{0.2238}M={\\text{[SO}}_{4}{}^{2-}][\/latex]<br \/>\n<em>K<\/em><sub>sp<\/sub> = 1.18 \u00d7 10<sup>\u201318<\/sup> = [Ag<sup>+<\/sup>]<sup>2<\/sup> [latex]{\\text{[SO}}_{4}{}^{2-}][\/latex] = 4<em>x<\/em><sup>2<\/sup>(<em>x<\/em> + 0.2238)<br \/>\n[latex]{x}^{2}=\\frac{1.18\\times {10}^{-18}}{4\\left(0.2238\\right)}=\\text{1.32}\\times {10}^{-18}[\/latex]<br \/>\n<em>x<\/em> = 1.15 \u00d7 10<sup>\u20139<\/sup>[Ag<sup>+<\/sup>] = 2<em>x<\/em> = 2.30 \u00d7 10<sup>\u20139<\/sup><em>M<\/em><\/li>\n<li>Check: [latex]\\frac{1.15\\times {10}^{-9}}{0.2238}\\times 100%=\\text{5.14}\\times {10}^{-7}[\/latex]; the condition is satisfied.<\/li>\n<\/ul>\n<\/li>\n<li>Find the concentration of OH<sup>\u2013<\/sup> from the pH:\n<ul>\n<li>pOH = 14.00 \u2013 11.45 = 2.55<br \/>\n[OH<sup>\u2013<\/sup>] = 2.8 \u00d7 10<sup>\u20133<\/sup><em>M<br \/>\n<\/em><em>K<\/em><sub>sp<\/sub> = 4.5 \u00d7 10<sup>\u201317<\/sup> = [Zn<sup>2+<\/sup>][OH<sup>\u2013<\/sup>]<sup>2<\/sup> = <em>x<\/em>(2<em>x<\/em> + 2.8 \u00d7 10<sup>\u20133<\/sup>)<sup>2<br \/>\n<\/sup>Assume that <em>x<\/em> is small when compared with 2.8 \u00d7 10<sup>\u20133<\/sup>:<br \/>\n[latex]x=\\frac{4.5\\times {10}^{-17}}{{\\left(2.8\\times {10}^{-3}\\right)}^{2}}=\\text{5.7}\\times 10 - 12\\text{}M={\\text{[Zn}}^{\\text{2+}}\\text{]}[\/latex]<\/li>\n<li>Check: [latex]\\frac{5.7\\times {10}^{-12}}{2.8\\times {10}^{-3}}\\times \\text{100%}=\\text{2.0}\\times {10}^{-7}%[\/latex]; <em>x<\/em> is less than 5% of [OH<sup>\u2013<\/sup>] and is, therefore, negligible. In each case the change in initial concentration of the common ion is less than 5%.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n<p>17. \u00a0(a) <em>K<\/em><sub>sp<\/sub> = 1.9 \u00d7 10<sup>\u20134<\/sup> = [Ti<sup>+<\/sup>][Cl<sup>\u2013<\/sup>]; Let <em>x<\/em> = [Cl<sup>\u2013<\/sup>]:<\/p>\n<p>1.9 \u00d7 10<sup>\u20134<\/sup> = (<em>x<\/em> = 0.025)<em>x<\/em><\/p>\n<p>Assume that <em>x<\/em> is small when compared with 0.025:<\/p>\n<p>[latex]x=\\frac{1.9\\times {10}^{-4}}{0.025}=7.6\\times {10}^{-3}\\text{}M[\/latex]<\/p>\n<p>Check: [latex]\\frac{7.6\\times {10}^{-3}}{0.025}\\times \\text{100%}=\\text{30%}[\/latex]<\/p>\n<p>This value is too large to drop <em>x<\/em>. Therefore solve by using the quadratic equation:<\/p>\n<p>[latex]\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}[\/latex]<\/p>\n<p><em>x<\/em><sup>2<\/sup> + 0.025<em>x<\/em> \u2013 1.9 \u00d7 10<sup>\u20134<\/sup> = 0<\/p>\n<p>[latex]\\begin{array}{l}{ }x=\\frac{-0.025\\pm \\sqrt{6.25\\times {10}^{-4}+7.6\\times {10}^{-4}}}{2}=\\frac{-0.025\\pm \\sqrt{1.385\\times {10}^{-3}}}{2}\\\\ =\\frac{-0.025\\pm 0.0372}{2}=0.00\\text{61}M\\end{array}[\/latex]<\/p>\n<p>(Use only the positive answer for physical sense.)<\/p>\n<p>[Ti<sup>+<\/sup>] = 0.025 + 0.0061 = 3.1 \u00d7 10<sup>\u20132<\/sup><em>M<\/em><\/p>\n<p>[Cl<sup>\u2013<\/sup>] = 6.1 \u00d7 10<sup>\u20133<\/sup><\/p>\n<p>(b) <em>K<\/em><sub>sp<\/sub> = 1.7 \u00d7 10<sup>\u20136<\/sup> = [Ba<sup>2+<\/sup>][F<sup>\u2013<\/sup>]<sup>2<\/sup>; Let <em>x<\/em> = [Ba<sup>2+<\/sup>]<\/p>\n<p>1.7 \u00d7 10<sup>\u20136<\/sup> = <em>x<\/em>(<em>x<\/em> + 0.0313)<sup>2<\/sup> = 1.7 \u00d7 10<sup>\u20133<\/sup><em>M<\/em><\/p>\n<p>Check: [latex]\\frac{1.7\\times {10}^{-3}}{0.0313}\\times \\text{100%}=\\text{5.5%}[\/latex]<\/p>\n<p>This value is too large to drop <em>x<\/em>, and the entire equation must be solved. One method to find the answer is to solve by successive approximations. Begin by choosing the value of <em>x<\/em> that has just been calculated:<\/p>\n<p><em>x<\/em>\u2032(5.4 \u00d7 10<sup>\u20135<\/sup> + 0.0313)<sup>2<\/sup> = 1.7 \u00d7 10<sup>\u20133<\/sup> or<\/p>\n<p>[latex]x\\prime =\\frac{1.7\\times {10}^{-6}}{1.089\\times {10}^{-3}}=\\text{1.6}\\times {10}^{-3}[\/latex]<\/p>\n<p>A third approximation using this last calculation is as follows:<\/p>\n<p><em>x<\/em>\u2032\u2032(1.7 \u00d7 10<sup>\u20133<\/sup> + 0.0313)<sup>2<\/sup> = 1.7 \u00d7 10<sup>\u20136<\/sup> or<\/p>\n<p>[latex]x\\prime \\prime =\\frac{1.7\\times {10}^{-6}}{1.089\\times {10}^{-3}}=\\text{1.6}\\times {10}^{-3}[\/latex]<\/p>\n<p>This value is well within 5% and is acceptable.<\/p>\n<p>[Ba<sup>2+<\/sup>] = 1.6 \u00d7 10<sup>\u20133<\/sup><em>M<\/em><\/p>\n<p>[F<sup>\u2013<\/sup>] = (1.6 \u00d7 10<sup>\u20133<\/sup> + 0.0313) = 0.0329 <em>M<\/em>;<\/p>\n<p>(c) Find the molar concentration of the Mg(NO<sub>3<\/sub>)<sub>2<\/sub>. The molar mass of Mg(NO<sub>3<\/sub>)<sub>2<\/sub> is 148.3149 g\/mol. The number of moles is [latex]\\frac{\\text{8.156 g}}{148.314{\\text{9 g mol}}^{-1}}=\\text{0.05499 mol}[\/latex]<\/p>\n<p>[latex]M=\\frac{\\text{0.05499 mol}}{\\text{2.250 L}}=\\text{0.02444 M}[\/latex]<\/p>\n<p>Let <em>x<\/em> = [latex]\\left[{\\text{C}}_{2}{\\text{O}}_{4}^{2-}\\right][\/latex] and assume that <em>x<\/em> is small when compared with 0.02444 <em>M<\/em>.<\/p>\n<p><em>K<\/em><sub>sp<\/sub> = 8.6 \u00d7 10<sup>\u20135<\/sup> = [Mg<sup>2+<\/sup>] [latex]\\left[{\\text{C}}_{2}{\\text{O}}_{4}^{2-}\\right][\/latex] = (<em>x<\/em>)(<em>x<\/em> + 0.02444)<\/p>\n<p>0.02444<em>x<\/em> = 8.6 \u00d7 10<sup>\u20135<\/sup><\/p>\n<p><em>x<\/em> = [latex]\\left[{\\text{C}}_{2}{\\text{O}}_{4}^{2-}\\right][\/latex] = 3.5 \u00d7 10<sup>\u20133<\/sup><\/p>\n<p>Check: [latex]\\frac{3.5\\times {10}^{-3}}{0.02444}\\times \\text{100%}=\\text{14%}[\/latex]<\/p>\n<p>This value is greater than 5%, so the quadratic equation must be used to solve for <em>x<\/em>:<\/p>\n<p>[latex]\\begin{array}{l}{}\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}\\\\{x}^{2}+0.02444x - 8.6\\times {10}^{-5}=0\\\\x=\\frac{-0.02444\\pm \\sqrt{5.973\\times {10}^{-4}+3.44\\times {10}^{-4}}}{2}=\\text{}\\frac{-0.02444\\pm 0.03068}{2}\\end{array}[\/latex]<\/p>\n<p>(Use only the positive answer for physical sense.)<\/p>\n<p><em>x<\/em>\u2032 = [latex]\\left[{\\text{C}}_{2}{\\text{O}}_{4}{}^{2-}\\right][\/latex] = 3.5 \u00d7 10<sup>\u20133<\/sup><em>M<\/em><\/p>\n<p>[Mg<sup>2+<\/sup>] = 3.1 \u00d7 10<sup>\u20133<\/sup> + 0.02444 = 0.0275 <em>M<\/em><\/p>\n<p>(d) pH = 12.700; pOH = 1.300<\/p>\n<p>[OH<sup>\u2013<\/sup>] = 0.0501 <em>M<\/em>; Let <em>x<\/em> = [Ca<sup>2+<\/sup>]<\/p>\n<p><em>K<\/em><sub>sp<\/sub> = 7.9 \u00d7 10<sup>\u20136<\/sup> = [Ca<sup>2+<\/sup>][OH<sup>\u2013<\/sup>]<sup>2<\/sup> = (<em>x<\/em>)(<em>x<\/em> + 0.050)<sup>2<\/sup><\/p>\n<p>Assume that <em>x<\/em> is small when compared with 0.050 <em>M<\/em>:<\/p>\n<p><em>x<\/em> = [Ca<sup>2+<\/sup>] = 3.15 \u00d7 10<sup>\u20133<\/sup> (one additional significant figure is carried)<\/p>\n<p>Check: [latex]\\frac{3.15\\times {10}^{-3}}{0.050}\\times \\text{100%}=\\text{6.28%}[\/latex]<\/p>\n<p>This value is greater than 5%, so a more exact method, such as successive approximations, must be used. Begin by choosing the value of <em>x<\/em> that has just been calculated:<\/p>\n<p><em>x<\/em>\u2032(3.15 \u00d7 10<sup>\u20133<\/sup> + 0.0501)<sup>2<\/sup> = 7.9 \u00d7 10<sup>\u20136<\/sup> or<\/p>\n<p>[latex]x\\prime =\\frac{7.9\\times {10}^{-6}}{2.836\\times {10}^{-3}}=\\text{2.8}\\times {10}^{-3}[\/latex]<\/p>\n<p>[Ca<sup>2+<\/sup>] = 2.8 \u00d7 10<sup>\u20133<\/sup><em>M<\/em><\/p>\n<p>[OH<sup>\u2013<\/sup>] = (2.8 \u00d7 10<sup>\u20133<\/sup> + 0.0501) = 0.053 \u00d7 10<sup>\u20132<\/sup><em>M<\/em><\/p>\n<p>In each case, the initial concentration of the common ion changes by more than 5%.<\/p>\n<p>19.\u00a0The changes in concentration are greater than 5% and thus exceed the maximum value for disregarding the change.<\/p>\n<p>21.\u00a0Ca(OH)<sub>2<\/sub>: [Ca<sup>2+<\/sup>][OH<sup>\u2013<\/sup>]<sup>2<\/sup> = 7.9 \u00d7 10<sup>\u20136<\/sup><\/p>\n<p>Let <em>x<\/em> be [Ca<sup>2+<\/sup>] = molar solubility; then [OH<sup>\u2013<\/sup>] = 2<em>x<\/em><\/p>\n<p><em>K<\/em><sub>sp<\/sub> = <em>x<\/em>(2<em>x<\/em>)<sup>2<\/sup> = 4<em>x<\/em><sup>3<\/sup> = 7.9 \u00d7 10<sup>\u20136<\/sup><\/p>\n<p><em>x<\/em><sup>3<\/sup> = 1.975 \u00d7 10<sup>\u20136<\/sup><\/p>\n<p><em>x<\/em> = 0.013 <em>M<\/em><\/p>\n<p>CaCO<sub>3<\/sub>: [Ca<sup>2+<\/sup>] [latex]{\\text{[CO}}_{3}^{2-}][\/latex] = 4.8 \u00d7 10<sup>\u20139<\/sup><\/p>\n<p>Let <em>x<\/em> be [Ca<sup>2+<\/sup>]; then [latex]{\\text{[CO}}_{3}^{2-}][\/latex] = [Ca<sup>2+<\/sup>] = molar solubility<\/p>\n<p><em>K<\/em><sub>sp<\/sub> = <em>x<\/em><sup>2<\/sup> = 4.8 \u00d7 10<sup>\u20139<\/sup><\/p>\n<p><em>x<\/em> = 6.9 \u00d7 10<sup>\u20135<\/sup><em>M<\/em><\/p>\n<p>CaSO<sub>4<\/sub>\u00b72H<sub>2<\/sub>O: [Ca<sup>2+<\/sup>] [latex]{\\text{[SO}}_{4}^{2-}]\\text{\\cdot }[\/latex] [H<sub>2<\/sub>O]<sup>2<\/sup> = 2.4 10<sup>\u20135<\/sup><\/p>\n<p>Let <em>x<\/em> be [Ca<sup>2+<\/sup>] = molar solubility = [latex]{\\text{[SO}}_{4}^{2-}][\/latex]; then [H<sub>2<\/sub>O] = 2<em>x<\/em><\/p>\n<p><em>K<\/em><sub>sp<\/sub> = (<em>x<\/em>)(<em>x<\/em>)(2<em>x<\/em>)<sup>2<\/sup> = 2.4 \u00d7 10<sup>\u20135<\/sup><\/p>\n<p><em>x<\/em><sup>4<\/sup> = 6.0 \u00d7 10<sup>\u20136<\/sup><\/p>\n<p><em>x<\/em> = 0.049 <em>M<\/em><\/p>\n<p>This value is more than three times the value given by the <a href=\"http:\/\/hbcponline.com\/faces\/contents\/ContentsSearch.xhtml\"><em>Handbook of Chemistry and Physics<\/em><\/a> of (0.014 <em>M<\/em>) and reflects the complex interaction of water within the precipitate:<\/p>\n<p>CaC<sub>2<\/sub>O<sub>4<\/sub>\u00b7H<sub>2<\/sub>O: [Ca<sup>2+<\/sup>] [latex]\\left[{\\text{C}}_{2}{\\text{O}}_{4}^{2-}\\right][\/latex] [H<sub>2<\/sub>O] = 2.27 \u00d7 10<sup>\u20139<\/sup><\/p>\n<p>Let <em>x<\/em> be [Ca<sup>2+<\/sup>] = molar solubility = [latex]\\left[{\\text{C}}_{2}{\\text{O}}_{4}^{2-}\\right][\/latex] = [H<sub>2<\/sub>O]<\/p>\n<p><em>x<\/em><sup>3<\/sup> = 2.27 \u00d7 10<sup>\u20139<\/sup><\/p>\n<p><em>x<\/em> = 1.3 \u00d7 10<sup>\u20133<\/sup><em>M<\/em><\/p>\n<p>In this case, the interaction of water is also complex and the solubility is considerably less than that calculated.<\/p>\n<p>Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub>: [Ca<sup>2+<\/sup>]<sup>3<\/sup> [latex]{\\left[{\\text{PO}}_{4}^{3-}\\right]}^{2}[\/latex] = 1 \u00d7 10<sup>\u201325<\/sup><\/p>\n<p>Upon solution there are three Ca<sup>2+<\/sup> and two [latex]{\\text{PO}}_{4}^{3-}[\/latex] ions. Let the concentration of Ca<sup>2+<\/sup> formed upon solution be <em>x<\/em>. Then [latex]\\frac{2}{3}x[\/latex] is the concentration of [latex]{\\text{PO}}_{4}{}^{\\text{3-}}:[\/latex]<\/p>\n<p>[latex]{x}^{3}{\\left(\\frac{2}{3}x\\right)}^{2}{x}^{3}=1\\times {10}^{-25}=\\text{0.4444}{x}^{5}[\/latex]<\/p>\n<p><em>x<\/em> = 1 \u00d7 10<sup>\u20135<\/sup><em>M<\/em> = [Ca<sup>2+<\/sup>]<\/p>\n<p>The solubility is then one-third the concentration of Ca<sup>2+<\/sup>, or 4 \u00d7 10<sup>\u20136<\/sup>.<\/p>\n<p>23. First, find the concentration in a saturated solution of CaSO<sub>4<\/sub>. Before placing the CaSO<sub>4<\/sub> in water, the concentrations of Ca<sup>2+<\/sup> and [latex]{\\text{SO}}_{4}^{2-}[\/latex] are 0. Let <em>x<\/em> be the change in concentration of Ca<sup>2+<\/sup>, which is equal to the concentration of [latex]{\\text{SO}}_{4}^{2-}:[\/latex]<\/p>\n<p><em>K<\/em><sub>sp<\/sub> = [Ca<sup>2+<\/sup>] [latex]\\left[{\\text{SO}}_{4}^{2-}\\right][\/latex] = 2.4 \u00d7 10<sup>\u20135<\/sup><\/p>\n<p><em>x<\/em> \u00d7 <em>x<\/em> = <em>x<\/em><sup>2<\/sup> = 2.4 \u00d7 10<sup>\u20135<\/sup><\/p>\n<p>[latex]x=\\sqrt{2.4\\times {10}^{-5}}[\/latex]<\/p>\n<p><em>x<\/em> = 4.9 \u00d7 10<sup>\u20133<\/sup><em>M<\/em> = [latex]\\left[{\\text{SO}}_{4}^{2-}\\right][\/latex] = [Ca<sup>2+<\/sup>]<\/p>\n<p>Since this concentration is higher than 2.60 \u00d7 10<sup>\u20133<\/sup><em>M<\/em>, \u201cgyp\u201d water does not meet the standards.<\/p>\n<p>25.\u00a0The amount of CaSO<sub>4<\/sub>\u00b72H<sub>2<\/sub>O that dissolves is limited by the presence of a substantial amount of [latex]{\\text{SO}}_{4}^{2-}[\/latex] already in solution from the 0.010 <em>M<\/em> [latex]{\\text{SO}}_{4}{}^{-}[\/latex]. This is a common-ion problem. Let <em>x<\/em> be the change in concentration of Ca<sup>2+<\/sup> and of [latex]{\\text{SO}}_{4}^{2-}[\/latex] that dissociates from CaSO<sub>4<\/sub>:<\/p>\n<p>[latex]{\\text{CaSO}}_{4}\\left(s\\right)\\longrightarrow {\\text{Ca}}^{\\text{2+}}\\left(aq\\right)+{\\text{SO}}_{4}{}^{\\text{2-}}\\left(aq\\right)[\/latex]<\/p>\n<p><em>K<\/em><sub>sp<\/sub> = [Ca<sup>2+<\/sup>] [latex]{\\text{[SO}}_{4}^{2-}][\/latex] = 2.4 \u00d7 10<sup>\u20135<\/sup><\/p>\n<p>Addition of 0.010 <em>M<\/em> [latex]{\\text{SO}}_{4}^{2-}[\/latex] generated from the complete dissociation of 0.010 <em>M<\/em> SO<sub>4<\/sub> gives<\/p>\n<p>[<em>x<\/em>][<em>x<\/em> + 0.010] = 2.4 \u00d7 10<sup>\u20135<\/sup>. Here, <em>x<\/em> cannot be neglected in comparison with 0.010 <em>M<\/em>; the quadratic equation must be used. In standard form:<\/p>\n<p><em>x<\/em><sup>2<\/sup> + 0.010<em>x<\/em> \u2013 2.4 \u00d7 10<sup>\u20135<\/sup> = 0<\/p>\n<p>[latex]x=\\frac{-0.01\\pm \\sqrt{1\\times {10}^{-4}+9.6\\times {10}^{-5}}}{2}=\\frac{-0.01\\pm 1.4\\times {10}^{-2}}{2}[\/latex]<\/p>\n<p>Only the positive value will give a meaningful answer:<\/p>\n<p><em>x<\/em> = 2.0 \u00d7 10<sup>\u20133<\/sup> = [Ca<sup>2+<\/sup>]<\/p>\n<p>This is also the concentration of CaSO<sub>4<\/sub>\u00b72H<sub>2<\/sub>O that has dissolved. The mass of the salt in 1 L is<\/p>\n<p>Mass (CaSO<sub>4<\/sub>\u00b72H<sub>2<\/sub>O) = 2.0 \u00d7 10<sup>\u20133<\/sup> mol\/L \u00d7 172.16 g\/mol = 0.34 g\/L<\/p>\n<p>Note that the presence of the common ion, [latex]{\\text{SO}}_{4}^{2-}[\/latex], has caused a decrease in the concentration of Ca<sup>2+<\/sup> that otherwise would be in solution:<\/p>\n<p>[latex]\\sqrt{2.4\\times {10}^{-5}}=\\text{4.9}\\times {10}^{-5}\\text{}M[\/latex]<\/p>\n<p>27.\u00a0In each of the following, allow <em>x<\/em> to be the molar concentration of the ion occurring only once in the formula.<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li><em>K<\/em><sub>sp<\/sub> = [Ag<sup>+<\/sup>][Cl<sup>\u2013<\/sup>] = 1.5 \u2013 10<sup>\u201316<\/sup> = [<em>x<\/em><sup>2<\/sup>], [<em>x<\/em>] = 1.2 \u2013 10<sup>\u20138<\/sup><em>M<\/em>, [Ag<sup>+<\/sup>] = [I<sup>\u2013<\/sup>] = 1.2 \u00d7 10<sup>\u20138<\/sup><em>M<\/em><\/li>\n<li><em>K<\/em><sub>sp<\/sub> = [latex]{{\\text{[Ag}}^{\\text{+}}]}^{2}\\left[{\\text{SO}}_{4}^{2-}\\right][\/latex] = 1.18 \u00d7 10<sup>\u20135<\/sup> = [2<em>x<\/em>]<sup>2<\/sup>[<em>x<\/em>], 4<em>x<\/em><sup>3<\/sup> = 1.18 \u00d7 10<sup>\u20135<\/sup>, <em>x<\/em> = 1.43 \u00d7 10<sup>\u20132<\/sup><em>M<br \/>\n<\/em>As there are 2 Ag<sup>+<\/sup> ions for each [latex]{\\text{SO}}_{4}^{2-}[\/latex] ion, [Ag<sup>+<\/sup>] = 2.86 \u00d7 10<sup>\u20132<\/sup><em>M<\/em>, [latex]{\\text{[SO}}_{4}^{2-}][\/latex] = 1.43 \u00d7 10<sup>\u20132<\/sup><em>M<\/em>; (c) <em>K<\/em>sp = [Mn<sup>2+<\/sup>]<sup>2<\/sup>[OH<sup>\u2013<\/sup>]<sup>2<\/sup> = 4.5 \u00d7 10<sup>\u201314<\/sup> = [<em>x<\/em>][2<em>x<\/em>]<sup>2<\/sup>, 4<em>x<\/em><sup>3<\/sup> = 4.5 \u00d7 10<sup>\u201314<\/sup>, <em>x<\/em> = 2.24 \u00d7 10<sup>\u20135<\/sup><em>M<\/em>.<\/li>\n<li>Since there are two OH<sup>\u2013<\/sup> ions for each Mn<sup>2+<\/sup> ion, multiplication of <em>x<\/em> by 2 gives 4.48 \u00d7 10<sup>\u20135<\/sup><em>M<\/em>. If the value of <em>x<\/em> is rounded to the correct number of significant figures, [Mn<sup>2+<\/sup>] = 2.2 \u00d7 10<sup>\u20135<\/sup><em>M<\/em>. [OH<sup>\u2013<\/sup>] = 4.5 \u00d7 10<sup>\u20135<\/sup><em>M<\/em>. If the value of <em>x<\/em> is rounded before determining the value of [OH<sup>\u2013<\/sup>], the resulting value of [OH<sup>\u2013<\/sup>] is 4.4 \u00d7 10<sup>\u20135<\/sup><em>M<\/em>. We normally maintain one additional figure in the calculator throughout all calculations before rounding.<\/li>\n<li><em>K<\/em><sub>sp<\/sub> = [Sr<sup>2+<\/sup>][OH<sup>\u2013<\/sup>]<sup>2<\/sup> = 3.2 \u00d7 10<sup>\u20134<\/sup> = [<em>x<\/em>][2<em>x<\/em>]<sup>2<\/sup>, 4<em>x<\/em><sup>3<\/sup> = 3.2 \u00d7 10<sup>\u20134<\/sup>, <em>x<\/em> = 4.3 \u00d7 10<sup>\u20132<\/sup><em>M<\/em>.<br \/>\nSubstitution gives [Sr<sup>2+<\/sup>] = 4.3 \u00d7 10<sup>\u20132<\/sup> M, [OH<sup>\u2013<\/sup>] = 8.6 \u00d7 10<sup>\u20132<\/sup><em>M<\/em><\/li>\n<li><em>K<\/em><sub>sp<\/sub> = [Mg<sup>2+<\/sup>]<sup>2<\/sup>[OH<sup>\u2013<\/sup>]<sup>2<\/sup> = 1.5 \u00d7 10<sup>\u201311<\/sup> = [<em>x<\/em>][2<em>x<\/em>]<sup>2<\/sup>, 4<em>x<\/em><sup>3<\/sup> = 1.5 \u00d7 10<sup>\u201311<\/sup>, <em>x<\/em> = 1.55 \u00d7 10<sup>\u20134<\/sup><em>M<\/em>, 2<em>x<\/em> = 3.1 \u00d7 10<sup>\u20134<\/sup>.<br \/>\nSubstitution and taking the correct number of significant figures gives [Mg<sup>2+<\/sup>] = 1.6 \u00d7 10<sup>\u20134<\/sup><em>M<\/em>, [OH\u2013] = 3.1 \u00d7 10<sup>\u20134<\/sup><em>M<\/em>. If the number is rounded first, the first value is still [Mg<sup>2+<\/sup>] = 1.6 \u00d7 10<sup>\u20134<\/sup><em>M<\/em>, but the second is [OH<sup>\u2013<\/sup>] = 3.2 \u00d7 10<sup>\u20134<\/sup><em>M<\/em>.<\/li>\n<\/ol>\n<p>29.\u00a0In each case the value of <em>K<\/em><sub>sp<\/sub> is found by multiplication of the concentrations raised to the ion\u2019s stoichiometric power. Molar units are not normally shown in the value of <em>K<\/em>.<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>TlCl: <em>K<\/em><sub>sp<\/sub> = (1.4 \u00d7 10<sup>\u20132<\/sup>)(1.2 \u00d7 10<sup>\u20132<\/sup>) = 2.0 \u00d7 10<sup>\u20134<\/sup><\/li>\n<li>Ce(IO<sub>3<\/sub>)<sub>4<\/sub>: <em>K<\/em><sub>sp<\/sub> = (1.8 \u00d7 10<sup>\u20134<\/sup>)(2.6 \u00d7 10<sup>\u201313<\/sup>)<sup>4<\/sup> = 5.1 \u00d7 10<sup>\u201317<\/sup><\/li>\n<li>Gd<sub>2<\/sub>(SO<sub>4<\/sub>)<sub>3<\/sub>: <em>K<\/em><sub>sp<\/sub> = (0.132)<sup>2<\/sup>(0.198)<sup>3<\/sup> = 1.35 \u00d7 10<sup>\u20134<\/sup><\/li>\n<li>Ag<sub>2<\/sub>SO<sub>4<\/sub>: <em>K<\/em><sub>sp<\/sub> = (2.40 \u00d7 10<sup>\u20132<\/sup>)<sup>2<\/sup>(2.05 \u00d7 10<sup>\u20132<\/sup>) = 1.18 \u00d7 10<sup>\u20135<\/sup><\/li>\n<li>BaSO<sub>4<\/sub>: <em>K<\/em><sub>sp<\/sub> = (0.500)(2.16 \u00d7 10<sup>\u201310<\/sup>) = 1.08 \u00d7 10<sup>\u201310<\/sup><\/li>\n<\/ol>\n<p>31.\u00a0(a) [latex]{\\text{CaCO}}_{3}:{\\text{CaCO}}_{3}\\left(s\\right)\\longrightarrow {\\text{Ca}}^{\\text{2+}}\\left(aq\\right)+{\\text{CO}}_{3}{}^{2-}\\left(aq\\right)[\/latex]<\/p>\n<p><em>K<\/em><sub>sp<\/sub> = [Ca<sup>2+<\/sup>] [latex]{\\text{[CO}}_{3}^{2-}][\/latex] = 4.8 \u00d7 10<sup>\u20139<\/sup><\/p>\n<p>test <em>K<\/em><sub>sp<\/sub> against <em>Q<\/em> = [Ca<sup>2+<\/sup>] [latex]{\\text{[CO}}_{3}^{2-}][\/latex]<\/p>\n<p><em>Q<\/em> = [Ca<sup>2+<\/sup>] [latex]{\\text{[CO}}_{3}^{2-}][\/latex] = (0.003)(0.003) = 9 \u00d7 10<sup>\u20136<\/sup><\/p>\n<p><em>K<\/em><sub>sp<\/sub> = 4.8 \u00d7 10<sup>\u20139<\/sup> &lt; 9 \u00d7 10<sup>\u20136<\/sup><\/p>\n<p>The ion product does exceed <em>K<\/em><sub>sp<\/sub>, so CaCO<sub>3<\/sub> does precipitate.<\/p>\n<p>(b) [latex]{\\text{Co(OH)}}_{2}:{\\text{Co(OH)}}_{2}\\left(s\\right)\\longrightarrow {\\text{Co}}^{\\text{2+}}\\left(aq\\right)+{\\text{2OH}}^{-}\\left(aq\\right)[\/latex]<\/p>\n<p><em>K<\/em><sub>sp<\/sub> = [Co<sup>2+<\/sup>][OH<sup>\u2013<\/sup>]<sup>2<\/sup> = 2 \u00d7 10<sup>\u201316<\/sup><\/p>\n<p>test <em>K<\/em><sub>sp<\/sub> against <em>Q<\/em> = [Co<sup>2+<\/sup>][OH<sup>\u2013<\/sup>]<sup>2<\/sup><\/p>\n<p><em>Q<\/em> = [Co<sup>2+<\/sup>][OH<sup>\u2013<\/sup>]<sup>2<\/sup> = (0.01)(1 \u00d7 10<sup>\u20137<\/sup>)<sup>2<\/sup> = 1 \u00d7 10<sup>\u201316<\/sup><\/p>\n<p><em>K<\/em><sub>sp<\/sub> = 2 \u00d7 10<sup>\u201316<\/sup> &gt; 1 \u00d7 10<sup>\u201316<\/sup><\/p>\n<p>The ion product does not exceed <em>K<\/em><sub>sp<\/sub>, so the compound does not precipitate.<\/p>\n<p>(c) CaHPO<sub>4<\/sub>: (<em>K<\/em><sub>sp<\/sub> = 5 \u00d7 10<sup>\u20136<\/sup>):<\/p>\n<p><em>Q<\/em> = [Ca<sup>2+<\/sup>] [latex]{\\text{[HPO}}_{4}^{2-}][\/latex] = (0.01)(2 \u00d7 10<sup>\u20136<\/sup>) = 2 \u00d7 10<sup>\u20138<\/sup> &lt; <em>K<\/em><sub>sp<\/sub><\/p>\n<p>The ion product does not exceed <em>K<\/em><sub>sp<\/sub>, so compound does not precipitate.<\/p>\n<p>(d) Pb<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub>: (<em>K<\/em><sub>sp<\/sub> = 3 \u00d7 10<sup>\u201344<\/sup>):<\/p>\n<p><em>Q<\/em> = [Pb<sup>2+<\/sup>]<sup>3<\/sup> [latex]{{\\text{[PO}}_{4}^{3-}]}^{2}[\/latex] = (0.01)<sup>3<\/sup>(1 \u00d7 10<sup>\u201313<\/sup>)<sup>2<\/sup> = 1 \u00d7 10<sup>\u201332<\/sup> &gt; <em>K<\/em><sub>sp<\/sub><\/p>\n<p>The ion product exceeds <em>K<\/em><sub>sp<\/sub>, so the compound precipitates.<\/p>\n<p>33.\u00a0Precipitation of [latex]{\\text{SO}}_{4}^{2-}[\/latex] will begin when the ion product of the concentration of the [latex]{\\text{SO}}_{4}^{2-}[\/latex] and Ba<sup>2+<\/sup> ions exceeds the <em>K<\/em><sub>sp<\/sub> of BaSO<sub>4<\/sub>.<\/p>\n<p><em>K<\/em><sub>sp<\/sub> = 1.08 \u00d7 10<sup>\u201310<\/sup> = [Ba<sup>2+<\/sup>] [latex]\\left[{\\text{SO}}_{4}^{2-}\\right][\/latex] = (0.0758) [latex]{\\text{[SO}}_{4}^{2-}][\/latex]<\/p>\n<p>[latex]\\left[\\text{SO}_{4}^{2-}\\right]=\\frac{1.08\\times{10}^{-10}}{0.0758}=\\text{1.42}\\times{10}^{-9}M[\/latex]<\/p>\n<p>35.\u00a0Precipitation of Ag<sub>3<\/sub>PO<sub>4<\/sub> will begin when the ion product of the concentrations of the Ag<sup>+<\/sup> and [latex]{\\text{PO}}_{4}{}^{\\text{3-}}[\/latex] ions exceeds <em>K<\/em><sub>sp<\/sub>:<\/p>\n<p>[latex]{\\text{Ag}}_{3}{\\text{PO}}_{4}\\left(s\\right)\\longrightarrow {\\text{3Ag}}^{\\text{+}}\\left(aq\\right)+{\\text{PO}}_{4}{}^{\\text{3-}}\\left(aq\\right)[\/latex]<\/p>\n<p><em>K<\/em><sub>sp<\/sub> = 1.8 \u00d7 10<sup>\u201318<\/sup> = [Ag<sup>+<\/sup>]<sup>3<\/sup> [latex]{\\text{[PO}}_{4}{}^{\\text{3-}}][\/latex] = (0.0125)<sup>3<\/sup> [latex]{\\text{[PO}}_{4}{}^{3-}][\/latex]<\/p>\n<p>[latex]\\left[\\text{PO}_{4}^{3-}\\right]=\\frac{1.08\\times{10}^{-18}}{\\left(0.0125\\right)^{3}}=\\text{9.2}\\times {10}^{-13}M[\/latex]<\/p>\n<p>37.\u00a0[latex]{\\text{Ag}}_{2}{\\text{CO}}_{3}\\left(s\\right)\\longrightarrow {\\text{2Ag}}^{\\text{+}}\\left(aq\\right)+{\\text{CO}}_{3}{}^{\\text{2-}}\\left(aq\\right)[\/latex]<\/p>\n<p>[Ag<sup>+<\/sup>]<sup>2<\/sup> [latex]\\left[{\\text{CO}}_{3}{}^{\\text{2-}}\\right][\/latex] = <em>K<\/em><sub>sp<\/sub> = 8.2 \u00d7 10<sup>\u201312<\/sup><\/p>\n<p>[Ag<sup>+<\/sup>]<sup>2<\/sup>(2.5 \u00d7 10<sup>\u20136<\/sup>) = 8.2 \u00d7 10<sup>\u201312<\/sup><\/p>\n<p>[latex]\\left[\\text{Ag}^{+}\\right]^{2}=\\frac{8.2\\times{10}^{-12}}{2.50\\times{10}^{-6}}=\\text{3.28}\\times{10}^{-6}[\/latex]<\/p>\n<p>[Ag<sup>+<\/sup>] = 1.8 \u00d7 10<sup>\u20133<\/sup><em>M<\/em><\/p>\n<p>39.\u00a0In the <em>K<\/em><sub>sp<\/sub> expression, substitute the concentration of Ca<sup>2+<\/sup> and solve for [F<sup>\u2013<\/sup>].<\/p>\n<p><em>K<\/em><sub>sp<\/sub> = 3.9 \u00d7 10<sup>\u201311<\/sup> = [Ca<sup>2+<\/sup>][F<sup>\u2013<\/sup>]<sup>2<\/sup> = (1.0 \u00d7 10<sup>\u20134<\/sup>)[F<sup>\u2013<\/sup>]<sup>2<\/sup><\/p>\n<p>[latex]{\\left[{\\text{F}}^{-}\\right]}^{2}=\\frac{3.9\\times {10}^{-11}}{1.0\\times {10}^{-4}}=\\text{3.9}\\times {10}^{-7}[\/latex]<\/p>\n<p>[F<sup>\u2013<\/sup>] = 6.2 \u00d7 10<sup>\u20134<\/sup><\/p>\n<p>41.\u00a0(a) 2.28 L; (b) 7.3 \u00d7 10<sup>\u20137<\/sup> g<\/p>\n<p>43.\u00a0When 99.9% of Cu<sup>2+<\/sup> has precipitated as CuS, then 0.1% remains in solution.<\/p>\n<p>[latex]\\frac{0.1}{100}[\/latex] \u00d7 0.010 mol\/L = 1 \u00d7 10<sup>\u20135<\/sup><em>M<\/em> = [Cu<sup>2+<\/sup>]<\/p>\n<p>[Cu<sup>2+<\/sup>][S<sup>2\u2013<\/sup>] = <em>K<\/em><sub>sp<\/sub> = 6.7 \u00d7 10<sup>\u201342<\/sup><\/p>\n<p>(1 \u00d7 10<sup>\u20135<\/sup>)[S<sup>2\u2013<\/sup>] = 6.7 \u00d7 10<sup>\u201342<\/sup><\/p>\n<p>[S<sup>2\u2013<\/sup>] = 7 \u00d7 10<sup>\u201337<\/sup><em>M<\/em><\/p>\n<p>[Cd<sup>2+<\/sup>][S<sup>2\u2013<\/sup>]<em>K<\/em><sub>sp<\/sub> = 2.8 \u00d7 10<sup>\u201335<\/sup><\/p>\n<p>[Cd<sup>2+<\/sup>](7 \u00d7 10<sup>\u201337<\/sup>) = 2.8 \u00d7 10<sup>\u201335<\/sup><\/p>\n<p>[Cd<sup>2+<\/sup>] = 4 \u00d7 10<sup>1<\/sup><em>M<\/em><\/p>\n<p>Thus [Cd<sup>2+<\/sup>] can increase to about 40 <em>M<\/em> before precipitation begins. [Cd<sup>2+<\/sup>] is only 0.010 <em>M<\/em>, so 100% of it is dissolved.<\/p>\n<p>45.\u00a0To compare ions of the same oxidation state, look for compounds with a common counter ion that have very different <em>K<\/em><sub>sp<\/sub> values, one of which has a relatively large <em>K<\/em><sub>sp<\/sub>\u2014that is, a compound that is somewhat soluble.<\/p>\n<p>(a) [latex]{\\text{Hg}}_{2}{}^{\\text{2+}}[\/latex] and Cu<sup>2+<\/sup>: Add [latex]{\\text{SO}}_{4}{}^{\\text{2-}}[\/latex]. CuSO<sub>4<\/sub> is soluble (see\u00a0<a class=\"target-chapter\" href=\".\/chapter\/solubility-products\/\" target=\"_blank\">Solubility Products<\/a>), but <em>K<\/em><sub>sp<\/sub> for Hg<sub>2<\/sub>SO<sub>4<\/sub> is 6.2 \u00d7 10<sup>\u20137<\/sup>. When only 0.1% [latex]{\\text{Hg}}_{2}{}^{\\text{2+}}[\/latex] remains in solution:<\/p>\n<p>[latex]\\left[{\\text{Hg}}_{2}{}^{\\text{2+}}\\right]=\\frac{0.1%}{100%}\\times \\text{0.10}=1\\times {10}^{-4}\\text{}M[\/latex]<\/p>\n<p>and<\/p>\n<p>[latex]\\left[{\\text{SO}}_{4}^{2-}\\right]=\\frac{{K}_{\\text{sp}}}{\\left[{\\text{Hg}}_{2}{}^{\\text{2+}}\\right]}=\\frac{6.2\\times {10}^{-7}}{1\\times {10}^{-4}}=\\text{6.2}\\times {10}^{-3}\\text{}M[\/latex];<\/p>\n<p>(b) [latex]{\\text{SO}}_{4}^{2-}[\/latex] and Cl<sup>\u2013<\/sup>: Add Ba<sup>2+<\/sup>. BaCl<sub>2<\/sub> is soluble (see the section on catalysis), but <em>K<\/em><sub>sp<\/sub> for BaSO<sub>4<\/sub> is 1.08 \u00d7 10<sup>\u201310<\/sup>. When only 0.1% [latex]{\\text{SO}}_{4}^{2-}[\/latex] remains in solution, [latex]{\\text{[SO}}_{4}^{2-}][\/latex] = 1 \u00d7 10<sup>\u20134<\/sup><em>M<\/em> and<\/p>\n<p>[latex]\\left[{\\text{Ba}}^{\\text{2+}}\\right]=\\frac{1.08\\times {10}^{-10}}{1\\times {10}^{-4}}=1\\times {10}^{-6}\\text{}M[\/latex];<\/p>\n<p>(c) Hg<sup>2+<\/sup> and Co<sup>2+<\/sup>: Add S<sup>2\u2013<\/sup>: For the least soluble form of CoS, <em>K<\/em><sub>sp<\/sub> = 6.7 \u00d7 10<sup>\u201329<\/sup> and for HgS, <em>K<\/em><sub>sp<\/sub> = 2 \u00d7 10<sup>\u201359<\/sup>. CoS will not begin to precipitate until:<\/p>\n<p>[Co<sup>2+<\/sup>][S<sup>2\u2013<\/sup>] = <em>K<\/em><sub>sp<\/sub> = 6.7 \u00d7 10<sup>\u201329<\/sup><\/p>\n<p>(0.10)[S<sup>2\u2013<\/sup>] = 6.7 \u00d7 10<sup>\u201329<\/sup><\/p>\n<p>[S<sup>2\u2013<\/sup>] = 6.7 \u00d7 10<sup>\u201328<\/sup><\/p>\n<p>At that concentration:<\/p>\n<p>[Hg<sup>2+<\/sup>](6.7 \u00d7 10<sup>\u201328<\/sup>) = 2 \u00d7 10<sup>\u201359<\/sup><\/p>\n<p>[Hg<sup>2+<\/sup>] = 3 \u00d7 10<sup>\u201332<\/sup><em>M<\/em><\/p>\n<p>That is, it is virtually 100% precipitated. For a saturated (0.10 <em>M<\/em>) H<sub>2<\/sub>S solution, the corresponding [latex]\\left[\\text{H}_{3}\\text{O}^{+}\\right][\/latex] is:<\/p>\n<p>[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]=\\frac{\\left[{\\text{H}}_{2}\\text{S}\\right]}{\\left[{\\text{S}}^{2-}\\right]}{K}_{\\text{a}}=\\frac{\\left(0.10\\right)}{\\left(6.7\\times {10}^{-28}\\right)}\\times \\text{1.0}\\times {10}^{-26}[\/latex]<\/p>\n<p>[latex]\\left[\\text{H}_{3}\\text{O}^{+}\\right]=1.5M[\/latex]<\/p>\n<p>A solution more basic than this would supply enough S<sup>2\u2013<\/sup> for CoS to precipitate.<\/p>\n<p>(d) Zn<sup>2+<\/sup> and Sr<sup>2+<\/sup>: Add OH<sup>\u2013<\/sup> until [OH<sup>\u2013<\/sup>] = 0.050 <em>M<\/em>. For Sr(OH)<sub>2<\/sub>\u00b78H<sub>2<\/sub>O, <em>K<\/em><sub>sp<\/sub> = 3.2 \u00d7 10<sup>\u20134<\/sup>. For Zn(OH)<sub>2<\/sub>, <em>K<\/em><sub>sp<\/sub> = 4.5 \u00d7 10<sup>\u201311<\/sup>. When Zn<sup>2+<\/sup> is 99.9% precipitated, then [Zn<sup>2+<\/sup>] = 1 \u00d7 10<sup>\u20134<\/sup><em>M<\/em> and<\/p>\n<p>[latex]{\\left[{\\text{OH}}^{-}\\right]}^{2}=\\frac{{K}_{\\text{sp}}}{\\left[{\\text{Zn}}^{\\text{2+}}\\right]}\\text{=}\\frac{4.5\\times {10}^{-11}}{1\\times {10}^{-4}}=\\text{4.5}\\times {10}^{-7}[\/latex]<\/p>\n<p>[OH<sup>\u2013<\/sup>] = 7 \u00d7 10<sup>\u20134<\/sup><em>M<\/em><\/p>\n<p>When Sr(OH)<sub>2<\/sub>\u00b78H<sub>2<\/sub>O just begins to precipitate:<\/p>\n<p>[latex]{\\left[{\\text{OH}}^{-}\\right]}^{2}=\\frac{{K}_{\\text{sp}}}{\\left[{\\text{Sr}}^{\\text{2+}}\\right]}=\\frac{3.2\\times {10}^{-4}}{0.10}=\\text{3.2}\\times {10}^{-3}[\/latex]<\/p>\n<p>[OH<sup>\u2013<\/sup>] = 0.057 <em>M<\/em><\/p>\n<p>If [OH<sup>\u2013<\/sup>] is maintained less than 0.056 <em>M<\/em>, then Zn<sup>2+<\/sup> will precipitate and Sr<sup>2+<\/sup> will not.<\/p>\n<p>(e) Ba<sup>2+<\/sup> and Mg<sup>2+<\/sup>: Add [latex]{\\text{SO}}_{4}^{2-}[\/latex]. MgSO<sub>4<\/sub> is soluble and BaSO<sub>4<\/sub> is not (<em>K<\/em><sub>sp<\/sub> = 2 \u00d7 10<sup>\u201310<\/sup>).<\/p>\n<p>(f) [latex]{\\text{CO}}_{3}^{2-}[\/latex] and OH<sup>\u2013<\/sup>: Add Ba<sup>2+<\/sup>. For Ba(OH)<sub>2<\/sub>, 8H<sub>2<\/sub>O, <em>K<\/em><sub>sp<\/sub> = 5.0 \u00d7 10<sup>\u20133<\/sup>; for BaCO<sub>3<\/sub>, <em>K<\/em><sub>sp<\/sub> = 8.1 \u00d7 10<sup>\u20139<\/sup>. When 99.9% of [latex]{\\text{CO}}_{3}^{2-}[\/latex] has been precipitated [latex]{\\text{[CO}}_{3}^{2-}][\/latex] = 1 \u00d7 10<sup>\u20134<\/sup><em>M<\/em> and<\/p>\n<p>[latex]\\left[\\text{Ba}^{2+}\\right]=\\frac{K_{\\text{sp}}}{\\left[\\text{CO}_{3}^{2-}\\right]}=\\frac{8.1\\times{10}^{-9}}{1\\times{10}^{-4}}=\\text{8.1}\\times{10}^{-5}M[\/latex]<\/p>\n<p>Ba(OH)<sub>2<\/sub>\u00b78H<sub>2<\/sub>O begins to precipitate when:<\/p>\n<p>[latex]{\\text{[Ba}}^{\\text{2+}}\\text{]}=\\frac{{K}_{\\text{sp}}}{{\\left[{\\text{OH}}^{-}\\right]}^{2}}=\\frac{5.0\\times {10}^{-3}}{{\\left(0.10\\right)}^{2}}=\\text{0.50}M[\/latex]<\/p>\n<p>As long as [Ba<sup>2+<\/sup>] is maintained at less than 0.50 <em>M<\/em>, BaCO<sub>3<\/sub> precipitates and Ba(OH)<sub>2<\/sub>\u00b78H<sub>2<\/sub>O does not.<\/p>\n<p>47.\u00a0Compare the concentration of Ag<sup>+<\/sup> as determined from the two solubility product expressions. The one requiring the smaller [Ag<sup>+<\/sup>] will precipitate first.<\/p>\n<p>For AgCl: <em>K<\/em><sub>sp<\/sub> = 1.8 \u00d7 10<sup>\u201310<\/sup> = [Ag<sup>+<\/sup>][Cl<sup>\u2013<\/sup>]<\/p>\n<p>[latex]\\left[{\\text{Ag}}^{\\text{+}}\\right]=\\frac{1.8\\times {10}^{-10}}{\\left[0.10\\right]}\\text{=1.8}\\times {10}^{-9}\\text{}M[\/latex]<\/p>\n<p>For AgI: <em>K<\/em><sub>sp<\/sub> = 1.5 \u00d7 10<sup>\u201316<\/sup> = [Ag<sup>+<\/sup>][I<sup>\u2013<\/sup>]<\/p>\n<p>[latex]\\left[{\\text{Ag}}^{\\text{+}}\\right]=\\frac{1.5\\times {10}^{-16}}{1.0\\times {10}^{-2}}=\\text{1.5}\\times {10}^{-9}\\text{}M[\/latex]<\/p>\n<p>As the value of [Ag<sup>+<\/sup>] is smaller for AgI, AgI will precipitate first.<\/p>\n<p>49.\u00a0The dissolution of Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub> yields:<\/p>\n<p>[latex]{\\text{Ca}}_{3}{{\\text{(PO}}_{4})}_{2}\\left(s\\right)\\rightleftharpoons {\\text{3Ca}}^{\\text{2+}}\\left(aq\\right)+{\\text{2PO}}_{4}^{3-}\\left(aq\\right)[\/latex]<\/p>\n<p>Given the concentration of Ca<sup>2+<\/sup> in solution, the maximum [latex]\\left[\\text{PO}_{4}^{3-}\\right][\/latex] can be calculated by using the <em>K<\/em><sub>sp<\/sub> expression for Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub>:<\/p>\n<p style=\"text-align: center;\"><em>K<\/em><sub>sp<\/sub> = 1 \u00d7 10<sup>\u201325<\/sup> = [Ca<sup>2+<\/sup>]<sup>3<\/sup> [latex]{{\\text{[PO}}_{4}{}^{3-]}}^{2}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]{\\left[{\\text{Ca}}^{\\text{2+}}\\right]}_{\\text{urine}}=\\frac{0.10\\text{}g\\left(\\frac{1\\text{}mol}{40.08\\text{}g}\\right)}{1.4\\text{L}}=\\text{1.8}\\times {10}^{-3}\\text{}M[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]{{\\text{[PO}}_{4}{}^{3-}]}^{2}=\\frac{1\\times {10}^{-25}}{{\\left(1.8\\times {10}^{-3}\\right)}^{3}}=\\text{1.7}\\times {10}^{-17}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]{{\\text{[PO}}_{4}{}^{3-}]}^{2}=4\\times {10}^{-9}\\text{}M[\/latex]<\/p>\n<p>51.\u00a0Calculate the amount of Mg<sup>2+<\/sup> present in sea water; then use <em>K<\/em><sub>sp<\/sub> to calculate the amount of Ca(OH)<sub>2<\/sub> required to precipitate the magnesium.<\/p>\n<p style=\"text-align: center;\">mass Mg = 1.00 \u00d7 10<sup>3<\/sup> L \u00d7 1000 cm<sup>3<\/sup>\/L \u00d7 1.026 g\/cm<sup>3<\/sup> \u00d7 1272 ppm \u00d7 10<sup>\u20136<\/sup> ppm<sup>\u20131<\/sup> = 1.305 \u00d7 10<sup>3<\/sup> g<\/p>\n<p>The concentration is 1.305 g\/L. If 99.9% is to be recovered 0.999 \u00d7 1.305 g\/L = 1.304 g\/L will be obtained. The molar concentration is:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{1.304{\\text{g L}}^{-1}}{24.305{\\text{g mol}}^{-1}}=\\text{0.05365}M[\/latex]<\/p>\n<p>As the Ca(OH)<sub>2<\/sub> reacts with Mg<sup>2+<\/sup> on a 1:1 mol basis, the amount of Ca(OH)<sub>2<\/sub> required to precipitate 99.9% of the Mg<sup>2+<\/sup> in 1 L is:<\/p>\n<p>0.05365 M \u00d7 74.09 g\/mol Ca(OH)<sub>2<\/sub> = 3.97 g\/L<\/p>\n<p>For treatment of 1000 L, 1000 L \u00d7 3.97 g\/L = 3.97 \u00d7 10<sup>3<\/sup> g = 3.97 kg. However, additional [OH<sup>\u2013<\/sup>] must be added to maintain the equilibrium:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{Mg(OH)}}_{2}\\left(s\\right)\\rightleftharpoons {\\text{Mg}}^{\\text{2+}}\\left(aq\\right)+{\\text{2OH}}^{-}\\left(aq\\right)\\left({K}_{\\text{sp}}=\\text{1.5}\\times {10}^{-11}\\right)[\/latex]<\/p>\n<p>When the initial 1.035 g\/L is reduced to 0.1% of the original, the molarity is calculated as:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{0.001\\times {\\text{1.305 g L}}^{-1}}{24.305{\\text{g mol}}^{-1}}=\\text{5.369}\\times {10}^{-5}\\text{}M[\/latex]<\/p>\n<p>The added amount of OH<sup>\u2013<\/sup> required is found from the solubility product:<\/p>\n<p style=\"text-align: center;\">[Mg<sup>2+<\/sup>][OH<sup>\u2013<\/sup>]<sup>2<\/sup> = (5.369 \u00d7 10<sup>\u20135<\/sup>)[OH<sup>\u2013<\/sup>]<sup>2<\/sup> = 1.5 \u00d7 10<sup>\u201311<\/sup> = <em>K<\/em><sub>sp<\/sub><\/p>\n<p style=\"text-align: center;\">[OH<sup>\u2013<\/sup>] = 5.29 \u00d7 10<sup>\u20134<\/sup><\/p>\n<p>Thus an additional [latex]\\frac{1}{2}[\/latex] \u00d7 5.29 \u00d7 10<sup>\u20134<\/sup> mol (2.65 \u00d7 10<sup>\u20134<\/sup> mol) of Ca(OH)<sub>2<\/sub> per liter is required to supply the OH<sup>\u2013<\/sup>. For 1000 L:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{mass Ca(OH)}}_{2}=2.65\\times {10}^{-4}{\\text{mol Ca(OH)}}_{2}{\\text{L}}^{-1}\\times \\text{1.00}\\times {10}^{3}\\text{L}\\times \\frac{74.0946\\text{g}}{{\\text{mol Ca(OH)}}_{2}}=\\text{20 g}[\/latex]<\/p>\n<p>The total Ca(OH)<sub>2<\/sub> required is 3.97 kg + 0.020 kg = 3.99 kg.<\/p>\n<p>53.\u00a0(a) <em>K<\/em><sub>sp<\/sub> = [La<sup>3+<\/sup>] [latex]{\\left[{\\text{IO}}_{3}{}^{-}\\right]}^{3}[\/latex] = [latex]\\left(\\frac{1}{3}\\times \\text{3.1}\\times {10}^{-3}\\right)[\/latex] (3.1 \u00d7 10<sup>\u20133<\/sup>)<sup>3<\/sup> = (0.0010)(3.0 \u00d7 10<sup>\u20138<\/sup>) = 3.0 \u00d7 10<sup>\u201311<\/sup>;<\/p>\n<p>(b) <em>K<\/em><sub>sp<\/sub> = [Cu<sup>2+<\/sup>] [latex]{\\left[{\\text{IO}}_{3}{}^{-}\\right]}^{2}[\/latex] = <em>x<\/em>(2<em>x<\/em>)<sup>2<\/sup> = 7.4 \u00d7 10<sup>\u20138<\/sup><\/p>\n<p>4<em>x<\/em><sup>3<\/sup> = 7.4 \u00d7 10<sup>\u20138<\/sup><\/p>\n<p><em>x<\/em><sup>3<\/sup> = 1.85 \u00d7 10<sup>\u20138<\/sup><\/p>\n<p><em>x<\/em> = 2.64 \u00d7 10<sup>\u20133<\/sup><\/p>\n<p>[Cu<sup>2+<\/sup>] = 2.6 \u00d7 10<sup>\u20133<\/sup><\/p>\n<p>[latex]\\left[{\\text{IO}}_{3}{}^{-}\\right][\/latex] = 2<em>x<\/em> = 5.3 \u00d7 10<sup>\u20133<\/sup><\/p>\n<p>55.\u00a01.8 \u00d7 10<sup>\u20135<\/sup> g Pb(OH)<sub>2<\/sub><\/p>\n<p>57.\u00a0[latex]{\\text{Mg(OH)}}_{2}\\text{(s)}\\rightleftharpoons {\\text{Mg}}^{\\text{2+}}+{\\text{2OH}}^{-}{K}_{\\text{sp}}=\\left[{\\text{Mg}}^{\\text{2+}}\\right]{\\left[{\\text{OH}}^{-}\\right]}^{2}[\/latex]<\/p>\n<p>7.1 \u00d7 10<sup>\u201312<\/sup> = (<em>x<\/em>)(2<em>x<\/em>)<sup>2<\/sup> = 4<em>x<\/em><sup>3<\/sup><\/p>\n<p><em>x<\/em> = 1.21 \u00d7 10<sup>\u20134<\/sup><em>M<\/em> = [Mg<sup>2+<\/sup>]<\/p>\n<p>[latex]\\frac{1.21\\times {10}^{-4}\\text{mol}}{\\text{L}}\\times 0.20\\text{0 L}\\times \\frac{{\\text{1 mol Mg(OH)}}_{2}}{{\\text{1 mol Mg}}^{\\text{2+}}}\\times \\frac{{\\text{58.3 g Mg(OH)}}_{2}}{{\\text{1 mol Mg(OH)}}_{2}}=\\text{1.14}\\times {10}^{-3}{\\text{Mg(OH)}}_{2}[\/latex]<\/p>\n<p>59.\u00a0SrCO<sub>3<\/sub> will form first, since it has the smallest <em>K<\/em><sub>sp<\/sub> value it is the least soluble. BaCO<sub>3<\/sub> will be the last to precipitate, it has the largest <em>K<\/em><sub>sp<\/sub> value.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Glossary<\/h2>\n<p><strong>common ion effect:<\/strong> effect on equilibrium when a substance with an ion in common with the dissolved species is added to the solution; causes a decrease in the solubility of an ionic species, or a decrease in the ionization of a weak acid or base<\/p>\n<p><strong>molar solubility: <\/strong>solubility of a compound expressed in units of moles per liter (mol\/L)<\/p>\n<p><strong>selective precipitation: <\/strong>process in which ions are separated using differences in their solubility with a given precipitating reagent<\/p>\n<p><strong>solubility product (<em>K<\/em><sub>sp<\/sub>): <\/strong>equilibrium constant for the dissolution of a slightly soluble electrolyte<\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-3557\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Chemistry. <strong>Provided by<\/strong>: OpenStax College. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/openstaxcollege.org\">http:\/\/openstaxcollege.org<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at https:\/\/openstaxcollege.org\/textbooks\/chemistry\/get<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17,"menu_order":24,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Chemistry\",\"author\":\"\",\"organization\":\"OpenStax College\",\"url\":\"http:\/\/openstaxcollege.org\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at https:\/\/openstaxcollege.org\/textbooks\/chemistry\/get\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-3557","chapter","type-chapter","status-publish","hentry"],"part":2983,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapters\/3557","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/wp\/v2\/users\/17"}],"version-history":[{"count":13,"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapters\/3557\/revisions"}],"predecessor-version":[{"id":6064,"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapters\/3557\/revisions\/6064"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/parts\/2983"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapters\/3557\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/wp\/v2\/media?parent=3557"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapter-type?post=3557"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/wp\/v2\/contributor?post=3557"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/wp\/v2\/license?post=3557"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}