{"id":3605,"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=3605"},"modified":"2016-10-19T21:29:41","modified_gmt":"2016-10-19T21:29:41","slug":"multiple-equilibria-2","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/chapter\/multiple-equilibria-2\/","title":{"raw":"Multiple Equilibria","rendered":"Multiple Equilibria"},"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>Describe examples of systems involving two (or more) simultaneous chemical equilibria<\/li>\r\n \t<li>Calculate reactant and product concentrations for multiple equilibrium systems<\/li>\r\n \t<li>Compare dissolution and weak electrolyte formation<\/li>\r\n<\/ul>\r\n<\/div>\r\nThere are times when one equilibrium reaction does not adequately describe the system being studied. Sometimes we have more than one type of equilibrium occurring at once (for example, an acid-base reaction and a precipitation reaction).\r\n\r\nThe ocean is a unique example of a system with <b>multiple equilibria<\/b>, or multiple states of solubility equilibria working simultaneously. Carbon dioxide in the air dissolves in sea water, forming carbonic acid (H<sub>2<\/sub>CO<sub>3<\/sub>). The carbonic acid then ionizes to form hydrogen ions and bicarbonate ions [latex]\\left({\\text{HCO}}_{3}{}^{-}\\right)[\/latex], which can further ionize into more hydrogen ions and carbonate ions [latex]\\left({\\text{CO}}_{3}{}^{2-}\\right):[\/latex]\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rll}{}{\\text{CO}}_{2}\\left(g\\right)&amp;\\rightleftharpoons&amp;{\\text{CO}}_{2}\\left(aq\\right)\\\\{\\text{CO}}_{2}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}&amp;\\rightleftharpoons&amp;{\\text{H}}_{2}{\\text{CO}}_{3}\\left(aq\\right)\\\\{\\text{H}}_{2}{\\text{CO}}_{3}\\left(aq\\right)&amp;\\rightleftharpoons&amp;{\\text{H}}^{\\text{+}}\\left(aq\\right)+{\\text{HCO}}_{3}{}^{-}\\left(aq\\right)\\\\{\\text{HCO}}_{3}{}^{-}\\left(aq\\right)&amp;\\rightleftharpoons&amp;{\\text{H}}^{\\text{+}}\\left(aq\\right)+{\\text{CO}}_{3}{}^{2-}\\left(aq\\right)\\end{array}[\/latex]<\/p>\r\nThe excess H<sup>+<\/sup> ions make seawater more acidic. Increased ocean acidification can then have negative impacts on reef-building coral, as they cannot absorb the calcium carbonate they need to grow and maintain their skeletons (Figure\u00a01). This in turn disrupts the local biosystem that depends upon the health of the reefs for its survival. If enough local reefs are similarly affected, the disruptions to sea life can be felt globally. The world\u2019s oceans are presently in the midst of a period of intense acidification, believed to have begun in the mid-nineteenth century, and which is now accelerating at a rate faster than any change to oceanic pH in the last 20 million years.\r\n\r\n[caption id=\"attachment_5431\" align=\"aligncenter\" width=\"975\"]<img class=\"size-full wp-image-5431\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09042406\/CNX_Chem_15_03_CoralReef.jpg\" alt=\"This figure contains two photographs of coral reefs. In a, a colorful reef that includes hues of purple and pink corals is shown in blue green water with fish swimming in the background. In b, grey-green mossy looking coral is shown in a blue aquatic environment. This photo does not have the colorful appearance or fish that were shown in figure a.\" width=\"975\" height=\"386\" \/> Figure\u00a01. Healthy coral reefs (a) support a dense and diverse array of sea life across the ocean food chain. But when coral are unable to adequately build and maintain their calcium carbonite skeletons because of excess ocean acidification, the unhealthy reef (b) is only capable of hosting a small fraction of the species as before, and the local food chain starts to collapse. (credit a: modification of work by NOAA Photo Library; credit b: modification of work by \u201cprilfish\u201d\/Flickr)[\/caption]\r\n\r\n<div class=\"textbox\">\r\n\r\nLearn more about ocean acidification and how it affects other marine creatures.\r\n\r\nhttps:\/\/www.youtube.com\/watch?v=kxPwbhFeZSw\r\n\r\n<\/div>\r\n<div class=\"textbox\">This site has detailed information about <a href=\"http:\/\/www.teachoceanscience.net\/teaching_resources\/education_modules\/coral_reefs_and_climate_change\/how_does_climate_change_affect_coral_reefs\/\" target=\"_blank\">how ocean acidification specifically affects coral reefs<\/a>.<\/div>\r\nSlightly soluble solids derived from weak acids generally dissolve in strong acids, unless their solubility products are extremely small. For example, we can dissolve CuCO<sub>3<\/sub>, FeS, and Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub> in HCl because their basic anions react to form weak acids (H<sub>2<\/sub>CO<sub>3<\/sub>, H<sub>2<\/sub>S, and [latex]{\\text{H}}_{2}{\\text{PO}}_{4}^{-}[\/latex] ). The resulting decrease in the concentration of the anion results in a shift of the equilibrium concentrations to the right in accordance with Le Ch\u00e2telier\u2019s principle.\r\n\r\nOf particular relevance to us is the dissolution of hydroxylapatite, Ca<sub>5<\/sub>(PO<sub>4<\/sub>)<sub>3<\/sub>OH, in acid. Apatites are a class of calcium phosphate minerals (Figure\u00a02); a biological form of hydroxylapatite is found as the principal mineral in the enamel of our teeth. A mixture of hydroxylapatite and water (or saliva) contains an equilibrium mixture of solid Ca<sub>5<\/sub>(PO<sub>4<\/sub>)<sub>3<\/sub>OH and dissolved Ca<sup>2+<\/sup>, [latex]{\\text{PO}}_{4}^{3-}[\/latex], and OH<sup>\u2013<\/sup> ions:\r\n<p style=\"text-align: center;\">[latex]{\\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)[\/latex]<\/p>\r\n\r\n\r\n[caption id=\"attachment_5434\" align=\"aligncenter\" width=\"650\"]<img class=\"wp-image-5434 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09042527\/CNX_Chem_15_03_Apatite.jpg\" alt=\"This figure includes an image of two large light blue apatite crystals in a mineral conglomerate that includes white, grey, and tan crystals. The blue apatite crystals have a dull, dusty, or powdered appearance.\" width=\"650\" height=\"383\" \/> Figure\u00a02. Crystal of the mineral hydroxylapatite, Ca<sub>5<\/sub>(PO<sub>4<\/sub>)<sub>3<\/sub>OH, is shown here. Pure apatite is white, but like many other minerals, this sample is colored because of the presence of impurities.[\/caption]\r\n\r\nWhen exposed to acid, phosphate ions react with hydronium ions to form hydrogen phosphate ions and ultimately, phosphoric acid:\r\n<p style=\"text-align: center;\">[latex]{\\text{PO}}_{4}{}^{3-}\\left(aq\\right)+{\\text{H}}_{3}{\\text{O}}^{+}\\longrightarrow {\\text{H}}_{2}{\\text{PO}}_{4}{}^{2-}+{\\text{H}}_{2}\\text{O}[\/latex]\r\n[latex]{\\text{PO}}_{4}{}^{2-}\\left(aq\\right)+{\\text{H}}_{3}{\\text{O}}^{+}\\longrightarrow {\\text{H}}_{2}{\\text{PO}}_{4}{}^{-}+{\\text{H}}_{2}\\text{O}[\/latex]\r\n[latex]{\\text{H}}_{2}{\\text{PO}}_{4}{}^{-}+{\\text{H}}_{3}{\\text{O}}^{+}\\longrightarrow {\\text{H}}_{3}{\\text{PO}}_{4}+{\\text{H}}_{2}\\text{O}[\/latex]<\/p>\r\nHydroxide ion reacts to form water:\r\n<p style=\"text-align: center;\">[latex]{\\text{OH}}^{-}\\left(aq\\right)+{\\text{H}}_{3}{\\text{O}}^{+}\\longrightarrow 2{\\text{H}}_{2}\\text{O}[\/latex]<\/p>\r\nThese reactions decrease the phosphate and hydroxide ion concentrations, and additional hydroxylapatite dissolves in an acidic solution in accord with Le Ch\u00e2telier\u2019s principle. Our teeth develop cavities when acid waste produced by bacteria growing on them causes the hydroxylapatite of the enamel to dissolve. Fluoride toothpastes contain sodium fluoride, NaF, or stannous fluoride [more properly named tin(II) fluoride], SnF<sub>2<\/sub>. They function by replacing the OH<sup>\u2013<\/sup> ion in hydroxylapatite with F<sup>\u2013<\/sup> ion, producing fluorapatite, Ca<sub>5<\/sub>(PO<sub>4<\/sub>)<sub>3<\/sub>F:\r\n<p style=\"text-align: center;\">[latex]\\text{NaF}+{\\text{Ca}}_{5}{\\left({\\text{PO}}_{4}\\right)}_{3}\\text{OH}\\rightleftharpoons {\\text{Ca}}_{5}{\\left({\\text{PO}}_{4}\\right)}_{3}\\text{F}+{\\text{Na}}^{+}+{\\text{OH}}^{-}[\/latex]<\/p>\r\nThe resulting Ca<sub>5<\/sub>(PO<sub>4<\/sub>)<sub>3<\/sub>F is slightly less soluble than Ca<sub>5<\/sub>(PO<sub>4<\/sub>)<sub>3<\/sub>OH, and F<sup>\u2013<\/sup> is a weaker base than OH<sup>\u2013<\/sup>. Both of these factors make the fluorapatite more resistant to attack by acids than hydroxylapatite. See the Chemistry in Everyday Life feature on the role of fluoride in preventing tooth decay for more information.\r\n<div class=\"textbox shaded\">\r\n<h3 data-type=\"title\">Role of Fluoride in Preventing Tooth Decay<\/h3>\r\nAs we saw previously, fluoride ions help protect our teeth by reacting with hydroxylapatite to form fluorapatite, Ca<sub>5<\/sub>(PO<sub>4<\/sub>)<sub>3<\/sub>F. Since it lacks a hydroxide ion, fluorapatite is more resistant to attacks by acids in our mouths and is thus less soluble, protecting our teeth. Scientists discovered that naturally fluorinated water could be beneficial to your teeth, and so it became common practice to add fluoride to drinking water. Toothpastes and mouthwashes also contain amounts of fluoride (Figure\u00a03).\r\n\r\n[caption id=\"attachment_5435\" align=\"aligncenter\" width=\"650\"]<img class=\"wp-image-5435 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09042617\/CNX_Chem_15_03_Toothpaste.jpg\" alt=\"A tube of toothpaste\" width=\"650\" height=\"269\" \/> Figure 3. Fluoride, found in many toothpastes, helps prevent tooth decay (credit: Kerry Ceszyk).[\/caption]\r\n\r\nUnfortunately, excess fluoride can negate its advantages. Natural sources of drinking water in various parts of the world have varying concentrations of fluoride, and places where that concentration is high are prone to certain health risks when there is no other source of drinking water. The most serious side effect of excess fluoride is the bone disease, skeletal fluorosis. When excess fluoride is in the body, it can cause the joints to stiffen and the bones to thicken. It can severely impact mobility and can negatively affect the thyroid gland. Skeletal fluorosis is a condition that over 2.7 million people suffer from across the world. So while fluoride can protect our teeth from decay, the US Environmental Protection Agency sets a maximum level of 4 ppm\u00a0(4 mg\/L)\u00a0of fluoride in drinking water in the US. Fluoride levels in water are not regulated in all countries, so fluorosis is a problem in areas with high levels of fluoride in the groundwater.\r\n\r\n<\/div>\r\nWhen acid rain attacks limestone or marble, which are calcium carbonates, a reaction occurs that is similar to the acid attack on hydroxylapatite. The hydronium ion from the acid rain combines with the carbonate ion from calcium carbonates and forms the hydrogen carbonate ion, a weak acid:\r\n<p style=\"text-align: center;\">[latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{CO}}_{3}{}^{2-}\\left(aq\\right)\\longrightarrow {\\text{HCO}}_{3}{}^{-}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/p>\r\nCalcium hydrogen carbonate, Ca(HCO<sub>3<\/sub>)<sub>2<\/sub>, is soluble, so limestone and marble objects slowly dissolve in acid rain.\r\n\r\nIf we add calcium carbonate to a concentrated acid, hydronium ion reacts with the carbonate ion according to the equation:\r\n<p style=\"text-align: center;\">[latex]2{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{CO}}_{3}{}^{2-}\\left(aq\\right)\\longrightarrow {\\text{H}}_{2}{\\text{CO}}_{3}\\left(aq\\right)+2{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/p>\r\n(Acid rain is usually not sufficiently acidic to cause this reaction; however, laboratory acids are.) The solution may become saturated with the weak electrolyte carbonic acid, which is unstable, and carbon dioxide gas can be evolved:\r\n<p style=\"text-align: center;\">[latex]{\\text{H}}_{2}{\\text{CO}}_{3}\\left(aq\\right)\\longrightarrow {\\text{CO}}_{2}\\left(g\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/p>\r\nThese reactions decrease the carbonate ion concentration, and additional calcium carbonate dissolves. If enough acid is present, the concentration of carbonate ion is reduced to such a low level that the reaction quotient for the dissolution of calcium carbonate remains less than the solubility product of calcium carbonate, even after all of the calcium carbonate has dissolved.\r\n<div class=\"textbox examples\">\r\n<h3>Example 1:\u00a0Prevention of Precipitation of Mg(OH)<sub>2<\/sub><\/h3>\r\nCalculate the concentration of ammonium ion that is required to prevent the precipitation of Mg(OH)<sub>2<\/sub> in a solution with [Mg<sup>2+<\/sup>] = 0.10 <em>M<\/em> and [NH<sub>3<\/sub>] = 0.10 <em>M<\/em>.\r\n[reveal-answer q=\"496110\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"496110\"]\r\n\r\nTwo equilibria are involved in this system:\r\n<ul>\r\n \t<li>Reaction (1): [latex]\\text{Mg}{\\left(\\text{OH}\\right)}_{2}\\left(s\\right)\\rightleftharpoons {\\text{Mg}}^{\\text{2+}}\\left(aq\\right)+2{\\text{OH}}^{-}\\left(aq\\right);{K}_{\\text{sp}}=1.5\\times {10}^{-11}[\/latex]<\/li>\r\n \t<li>Reaction (2): [latex]{\\text{NH}}_{3}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{NH}}_{4}{}^{\\text{+}}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right);{K}_{\\text{sp}}=1.8\\times {10}^{-5}[\/latex]<\/li>\r\n<\/ul>\r\nTo prevent the formation of solid Mg(OH)<sub>2<\/sub>, we must adjust the concentration of OH<sup>\u2013<\/sup> so that the reaction quotient for Equation (1), <em>Q<\/em> = [Mg<sup>2+<\/sup>][OH<sup>\u2013<\/sup>]<sup>2<\/sup>, is less than <em>K<\/em><sub>sp<\/sub> for Mg(OH)<sub>2<\/sub>. (To simplify the calculation, we determine the concentration of OH<sup>\u2013<\/sup> when <em>Q<\/em> = <em>K<\/em><sub>sp<\/sub>.) [OH<sup>\u2013<\/sup>] can be reduced by the addition of [latex]{\\text{NH}}_{4}^{+}[\/latex], which shifts Reaction (2) to the left and reduces [OH<sup>\u2013<\/sup>].\r\n<ol>\r\n \t<li><em>We determine the [OH<sup>\u2013<\/sup>] at which Q = K<sub>sp<\/sub> when [Mg<sup>2+<\/sup>] = 0.10<\/em> M<em>:<\/em>\r\n[latex]Q=\\left[{\\text{Mg}}^{\\text{2+}}\\right]{\\left[{\\text{OH}}^{-}\\right]}^{2}=\\left(0.10\\right){\\left[{\\text{OH}}^{-}\\right]}^{2}=1.5\\times {10}^{-11}[\/latex][latex]\\left[{\\text{OH}}^{-}\\right]=1.2\\times {10}^{-5}M[\/latex]Solid Mg(OH)<sub>2<\/sub> will not form in this solution when [OH<sup>\u2013<\/sup>] is less than 1.2 [latex]\\times [\/latex] 10<sup>\u20135<\/sup><em>M<\/em>.<\/li>\r\n \t<li><em>We calculate the<\/em> [latex]\\left[N{H}_{4}{}^{+}\\right][\/latex] <em>needed to decrease [OH<sup>\u2013<\/sup>] to 1.2 \u00d7 10<sup>\u20135<\/sup><\/em> M <em>when [NH<sub>3<\/sub>] = 0.10.<\/em>[latex]{K}_{\\text{b}}=\\frac{\\left[{\\text{NH}}_{4}{}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\text{NH}}_{3}\\right]}=\\frac{\\left[{\\text{NH}}_{4}{}^{+}\\right]\\left(1.2\\times {10}^{-5}\\right)}{0.10}=1.8\\times {10}^{-5}[\/latex][latex]\\left[{\\text{NH}}_{4}{}^{+}\\right]=0.15M[\/latex]<\/li>\r\n<\/ol>\r\nWhen [latex]\\left[{\\text{NH}}_{4}{}^{+}\\right][\/latex] equals 0.15 <em>M<\/em>, [OH<sup>\u2013<\/sup>] will be 1.2 [latex]\\times [\/latex] 10<sup>\u20135<\/sup><em>M<\/em>. Any [latex]\\left[{\\text{NH}}_{4}{}^{+}\\right][\/latex] greater than 0.15 <em>M<\/em> will reduce [OH<sup>\u2013<\/sup>] below 1.2 [latex]\\times [\/latex] 10<sup>\u20135<\/sup><em>M<\/em> and prevent the formation of Mg(OH)<sub>2<\/sub>.\r\n\r\n[\/hidden-answer]\r\n<h4 id=\"fs-idm329760\">Check Your Learning<\/h4>\r\nConsider the two equilibria:\r\n<p style=\"text-align: center;\">[latex]\\text{ZnS}\\left(s\\right)\\rightleftharpoons {\\text{Zn}}^{\\text{2+}}\\left(aq\\right)+{\\text{S}}^{2-}\\left(aq\\right);{K}_{\\text{sp}}=1\\times {10}^{-27}[\/latex]\r\n[latex]2{\\text{H}}_{2}\\text{O}\\left(l\\right)+{\\text{H}}_{2}\\text{S}\\left(aq\\right)\\rightleftharpoons 2{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{S}}^{2-}\\left(aq\\right);K=1.0\\times {10}^{\\text{-26}}[\/latex]<\/p>\r\nand calculate the concentration of hydronium ion required to prevent the precipitation of ZnS in a solution that is 0.050 <em>M<\/em> in Zn<sup>2+<\/sup> and saturated with H<sub>2<\/sub>S (0.10 <em>M<\/em> H<sub>2<\/sub>S).\r\n\r\n[reveal-answer q=\"389689\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"389689\"]\r\n\r\n[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\text{&gt;}0.2M[\/latex]\r\n\r\n([S<sup>2\u2013<\/sup>] is less than 2 [latex]\\times [\/latex] 10<sup>\u201326<\/sup><em>M<\/em> and precipitation of ZnS does not occur.)\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nTherefore, precise calculations of the solubility of solids from the solubility product are limited to cases in which the only significant reaction occurring when the solid dissolves is the formation of its ions.\r\n<div class=\"textbox examples\">\r\n<h3>Example 2:\u00a0Multiple Equilibria<\/h3>\r\nUnexposed silver halides are removed from photographic film when they react with sodium thiosulfate (Na<sub>2<\/sub>S<sub>2<\/sub>O<sub>3<\/sub>, called hypo) to form the complex ion [latex]\\text{Ag}{\\left({\\text{S}}_{2}{\\text{O}}_{3}\\right)}_{2}{}^{3-}[\/latex] (<em>K<\/em><sub>f<\/sub> = 4.7 [latex]\\times [\/latex] 10<sup>13<\/sup>). The reaction with silver bromide is:\r\n\r\n<img class=\"aligncenter size-full wp-image-5437\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09042721\/CNX_Chem_15_03_AgBr_img.jpg\" alt=\"A chemical reaction is shown using structural formulas. On the left, A g superscript plus is followed by a plus sign, the number 2, and a structure in brackets. The structure is composed of a central S atom which has O atoms single bonded above, right, and below. A second S atom is single bonded to the left. Each of these bonded atoms has 6 dots around it. Outside the brackets is a superscript 2 negative. Following a bidirectional arrow is a structure in brackets with a central A g atom. To the left and right, S atoms are single bonded to the A g atom. Each of these S atoms has four dots around it, and an S atom connected with a single bond moving out from the central A g atom, forming the ends of the structure. Each of these atoms has three O atoms attached with single bonds above, below, and at the end of the structure. Each O atom has six dots around it. Outside the brackets is a superscript 3 negative.\" width=\"1000\" height=\"176\" \/>\r\n\r\n&nbsp;\r\n\r\nWhat mass of Na<sub>2<\/sub>S<sub>2<\/sub>O<sub>3<\/sub> is required to prepare 1.00 L of a solution that will dissolve 1.00 g of AgBr by the formation of [latex]\\text{Ag}{\\left({\\text{S}}_{2}{\\text{O}}_{3}\\right)}_{2}{}^{3-}?[\/latex]\r\n\r\n[reveal-answer q=\"693371\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"693371\"]\r\n\r\nTwo equilibria are involved when AgBr dissolves in a solution containing the [latex]{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}[\/latex] ion:\r\n<ul>\r\n \t<li>Reaction (1): [latex]\\text{AgBr}\\left(s\\right)\\rightleftharpoons {\\text{Ag}}^{\\text{+}}\\left(aq\\right)+{\\text{Br}}^{-}\\left(aq\\right);{K}_{\\text{sp}}=3.3\\times {10}^{\\text{-13}}[\/latex]<\/li>\r\n \t<li>Reaction (2): [latex]{\\text{Ag}}^{\\text{+}}\\left(aq\\right)+{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}\\left(aq\\right)\\rightleftharpoons \\text{Ag}{\\left({\\text{S}}_{2}{\\text{O}}_{3}\\right)}_{2}{}^{3-}\\left(aq\\right);{K}_{\\text{f}}=4.7\\times {10}^{13}[\/latex]<\/li>\r\n<\/ul>\r\nIn order for 1.00 g of AgBr to dissolve, the [Ag<sup>+<\/sup>] in the solution that results must be low enough for <em>Q<\/em> for Reaction (1) to be smaller than <em>K<\/em><sub>sp<\/sub> for this reaction. We reduce [Ag<sup>+<\/sup>] by adding [latex]{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}[\/latex] and thus cause Reaction (2) to shift to the right. We need the following steps to determine what mass of Na<sub>2<\/sub>S<sub>2<\/sub>O<sub>3<\/sub> is needed to provide the necessary [latex]{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}[\/latex] .\r\n<ol>\r\n \t<li><em>We calculate the [Br<sup>\u2013<\/sup>] produced by the complete dissolution of<\/em> <em>1.00 g of AgBr (5.33<\/em> [latex]\\times [\/latex] <em>10<sup>\u20133<\/sup> mol AgBr) in 1.00 L of solution:\r\n<\/em>[latex]\\left[{\\text{Br}}^{-}\\right]=5.33\\times {10}^{\\text{-3}}M[\/latex]<\/li>\r\n \t<li><em>We use [Br<sup>\u2013<\/sup>] and K<sub>sp<\/sub> to determine the maximum possible concentration of Ag<sup>+<\/sup> that can be present without causing reprecipitation of AgBr:<\/em>[latex]\\left[{\\text{Ag}}^{+}\\right]=6.2\\times {10}^{-11}M[\/latex]<\/li>\r\n \t<li><em>We determine the<\/em> [latex]\\left[{S}_{2}{O}_{3}{}^{2-}\\right][\/latex] <em>required to make [Ag<sup>+<\/sup>] = 6.2<\/em> [latex]\\times [\/latex] <em>10<sup>\u201311<\/sup> M<\/em> <em>after the remaining Ag<sup>+<\/sup> ion has reacted with<\/em> [latex]{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}[\/latex] <em>according to the equation:<\/em>[latex]{\\text{Ag}}^{+}+2{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}\\rightleftharpoons \\text{Ag}{\\left({\\text{S}}_{2}{\\text{O}}_{3}\\right)}_{2}{}^{\\text{3-}}{K}_{\\text{f}}=4.7\\times {10}^{13}[\/latex]Because 5.33 [latex]\\times [\/latex] 10<sup>\u20133<\/sup> mol of AgBr dissolves:\r\n\r\n[latex]\\left(5.33\\times {10}^{-3}\\right)-\\left(6.2\\times {10}^{\\text{-11}}\\right)=5.33\\times {10}^{-3}\\text{mol}\\text{Ag}{\\left({\\text{S}}_{2}{\\text{O}}_{3}\\right)}_{2}{}^{3-}[\/latex]\r\n\r\nThus, at equilibrium: [latex]\\left[\\text{Ag}{\\left({\\text{S}}_{2}{\\text{O}}_{3}\\right)}_{2}{}^{3-}\\right][\/latex] = 5.33 [latex]\\times [\/latex] 10<sup>\u20133<\/sup><em>M<\/em>, [Ag<sup>+<\/sup>] = 6.2 [latex]\\times [\/latex] 10<sup>\u201311<\/sup><em>M<\/em>, and <em>Q<\/em> = <em>K<\/em><sub>f<\/sub> = 4.7 [latex]\\times [\/latex] 10<sup>13<\/sup>:\r\n\r\n[latex]{K}_{\\text{f}}=\\frac{\\left[\\text{Ag}{\\left({\\text{S}}_{2}{\\text{O}}_{3}\\right)}_{2}{}^{\\text{3-}}\\right]}{\\left[{\\text{Ag}}^{+}\\right]{\\left[{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}\\right]}^{2}}=4.7\\times {10}^{13}[\/latex]\r\n\r\n[latex]\\left[{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}\\right]=1.4\\times {10}^{-3}M[\/latex]\r\n\r\nWhen [latex]\\left[{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}\\right][\/latex] is 1.4 [latex]\\times [\/latex] 10<sup>\u20133<\/sup><em>M<\/em>, [Ag<sup>+<\/sup>] is 6.2 [latex]\\times [\/latex] 10<sup>\u201311<\/sup><em>M<\/em> and all AgBr remains dissolved.<\/li>\r\n \t<li><em>We determine the total number of moles of<\/em> [latex]{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}[\/latex] <em>that must be added to the solution.<\/em> This equals the amount that reacts with Ag<sup>+<\/sup> to form [latex]\\text{Ag}{\\left({\\text{S}}_{2}{\\text{O}}_{3}\\right)}_{2}{}^{3-}[\/latex] plus the amount of free [latex]{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}[\/latex] in solution at equilibrium. To form 5.33 [latex]\\times [\/latex] 10<sup>\u20133<\/sup> mol of [latex]\\text{Ag}{\\left({\\text{S}}_{2}{\\text{O}}_{3}\\right)}_{2}{}^{3-}[\/latex] requires 2 [latex]\\times [\/latex] (5.33 [latex]\\times [\/latex] 10<sup>\u20133<\/sup>) mol of [latex]{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}[\/latex]. In addition, 1.4 [latex]\\times [\/latex] 10<sup>\u20133<\/sup> mol of unreacted [latex]{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}[\/latex] is present (Step 3). Thus, the total amount of [latex]{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}[\/latex] that must be added is:[latex]2\\times \\left(5.33\\times {10}^{-3}\\text{mol}{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}\\right)+1.4\\times {10}^{-3}\\text{mol}{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}=1.21\\times {10}^{-2}\\text{mol}{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}[\/latex]<\/li>\r\n \t<li><em>We determine the mass of Na<sub>2<\/sub>S<sub>2<\/sub>O<sub>3<\/sub> required to give 1.21 [latex]\\times [\/latex] 10<sup>\u20132<\/sup> mol<\/em> [latex]{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}[\/latex] <em>using the<\/em> <em>molar mass of Na<sub>2<\/sub>S<sub>2<\/sub>O<sub>3<\/sub>:<\/em>[latex]1.21\\times {10}^{\\text{-2}}\\text{mol}{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}\\times \\frac{158.1\\text{g}{\\text{Na}}_{2}{\\text{S}}_{2}{\\text{O}}_{3}}{1\\text{mol}{\\text{Na}}_{2}{\\text{S}}_{2}{\\text{O}}_{3}}=1.9\\text{g}{\\text{Na}}_{2}{\\text{S}}_{2}{\\text{O}}_{3}[\/latex]Thus, 1.00 L of a solution prepared from 1.9 g Na<sub>2<\/sub>S<sub>2<\/sub>O<sub>3<\/sub> dissolves 1.0 g of AgBr.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n<h4>Check Your Learning<\/h4>\r\nAgCl(s), silver chloride, is well known to have a very low solubility: Ag(<em>s<\/em>) \u21cc Ag<sup>+<\/sup>(<em>aq<\/em>) + Cl<sup>\u2212<\/sup>(<em>aq<\/em>), <em>K<sub>sp<\/sub><\/em> = 1.77 \u00d7 10<sup>\u201310<\/sup>. Adding ammonia significantly increases the solubility of AgCl because a complex ion is formed: Ag<sup>+<\/sup>(<em>aq<\/em>) + 2NH<sub>3<\/sub>(<em>aq<\/em>) \u21cc Ag(NH<sub>3<\/sub>)<sub>2<\/sub><sup>+<\/sup>(<em>aq<\/em>), <em>K<sub>f<\/sub><\/em> = 1.7 \u00d7 10<sup>7<\/sup>. What mass of NH<sub>3<\/sub> is required to prepare 1.00 L of solution that will dissolve 2.00 g of AgCl by formation of Ag(NH<sub>3<\/sub>)<sub>2<\/sub><sup>+<\/sup>?\r\n\r\n[reveal-answer q=\"88870\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"88870\"]1.00 L of a solution prepared with 4.84 g NH<sub>3<\/sub> dissolves 2.0 g of AgCl.[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Dissolution versus Weak Electrolyte Formation<\/h2>\r\nWe can determine how to shift the concentration of ions in the equilibrium between a slightly soluble solid and a solution of its ions by applying Le Ch\u00e2telier\u2019s principle. For example, one way to control the concentration of manganese(II) ion, Mn<sup>2+<\/sup>, in a solution is to adjust the pH of the solution and, consequently, to manipulate the equilibrium between the slightly soluble solid manganese(II) hydroxide, manganese(II) ion, and hydroxide ion:\r\n<p style=\"text-align: center;\">[latex]\\text{Mn}{\\left(\\text{OH}\\right)}_{2}\\left(s\\right)\\rightleftharpoons {\\text{Mn}}^{\\text{2+}}\\left(aq\\right)+2{\\text{OH}}^{-}\\left(aq\\right){K}_{\\text{sp}}=\\left[{\\text{Mn}}^{\\text{2+}}\\right]{\\left[{\\text{OH}}^{-}\\right]}^{2}[\/latex]<\/p>\r\nThis could be important to a laundry because clothing washed in water that has a manganese concentration exceeding 0.1 mg per liter may be stained by the manganese. We can reduce the concentration of manganese by increasing the concentration of hydroxide ion. We could add, for example, a small amount of NaOH or some other base such as the silicates found in many laundry detergents. As the concentration of OH<sup>\u2013<\/sup> ion increases, the equilibrium responds by shifting to the left and reducing the concentration of Mn<sup>2+<\/sup> ion while increasing the amount of solid Mn(OH)<sub>2<\/sub> in the equilibrium mixture, as predicted by Le Ch\u00e2telier\u2019s principle.\r\n<div class=\"textbox examples\">\r\n<h3>Example 3:\u00a0Solubility Equilibrium of a Slightly Soluble Solid<\/h3>\r\nWhat is the effect on the amount of solid Mg(OH)<sub>2<\/sub> that dissolves and the concentrations of Mg<sup>2+<\/sup> and OH<sup>\u2013<\/sup> when each of the following are added to a mixture of solid Mg(OH)<sub>2<\/sub> in water at equilibrium?\r\n<ol>\r\n \t<li>MgCl<sub>2<\/sub><\/li>\r\n \t<li>KOH<\/li>\r\n \t<li>an acid<\/li>\r\n \t<li>NaNO<sub>3<\/sub><\/li>\r\n \t<li>Mg(OH)<sub>2<\/sub><\/li>\r\n<\/ol>\r\n[reveal-answer q=\"661709\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"661709\"]\r\n\r\nThe equilibrium among solid Mg(OH)<sub>2<\/sub> and a solution of Mg<sup>2+<\/sup> and OH<sup>\u2013<\/sup> is:\r\n<p style=\"text-align: center;\">[latex]\\text{Mg}{\\left(\\text{OH}\\right)}_{2}\\left(s\\right)\\rightleftharpoons {\\text{Mg}}^{\\text{2+}}\\left(aq\\right)+2{\\text{OH}}^{-}\\left(aq\\right)[\/latex].<\/p>\r\n\r\n<ol>\r\n \t<li>The reaction shifts to the left to relieve the stress produced by the additional Mg<sup>2+<\/sup> ion, in accordance with Le Ch\u00e2telier\u2019s principle. In quantitative terms, the added Mg<sup>2+<\/sup> causes the reaction quotient to be larger than the solubility product (<em>Q<\/em> &gt; <em>K<\/em><sub>sp<\/sub>), and Mg(OH)<sub>2<\/sub> forms until the reaction quotient again equals <em>K<\/em><sub>sp<\/sub>. At the new equilibrium, [OH<sup>\u2013<\/sup>] is less and [Mg<sup>2+<\/sup>] is greater than in the solution of Mg(OH)<sub>2<\/sub> in pure water. More solid Mg(OH)<sub>2<\/sub> is present.<\/li>\r\n \t<li>The reaction shifts to the left to relieve the stress of the additional OH<sup>\u2013<\/sup> ion. Mg(OH)<sub>2<\/sub> forms until the reaction quotient again equals <em>K<\/em><sub>sp<\/sub>. At the new equilibrium, [OH<sup>\u2013<\/sup>] is greater and [Mg<sup>2+<\/sup>] is less than in the solution of Mg(OH)<sub>2<\/sub> in pure water. More solid Mg(OH)<sub>2<\/sub> is present.<\/li>\r\n \t<li>The concentration of OH<sup>\u2013<\/sup> is reduced as the OH<sup>\u2013<\/sup> reacts with the acid. The reaction shifts to the right to relieve the stress of less OH<sup>\u2013<\/sup> ion. In quantitative terms, the decrease in the OH<sup>\u2013<\/sup> concentration causes the reaction quotient to be smaller than the solubility product (<em>Q<\/em> &lt; <em>K<\/em><sub>sp<\/sub>), and additional Mg(OH)<sub>2<\/sub> dissolves until the reaction quotient again equals <em>K<\/em><sub>sp<\/sub>. At the new equilibrium, [OH<sup>\u2013<\/sup>] is less and [Mg<sup>2+<\/sup>] is greater than in the solution of Mg(OH)<sub>2<\/sub> in pure water. More Mg(OH)<sub>2<\/sub> is dissolved.<\/li>\r\n \t<li>NaNO<sub>3<\/sub> contains none of the species involved in the equilibrium, so we should expect that it has no appreciable effect on the concentrations of Mg<sup>2+<\/sup> and OH<sup>\u2013<\/sup>. (As we have seen previously, dissolved salts change the activities of the ions of an electrolyte. However, the salt effect is generally small, and we shall neglect the slight errors that may result from it.)<\/li>\r\n \t<li>The addition of solid Mg(OH)<sub>2<\/sub> has no effect on the solubility of Mg(OH)<sub>2<\/sub> or on the concentration of Mg<sup>2+<\/sup> and OH<sup>\u2013<\/sup>. The concentration of Mg(OH)<sub>2<\/sub> does not appear in the equation for the reaction quotient:\r\n[latex]Q=\\left[{\\text{Mg}}^{\\text{2+}}\\right]{\\left[{\\text{OH}}^{-}\\right]}^{2}[\/latex]\r\nThus, changing the amount of solid magnesium hydroxide in the mixture has no effect on the value of <em>Q<\/em>, and no shift is required to restore <em>Q<\/em> to the value of the equilibrium constant.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n<h4 id=\"fs-idm701008\">Check Your Learning<\/h4>\r\nWhat is the effect on the amount of solid NiCO<sub>3<\/sub> that dissolves and the concentrations of Ni<sup>2+<\/sup> and [latex]{\\text{CO}}_{3}{}^{2-}[\/latex] when each of the following are added to a mixture of the slightly soluble solid NiCO<sub>3<\/sub> and water at equilibrium?\r\n<ol>\r\n \t<li>Ni(NO<sub>3<\/sub>)<sub>2<\/sub><\/li>\r\n \t<li>KClO<sub>4<\/sub><\/li>\r\n \t<li>NiCO<sub>3<\/sub><\/li>\r\n \t<li>K<sub>2<\/sub>CO<sub>3<\/sub><\/li>\r\n \t<li>HNO<sub>3<\/sub> (reacts with carbonate giving [latex]{\\text{HCO}}_{3}{}^{-}[\/latex] or H<sub>2<\/sub>O and CO<sub>2<\/sub>)<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"267299\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"267299\"]\r\n<ol>\r\n \t<li>mass of NiCO<sub>3<\/sub>(s) increases, [Ni<sup>2+<\/sup>] increases, [latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right][\/latex] decreases<\/li>\r\n \t<li>no appreciable effect<\/li>\r\n \t<li>no effect except to increase the amount of solid NiCO<sub>3<\/sub><\/li>\r\n \t<li>mass of NiCO<sub>3<\/sub>(s) increases, [Ni<sup>2+<\/sup>] decreases, [latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right][\/latex] increases<\/li>\r\n \t<li>mass of NiCO<sub>3<\/sub>(s) decreases, [Ni<sup>2+<\/sup>] increases, [latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right][\/latex] decreases<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Key Concepts and Summary<\/h3>\r\nSeveral systems we encounter consist of multiple equilibria, systems where two or more equilibria processes are occurring simultaneously. Some common examples include acid rain, fluoridation, and dissolution of carbon dioxide in sea water. When looking at these systems, we need to consider each equilibrium separately and then combine the individual equilibrium constants into one solubility product or reaction quotient expression using the tools from the first equilibrium chapter. Le Ch\u00e2telier\u2019s principle also must be considered, as each reaction in a multiple equilibria system will shift toward reactants or products based on what is added to the initial reaction and how it affects each subsequent equilibrium reaction.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Exercises<\/h3>\r\n<ol>\r\n \t<li id=\"fs-idm301296\">A saturated solution of a slightly soluble electrolyte in contact with some of the solid electrolyte is said to be a system in equilibrium. Explain. Why is such a system called a heterogeneous equilibrium?<\/li>\r\n \t<li>Calculate the equilibrium concentration of Ni<sup>2+<\/sup> in a 1.0-<em>M<\/em> solution [Ni(NH<sub>3<\/sub>)<sub>6<\/sub>](NO<sub>3<\/sub>)<sub>2<\/sub>.<\/li>\r\n \t<li>Calculate the equilibrium concentration of Zn<sup>2+<\/sup> in a 0.30-<em>M<\/em> solution of [latex]\\text{Zn}{\\left(\\text{CN}\\right)}_{4}{}^{2-}[\/latex].<\/li>\r\n \t<li>Calculate the equilibrium concentration of Cu<sup>2+<\/sup> in a solution initially with 0.050 <em>M<\/em> Cu<sup>2+<\/sup> and 1.00 <em>M<\/em> NH<sub>3<\/sub>.<\/li>\r\n \t<li>Calculate the equilibrium concentration of Zn<sup>2+<\/sup> in a solution initially with 0.150 <em>M<\/em> Zn<sup>2+<\/sup> and 2.50 <em>M<\/em> CN<sup>\u2013<\/sup>.<\/li>\r\n \t<li>Calculate the Fe<sup>3+<\/sup> equilibrium concentration when 0.0888 mole of K<sub>3<\/sub>[Fe(CN)<sub>6<\/sub>] is added to a solution with 0.0.00010 <em>M<\/em> CN<sup>\u2013<\/sup>.<\/li>\r\n \t<li>Calculate the Co<sup>2+<\/sup> equilibrium concentration when 0.100 mole of [Co(NH<sub>3<\/sub>)<sub>6<\/sub>](NO<sub>3<\/sub>)<sub>2<\/sub> is added to a solution with 0.025 <em>M<\/em> NH<sub>3<\/sub>. Assume the volume is 1.00 L.<\/li>\r\n \t<li>The equilibrium constant for the reaction [latex]{\\text{Hg}}^{\\text{2+}}\\left(aq\\right)+2{\\text{Cl}}^{-}\\left(aq\\right)\\rightleftharpoons {\\text{HgCl}}_{2}\\left(aq\\right)[\/latex] is 1.6 [latex]\\times [\/latex] 10<sup>13<\/sup>. Is HgCl<sub>2<\/sub> a strong electrolyte or a weak electrolyte? What are the concentrations of Hg<sup>2+<\/sup> and Cl<sup>\u2013<\/sup> in a 0.015-<em>M<\/em> solution of HgCl<sub>2<\/sub>?<\/li>\r\n \t<li>Calculate the molar solubility of Sn(OH)<sub>2<\/sub> in a buffer solution containing equal concentrations of NH<sub>3<\/sub> and [latex]{\\text{NH}}_{4}{}^{+}[\/latex].<\/li>\r\n \t<li>Calculate the molar solubility of Al(OH)<sub>3<\/sub> in a buffer solution with 0.100 <em>M<\/em> NH<sub>3<\/sub> and 0.400 <em>M<\/em> [latex]{\\text{NH}}_{4}{}^{+}[\/latex].<\/li>\r\n \t<li>What is the molar solubility of CaF<sub>2<\/sub> in a 0.100-<em>M<\/em> solution of HF? <em>K<\/em><sub>a<\/sub> for HF = 7.2 [latex]\\times [\/latex] 10<sup>\u20134<\/sup>.<\/li>\r\n \t<li>What is the molar solubility of BaSO<sub>4<\/sub> in a 0.250-<em>M<\/em> solution of NaHSO<sub>4<\/sub>? <em>K<\/em><sub>a<\/sub> for [latex]{\\text{HSO}}_{4}{}^{-}[\/latex] = 1.2 [latex]\\times [\/latex] 10<sup>\u20132<\/sup>.<\/li>\r\n \t<li>What is the molar solubility of Tl(OH)<sub>3<\/sub> in a 0.10-<em>M<\/em> solution of NH<sub>3<\/sub>?<\/li>\r\n \t<li>What is the molar solubility of Pb(OH)<sub>2<\/sub> in a 0.138-<em>M<\/em> solution of CH<sub>3<\/sub>NH<sub>2<\/sub>?<\/li>\r\n \t<li>A solution of 0.075 <em>M<\/em> CoBr<sub>2<\/sub> is saturated with H<sub>2<\/sub>S ([H<sub>2<\/sub>S] = 0.10 <em>M<\/em>). What is the minimum pH at which CoS begins to precipitate?\r\n[latex]\\text{CoS}\\left(s\\right)\\rightleftharpoons {\\text{Co}}^{\\text{2+}}\\left(aq\\right)+{\\text{S}}^{2-}\\left(aq\\right){K}_{\\text{sp}}=4.5\\times {10}^{-27}[\/latex][latex]{\\text{H}}_{2}\\text{S}\\left(aq\\right)+2{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons 2{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{S}}^{2-}\\left(aq\\right)K=1.0\\times {10}^{\\text{-26}}[\/latex]<\/li>\r\n \t<li>A 0.125-<em>M<\/em> solution of Mn(NO<sub>3<\/sub>)<sub>2<\/sub> is saturated with H<sub>2<\/sub>S ([H<sub>2<\/sub>S] = 0.10 <em>M<\/em>). At what pH does MnS begin to precipitate?[latex]\\text{MnS}\\left(s\\right)\\rightleftharpoons {\\text{Mn}}^{\\text{2+}}\\left(aq\\right)+{\\text{S}}^{2-}\\left(aq\\right){K}_{\\text{sp}}=4.3\\times {10}^{-22}[\/latex][latex]{\\text{H}}_{2}\\text{S}\\left(aq\\right)+2{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons 2{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{S}}^{2-}\\left(aq\\right)K=1.0\\times {10}^{-26}[\/latex]<\/li>\r\n \t<li>Calculate the molar solubility of BaF<sub>2<\/sub> in a buffer solution containing 0.20 <em>M<\/em> HF and 0.20 <em>M<\/em> NaF.<\/li>\r\n \t<li>Calculate the molar solubility of CdCO<sub>3<\/sub> in a buffer solution containing 0.115 <em>M<\/em> Na<sub>2<\/sub>CO<sub>3<\/sub> and 0.120 <em>M<\/em> NaHCO<sub>3<\/sub><\/li>\r\n \t<li>To a 0.10-<em>M<\/em> solution of Pb(NO<sub>3<\/sub>)<sub>2<\/sub> is added enough HF(<em>g<\/em>) to make [HF] = 0.10 <em>M<\/em>.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Does PbF<sub>2<\/sub> precipitate from this solution? Show the calculations that support your conclusion.<\/li>\r\n \t<li>What is the minimum pH at which PbF<sub>2<\/sub> precipitates?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Calculate the concentration of Cd<sup>2+<\/sup> resulting from the dissolution of CdCO<sub>3<\/sub> in a solution that is 0.250 <em>M<\/em> in CH<sub>3<\/sub>CO<sub>2<\/sub>H, 0.375 <em>M<\/em> in NaCH<sub>3<\/sub>CO<sub>2<\/sub>, and 0.010 <em>M<\/em> in H<sub>2<\/sub>CO<sub>3<\/sub>.<\/li>\r\n \t<li>Both AgCl and AgI dissolve in NH<sub>3<\/sub>.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>What mass of AgI dissolves in 1.0 L of 1.0 <em>M<\/em> NH<sub>3<\/sub>?<\/li>\r\n \t<li>What mass of AgCl dissolves in 1.0 L of 1.0 <em>M<\/em> NH<sub>3<\/sub>?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Calculate the volume of 1.50 <em>M<\/em> CH<sub>3<\/sub>CO<sub>2<\/sub>H required to dissolve a precipitate composed of 350 mg each of CaCO<sub>3<\/sub>, SrCO<sub>3<\/sub>, and BaCO<sub>3<\/sub>.<\/li>\r\n \t<li>Even though Ca(OH)<sub>2<\/sub> is an inexpensive base, its limited solubility restricts its use. What is the pH of a saturated solution of Ca(OH)<sub>2<\/sub>?<\/li>\r\n \t<li>What mass of NaCN must be added to 1 L of 0.010 <em>M<\/em> Mg(NO<sub>3<\/sub>)<sub>2<\/sub> in order to produce the first trace of Mg(OH)<sub>2<\/sub>?<\/li>\r\n \t<li>Magnesium hydroxide and magnesium citrate function as mild laxatives when they reach the small intestine. Why do magnesium hydroxide and magnesium citrate, two very different substances, have the same effect in your small intestine. (Hint: The contents of the small intestine are basic.)<\/li>\r\n \t<li>The following question is taken from a Chemistry Advanced Placement Examination and is used with the permission of the Educational Testing Service. Solve the following problem: [latex]{\\text{MgF}}_{2}\\left(s\\right)\\rightleftharpoons {\\text{Mg}}^{\\text{2+}}\\left(aq\\right)+2{\\text{F}}^{-}\\left(aq\\right)[\/latex]In a saturated solution of MgF<sub>2<\/sub> at 18 \u00b0C, the concentration of Mg<sup>2+<\/sup> is 1.21 [latex]\\times [\/latex] 10<sup>\u20133<\/sup><em>M<\/em>. The equilibrium is represented by the equation above.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Write the expression for the solubility-product constant, <em>K<\/em><sub>sp<\/sub>, and calculate its value at 18 \u00b0C.<\/li>\r\n \t<li>Calculate the equilibrium concentration of Mg<sup>2+<\/sup> in 1.000 L of saturated MgF<sub>2<\/sub> solution at 18 \u00b0C to which 0.100 mol of solid KF has been added. The KF dissolves completely. Assume the volume change is negligible.<\/li>\r\n \t<li>Predict whether a precipitate of MgF<sub>2<\/sub> will form when 100.0 mL of a 3.00 [latex]\\times [\/latex] 10<sup>\u20133<\/sup>-<em>M<\/em> solution of Mg(NO<sub>3<\/sub>)<sub>2<\/sub> is mixed with 200.0 mL of a 2.00 [latex]\\times [\/latex] 10<sup>\u20133<\/sup>-<em>M<\/em> solution of NaF at 18 \u00b0C. Show the calculations to support your prediction.<\/li>\r\n \t<li>At 27 \u00b0C the concentration of Mg<sup>2+<\/sup> in a saturated solution of MgF<sub>2<\/sub> is 1.17 [latex]\\times [\/latex] 10<sup>\u20133<\/sup><em>M<\/em>. Is the dissolving of MgF<sub>2<\/sub> in water an endothermic or an exothermic process? Give an explanation to support your conclusion.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Which of the following compounds, when dissolved in a 0.01-<em>M<\/em> solution of HClO<sub>4<\/sub>, has a solubility greater than in pure water: AgBr, BaF<sub>2<\/sub>, Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>3<\/sub>, ZnS, PbI<sub>2<\/sub>? Explain your answer.<\/li>\r\n \t<li>What is the effect on the amount of solid Mg(OH)<sub>2<\/sub> that dissolves and the concentrations of Mg<sup>2+<\/sup> and OH<sup>\u2013<\/sup> when each of the following are added to a mixture of solid Mg(OH)<sub>2<\/sub> and water at equilibrium?\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>MgCl<sub>2<\/sub><\/li>\r\n \t<li>KOH<\/li>\r\n \t<li>HClO<sub>4<\/sub><\/li>\r\n \t<li>NaNO<sub>3<\/sub><\/li>\r\n \t<li>Mg(OH)<sub>2<\/sub><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>What is the effect on the amount of CaHPO<sub>4<\/sub> that dissolves and the concentrations of Ca<sup>2+<\/sup> and [latex]{\\text{HPO}}_{4}{}^{-}[\/latex] when each of the following are added to a mixture of solid CaHPO<sub>4<\/sub> and water at equilibrium?\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>CaCl<sub>2<\/sub><\/li>\r\n \t<li>HCl<\/li>\r\n \t<li>KClO<sub>4<\/sub><\/li>\r\n \t<li>NaOH<\/li>\r\n \t<li>CaHPO<sub>4<\/sub><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Identify all chemical species present in an aqueous solution of Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub> and list these species in decreasing order of their concentrations. (Hint: Remember that the [latex]{\\text{PO}}_{4}{}^{\\text{3-}}[\/latex] ion is a weak base.)<\/li>\r\n \t<li>A volume of 50 mL of 1.8 <em>M<\/em> NH<sub>3<\/sub> is mixed with an equal volume of a solution containing 0.95 g of MgCl<sub>2<\/sub>. What mass of NH<sub>4<\/sub>Cl must be added to the resulting solution to prevent the precipitation of Mg(OH)<sub>2<\/sub>?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"348957\"]Show Selected Answers[\/reveal-answer]\r\n[hidden-answer a=\"348957\"]\r\n\r\n2. [latex]{\\text{Ni}}^{\\text{2+}}\\left(aq\\right)+6{\\text{NH}}_{3}\\left(aq\\right)\\rightleftharpoons {\\left[\\text{Ni}{\\left({\\text{NH}}_{3}\\right)}_{6}\\right]}^{\\text{2+}}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,{;}\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{f}}=1.8\\times {10}^{8}[\/latex]\r\n\r\nLet <em>x<\/em> be the change in concentration as Ni<sup>2+<\/sup> dissociates. Because the initial Ni<sup>2+<\/sup> concentration is 0, the concentration at any times is <em>x<\/em>:\r\n<p style=\"text-align: center;\">[latex]1.8\\times {10}^{8}=\\frac{{\\left[\\text{Ni}{\\left({\\text{NH}}_{3}\\right)}_{6}\\right]}^{\\text{2+}}}{\\left[{\\text{Ni}}^{\\text{2+}}\\right]{\\left[{\\text{NH}}_{3}\\right]}^{6}}=\\frac{\\left(1.0-x\\right)}{x{\\left(6x\\right)}^{6}}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">1.8 [latex]\\times [\/latex] 10<sup>8<\/sup>(46656<em>x<\/em><sup>7<\/sup>) = 1.0 \u2013 <em>x<\/em><\/p>\r\n<p style=\"text-align: center;\">8.40 [latex]\\times [\/latex] 10<sup>12<\/sup>(<em>x<\/em><sup>2<\/sup>) = 1.0 \u2013 <em>x<\/em><\/p>\r\nSince <em>x<\/em> is small in comparison with 1.0, drop <em>x<\/em>:\r\n<p style=\"text-align: center;\">8.40 [latex]\\times [\/latex] 10<sup>12<\/sup>(<em>x<\/em><sup>7<\/sup>) = 1.0<\/p>\r\n<p style=\"text-align: center;\"><em>x<\/em><sup>7<\/sup> = 1.19 [latex]\\times [\/latex] 10<sup>\u201313<\/sup><\/p>\r\n<p style=\"text-align: center;\"><em>x<\/em> = 0.014 <em>M<\/em><\/p>\r\n4.\u00a0Assume that all Cu<sup>2+<\/sup> forms the complex whose concentration is 0.050 <em>M<\/em> and the remaining NH<sub>3<\/sub> has a concentration of 1.00 <em>M<\/em> \u2013 4(0.050 <em>M<\/em>) = 0.80 <em>M<\/em>. The complex dissociates:\r\n<p style=\"text-align: center;\">[latex]{\\left[\\text{Cu}{\\left({\\text{NH}}_{3}\\right)}_{4}\\right]}^{\\text{2+}}\\rightleftharpoons \\left[{\\text{Cu}}^{\\text{2+}}\\right]+4\\left[{\\text{NH}}_{3}\\right][\/latex]<\/p>\r\nLet <em>x<\/em> be the change in concentration of Cu<sup>2+<\/sup> that dissociates:\r\n<table>\r\n<thead>\r\n<tr>\r\n<th><\/th>\r\n<th>[Cu(NH<sub>3<\/sub>)<sub>4<\/sub><sup>2+<\/sup>]<\/th>\r\n<th>[Cu<sup>2+<\/sup>]<\/th>\r\n<th>[NH<sub>3<\/sub>]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<th>Initial concentration (<em>M<\/em>)<\/th>\r\n<td>0.050<\/td>\r\n<td>0<\/td>\r\n<td>0.80<\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Equilibrium (<em>M<\/em>)<\/th>\r\n<td>0.050\u00a0\u2212\u00a0<em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<td>4<em>x<\/em> + 0.80<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p style=\"text-align: center;\">[latex]\\frac{\\left[\\text{Cu}{\\left({\\text{NH}}_{3}\\right)}_{4}{}^{\\text{2+}}\\right]}{\\left[{\\text{Cu}}^{\\text{2+}}\\right]{\\left[{\\text{NH}}_{3}\\right]}^{4}}=1.2\\times {10}^{12}=\\frac{0.050-x}{x{\\left(4x+0.80\\right)}^{4}}[\/latex]<\/p>\r\nAssume that 4<em>x<\/em> is small when compared with 0.80 and that <em>x<\/em> is small when compared with 0.050:\r\n<p style=\"text-align: center;\">(0.80)<sup>4<\/sup> [latex]\\times [\/latex] 1.2 [latex]\\times [\/latex] 10<sup>12<\/sup><em>x<\/em> = 0.050<\/p>\r\n<p style=\"text-align: center;\"><em>x<\/em> = 1.0 [latex]\\times [\/latex] 10<sup>\u201313<\/sup><em>M<\/em><\/p>\r\n6.\u00a0Set up a table listing initial and equilibrium concentrations for the reaction:\r\n<p style=\"text-align: center;\">[latex]{\\text{Fe}}^{\\text{3+}}+6{\\text{CN}}^{-}\\rightleftharpoons {\\left(\\text{Fe}{\\left(\\text{CN}\\right)}_{6}\\right]}^{3-}{K}_{\\text{f}}=1\\times {10}^{44}[\/latex]<\/p>\r\nLet <em>x<\/em> be the concentration of Fe<sup>3+<\/sup> that dissociates when 0.0888 mol dissolves in 1.00 L of 0.00010 <em>M<\/em> CN<sup>\u2013<\/sup>. Assume no volume change upon dissolution:\r\n<table>\r\n<thead>\r\n<tr>\r\n<th><\/th>\r\n<th>[Fe(CN)<sub>6<\/sub><sup>3\u2212<\/sup>]<\/th>\r\n<th>[Fe<sup>3+<\/sup>]<\/th>\r\n<th>[CN<sup>\u2212<\/sup>]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<th>Initial concentration (<em>M<\/em>)<\/th>\r\n<td>0.0888<\/td>\r\n<td>0<\/td>\r\n<td>0.00010<\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Equilibrium (<em>M<\/em>)<\/th>\r\n<td>0.0888\u00a0\u2212\u00a0<em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<td>0.00010\u00a0\u2212 6<em>x<\/em><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p style=\"text-align: center;\">[latex]\\frac{\\left[\\text{Fe}{\\left(\\text{CN}\\right)}_{6}{}^{3-}\\right]}{\\left[{\\text{Fe}}^{\\text{3+}}\\right]{\\left[{\\text{CN}}^{-}\\right]}^{6}}=\\frac{0.0888-x}{x{\\left(0.000010 - 6x\\right)}^{6}}=1\\times {10}^{44}[\/latex]<\/p>\r\nAssume that <em>x<\/em> is small when compared with the terms from which it is subtracted:\r\n<p style=\"text-align: center;\">0.0888 = (0.00010)<sup>6<\/sup>(<em>x<\/em>)(1 [latex]\\times [\/latex] 10<sup>44<\/sup>)<\/p>\r\n<p style=\"text-align: center;\">[latex]x=\\frac{0.0888}{1\\times {10}^{26}}=9\\times {10}^{-22}M[\/latex]<\/p>\r\n8.\u00a0Let <em>x<\/em> be the change in the number of moles of Hg<sup>2+<\/sup> that form per liter:\r\n<table>\r\n<thead>\r\n<tr>\r\n<th><\/th>\r\n<th>[HgCl<sub>2<\/sub>]<\/th>\r\n<th>[Hg<sup>2+<\/sup>]<\/th>\r\n<th>[Cl<sup>\u2212<\/sup>]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<th>Initial concentration (<em>M<\/em>)<\/th>\r\n<td>0.015<\/td>\r\n<td>0<\/td>\r\n<td>0<\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Equilibrium (<em>M<\/em>)<\/th>\r\n<td>0.015\u00a0\u2212\u00a0<em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<td>2<em>x<\/em><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\frac{\\left[{\\text{HgCl}}_{2}\\right]}{\\left[{\\text{Hg}}^{\\text{2+}}\\right]{\\left[{\\text{Cl}}^{-}\\right]}^{2}}=K=1.6\\times {10}^{13}\\\\ \\frac{0.015-x}{\\left(x\\right){\\left(2x\\right)}^{2}}\\approx \\frac{0.015}{4{x}^{3}}=1.6\\times {10}^{13}\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: center;\"><em>x<\/em><sup>3<\/sup> = 2.3 [latex]\\times [\/latex] 10<sup>\u201316<\/sup><\/p>\r\n<p style=\"text-align: center;\"><em>x<\/em> = 6.2 [latex]\\times [\/latex] 10<sup>\u20136<\/sup><em>M<\/em> = [Hg<sup>2+<\/sup>]<\/p>\r\n<p style=\"text-align: center;\">2x = 1.2 [latex]\\times [\/latex] 10<sup>\u20135<\/sup>] <em>M<\/em> = [Cl<sup>\u2013<\/sup>]<\/p>\r\nThe substance is a weak electrolyte because very little of the initial 0.015 <em>M<\/em> HgCl<sub>2<\/sub> dissolved.\r\n\r\n10.\u00a0[latex]{K}_{\\text{b}}=\\frac{\\left[{\\text{NH}}_{4}{}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\text{NH}}_{3}\\right]}=\\frac{\\left(0.400\\right)\\left[{\\text{OH}}^{-}\\right]}{\\left(0.100\\right)}=1.8\\times {10}^{-5}[\/latex]\r\n\r\n[latex]\\left[{\\text{OH}}^{-}\\right]=\\frac{\\left(0.100\\right)\\left(1.8\\times {10}^{-5}\\right)}{0.0400}\\text{=}4.5\\times {10}^{-5}[\/latex]\r\n\r\n<em>K<\/em><sub>sp<\/sub> = [Al<sup>3+<\/sup>][OH<sup>\u2013<\/sup>]<sup>3<\/sup> = [Al<sup>3+<\/sup>](4.5 [latex]\\times [\/latex] 10<sup>\u20135<\/sup>)<sup>3<\/sup> = 1.9 [latex]\\times [\/latex] 10<sup>\u201333<\/sup>\r\n\r\n[Al<sup>3+<\/sup>] = 2.1 [latex]\\times [\/latex] 10<sup>\u201320<\/sup> (molar solubility)\r\n\r\n12.\u00a0Find the amount of [latex]{\\text{SO}}_{4}{}^{2-}[\/latex] present from <em>K<\/em><sub>a<\/sub> for the equilibrium:\r\n<p style=\"text-align: center;\">[latex]{\\text{HSO}}_{4}{}^{-}+{\\text{H}}_{2}\\text{O}\\longrightarrow {\\text{H}}_{3}{\\text{O}}^{+}+{\\text{SO}}_{4}{}^{2-}[\/latex]<\/p>\r\nLet <em>x<\/em> be the change in [latex]\\left[{\\text{SO}}_{4}{}^{2-}\\right][\/latex] :\r\n<p style=\"text-align: center;\">[latex]{K}_{\\text{a}}=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{SO}}_{4}{}^{2-}\\right]}{\\left[{\\text{HSO}}_{4}{}^{-}\\right]}=\\frac{{x}^{2}}{0.250-x}=1.2\\times {10}^{-2}[\/latex]<\/p>\r\nBecause <em>K<\/em><sub>a<\/sub> is too large to disregard <em>x<\/em> in the expression 0.250 \u2013 <em>x<\/em>, we must solve the quadratic equation:\r\n<p style=\"text-align: center;\"><em>x<\/em><sup>2<\/sup> + 1.2 [latex]\\times [\/latex] 10<sup>\u20132<\/sup><em>x<\/em> \u2013 0.250(1.2 [latex]\\times [\/latex] 10<sup>\u20132<\/sup>) = 0<\/p>\r\n<p style=\"text-align: center;\">[latex]x=\\frac{-1.2\\times {10}^{-2}\\pm \\sqrt{{\\left(1.2\\times {10}^{-2}\\right)}^{2}+4\\left(3.0\\times {10}^{\\text{-3}}\\right)}}{2}=\\frac{-1.2\\times {10}^{-2}\\pm 0.11}{2}=0.049M[\/latex]<\/p>\r\n<p style=\"text-align: center;\"><em>K<\/em><sub>sp<\/sub> = [Ba<sup>2+<\/sup>] [latex]\\left[{\\text{SO}}_{4}{}^{2-}\\right][\/latex] = [Ba<sup>2+<\/sup>](0.049) = 1.08 [latex]\\times [\/latex] 10<sup>\u201310<\/sup><\/p>\r\n<p style=\"text-align: center;\">[Ba<sup>2+<\/sup>] = 2.2 [latex]\\times [\/latex] 10<sup>\u20139<\/sup> (molar solubility)<\/p>\r\n14.\u00a0[latex]{\\text{CH}}_{3}{\\text{NH}}_{2}+{\\text{H}}_{2}\\text{O}\\rightleftharpoons {\\text{CH}}_{3}{\\text{NH}}_{3}{}^{+}+{\\text{OH}}^{-}[\/latex]\r\n<p style=\"text-align: center;\">[latex]{K}_{\\text{b}}=\\frac{\\left[{\\text{CH}}_{3}{\\text{NH}}_{3}{}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\text{CH}}_{3}{\\text{NH}}_{2}\\right]}=\\frac{\\left(x\\right)\\left(x\\right)}{0.138-x}=4.4\\times {10}^{-4}[\/latex]<\/p>\r\nSolve the quadratic equation using the quadratic formula:\r\n<p style=\"text-align: center;\"><em>x<\/em><sup>2<\/sup> + 4.4 [latex]\\times [\/latex] 10<sup>\u20134<\/sup><em>x<\/em> \u2013 0.138(4.4 [latex]\\times [\/latex] 10<sup>\u20134<\/sup>) = 0<\/p>\r\n<p style=\"text-align: center;\">[latex]x=\\frac{-4.4\\times {10}^{-4}\\pm \\sqrt{{\\left(4.4\\times {10}^{-4}\\right)}^{2}+4\\left(6.07\\times {10}^{\\text{-5}}\\right)}}{2}=\\frac{-4.4\\times {10}^{\\text{-4}}\\pm \\sqrt{2.43\\times {10}^{-4}}}{2}=\\frac{0.0152}{2}=7.6\\times {10}^{-3}M[\/latex]<\/p>\r\n<p style=\"text-align: center;\"><em>K<\/em><sub>sp<\/sub> = [Pb<sup>2+<\/sup>][OH<sup>\u2013<\/sup>]<sup>2<\/sup> = [Pb<sup>2+<\/sup>](7.6 [latex]\\times [\/latex] 10<sup>\u20133<\/sup>)<sup>2<\/sup> = 2.8 [latex]\\times [\/latex] 10<sup>\u201316<\/sup><\/p>\r\n<p style=\"text-align: center;\">[Pb<sup>2+<\/sup>] = 4.8 [latex]\\times [\/latex] 10<sup>\u201312<\/sup> (molar solubility)<\/p>\r\n16.\u00a0Two equilibria are in competition for the ions and must be considered simultaneously. Precipitation of MnS will occur when the concentration of S<sup>2\u2013<\/sup> in conjunction with 0.125 <em>M<\/em> Mn<sup>2+<\/sup> exceeds the <em>K<\/em><sub>sp<\/sub> of MnS. The [S<sup>2\u2013<\/sup>] must come from the ionization of H<sub>2<\/sub>S as defined by the equilibrium:\r\n<p style=\"text-align: center;\">[latex]{\\text{H}}_{2}\\text{S}\\left(aq\\right)+2{\\text{H}}_{2}\\text{O}\\left(l\\right)\\longrightarrow 2{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{S}}^{2-}\\left(aq\\right)[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\frac{{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}^{2}\\left[{\\text{S}}^{2-}\\right]}{\\left[{\\text{H}}_{2}\\text{S}\\right]}={K}_{1}{K}_{2}\\left({\\text{H}}_{2}\\text{S}\\right)=1.0\\times {10}^{\\text{-26}}[\/latex]<\/p>\r\nAs a saturated solution of H<sub>2<\/sub>S is 0.10 <em>M<\/em>, this later expression becomes:\r\n<p style=\"text-align: center;\">[latex]{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}^{2}\\left[{\\text{S}}^{2-}\\right]=1.0\\times {10}^{-27}[\/latex]<\/p>\r\nFrom the equilibrium of MnS, the minimum concentration of S<sup>2\u2013<\/sup> required to cause precipitation is calculated as:\r\n<p style=\"text-align: center;\">[latex]\\text{MnS}\\left(s\\right)\\longrightarrow {\\text{Mn}}^{\\text{2+}}\\left(aq\\right)+{\\text{S}}^{2-}\\left(aq\\right)[\/latex]<\/p>\r\n<p style=\"text-align: center;\"><em>K<\/em><sub>sp<\/sub> = [Mn<sup>2+<\/sup>][S<sup>2\u2013<\/sup>] = 4.3 [latex]\\times [\/latex] 10<sup>\u201322<\/sup><\/p>\r\n<p style=\"text-align: center;\">[latex]\\left[{\\text{S}}^{2-}\\right]=\\frac{4.3\\times {10}^{-22}}{0.125}=3.44\\times {10}^{-21}[\/latex]<\/p>\r\nThis amount of S<sup>2\u2013<\/sup> will exist in solution at a pH defined by the H<sub>2<\/sub>S equilibrium:\r\n<ul>\r\n \t<li>[latex]{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}^{2}\\left(3.44\\times {10}^{-21}\\right)=1.0\\times {10}^{-27}[\/latex]<\/li>\r\n \t<li>[latex]{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}^{2}=2.91\\times {10}^{-7}[\/latex]<\/li>\r\n \t<li>[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]=5.39\\times {10}^{-4}M[\/latex]<\/li>\r\n \t<li>[latex]\\text{pH}=-\\text{log}\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]=3.27[\/latex]<\/li>\r\n<\/ul>\r\n18.\u00a0Three equilibria are involved:\r\n<ul>\r\n \t<li>[latex]{\\text{H}}_{2}{\\text{CO}}_{3}+{\\text{H}}_{2}\\text{O}\\longrightarrow {\\text{H}}_{3}{\\text{O}}^{+}+{\\text{HCO}}_{3}{}^{-}{K}_{{\\text{a}}_{1}}=4.3\\times {10}^{-7}[\/latex]<\/li>\r\n \t<li>[latex]{\\text{HCO}}_{3}{}^{-}+{\\text{H}}_{2}\\text{O}\\longrightarrow {\\text{H}}_{3}{\\text{O}}^{+}+{\\text{CO}}_{3}{}^{2-}{K}_{{\\text{a}}_{2}}=7\\times {10}^{-11}[\/latex]<\/li>\r\n \t<li>[latex]{\\text{CdCO}}_{3}\\longrightarrow {\\text{Cd}}^{\\text{2+}}+{\\text{CO}}_{3}{}^{2-}{K}_{\\text{sp}}=2.5\\times {10}^{-14}[\/latex]<\/li>\r\n<\/ul>\r\nFirst, find the pH of the buffer from the Henderson-Hasselbach equation. Then find [H<sub>3<\/sub>O<sup>+<\/sup>]:\r\n<p style=\"text-align: center;\">[latex]\\text{pH}=\\text{p}{K}_{\\text{a}}+\\mathrm{log}\\frac{\\left[{\\text{CO}}_{3}{}^{2-}\\right]}{\\left[{\\text{HCO}}_{3}{}^{-}\\right]}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">10.155 + log [latex]\\frac{0.115}{0.120}[\/latex] = 10.155 \u2013 0.018 = 10.137<\/p>\r\nIn this case, several more significant figures are carried than justified so that the value of the log ratio is meaningful:\r\n<p style=\"text-align: center;\">[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]=7.3\\times {10}^{-11}[\/latex]<\/p>\r\nNow, find the concentration of [latex]{\\text{CO}}_{3}{}^{2-}[\/latex] present in the buffer solution. Next, using [latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right][\/latex] and <em>K<\/em><sub>sp<\/sub>, calculate the concentration of Cd<sup>2+<\/sup>. This latter value represents the molar solubility. From [latex]{K}_{{a}_{1}}[\/latex] determine [H<sub>2<\/sub>CO<sub>3<\/sub>]:\r\n<p style=\"text-align: center;\">[latex]{K}_{{\\text{a}}_{1}}=4.3\\times {10}^{-7}=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{HCO}}_{3}{}^{-}\\right]}{\\left[{\\text{H}}_{2}{\\text{CO}}_{3}\\right]}=\\frac{\\left(7.3\\times {10}^{-11}\\right)\\left(0.120\\right)}{\\left[{\\text{H}}_{2}{\\text{CO}}_{3}\\right]}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[H<sub>2<\/sub>CO<sub>3<\/sub>] = 2.05 [latex]\\times [\/latex] 10<sup>\u20135<\/sup><\/p>\r\nFrom [latex]{K}_{{\\text{a}}_{1}}{K}_{{\\text{a}}_{2}}[\/latex] , find the concentration of [latex]{\\text{CO}}_{3}{}^{2-}[\/latex] :\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rll}{}{K}_{{\\text{a}}_{1}}{K}_{{\\text{a}}_{2}}&amp;=&amp;\\left(4.3\\times 10 - 7\\right)\\left(7\\times 10 - 11\\right)=\\frac{\\left[{\\text{HCO}}_{3}{}^{-}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}{\\left[{\\text{H}}_{2}{\\text{CO}}_{3}\\right]}\\times \\frac{\\left[{\\text{CO}}_{3}{}^{2-}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}{\\left[{\\text{H}}_{2}{\\text{CO}}_{3}\\right]}=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{CO}}_{3}{}^{2-}\\right]}{\\left[{\\text{H}}_{2}{\\text{CO}}_{3}\\right]}=\\frac{{\\left(7.3\\times {10}^{-11}\\right)}^{2}\\times \\left[{\\text{CO}}_{3}{}^{2-}\\right]}{2.04\\times {10}^{-5}}\\\\&amp; =&amp;3\\times {10}^{-17}\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right][\/latex] = 0.115 <em>M<\/em><\/p>\r\n<p style=\"text-align: center;\"><em>K<\/em><sub>sp<\/sub> = [Cd<sup>2+<\/sup>] [latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right][\/latex] = [Cd<sup>2+<\/sup>](0.115) = 3 [latex]\\times [\/latex] 10<sup>\u201313<\/sup><\/p>\r\n<p style=\"text-align: center;\">[latex]\\left[{\\text{Cd}}^{\\text{2+}}\\right]=\\frac{3\\times {10}^{-13}}{0.115}=3\\times {10}^{-12}M[\/latex]<\/p>\r\n20.\u00a0For <em>K<\/em><sub>sp<\/sub>, (CdCO<sub>3<\/sub>) = [Cd<sup>2+<\/sup>] [latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right][\/latex] = 2.5 [latex]\\times [\/latex] 10<sup>\u201314<\/sup>, the amount of [latex]{\\text{CO}}_{3}{}^{2-}[\/latex] is governed by <em>K<\/em><sub>a<\/sub> of H<sub>2<\/sub>CO<sub>3<\/sub>, 4.3 [latex]\\times [\/latex] 10<sup>\u20137<\/sup>, and <em>K<\/em><sub>a<\/sub> of [latex]{\\text{HCO}}_{3}{}^{-}[\/latex] , 7 [latex]\\times [\/latex] 10<sup>\u201311<\/sup>, and [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] by <em>K<\/em><sub>a<\/sub> of acetic acid. First, calculate the [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] from <em>K<\/em><sub>a<\/sub> of acetic acid (HOAc):\r\n<p style=\"text-align: center;\">[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]=\\frac{{K}_{\\text{a}}\\left[\\text{HOAc}\\right]}{\\left[{\\text{OAc}}^{-}\\right]}=\\frac{1.8\\times {10}^{-5}\\left(0.250\\right)}{\\left(0.375\\right)}=1.2\\times {10}^{-5}M[\/latex]<\/p>\r\nFrom this and <em>K<\/em><sub>a<\/sub> for H<sub>2<\/sub>CO<sub>3<\/sub>, calculate [latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right][\/latex] present. As [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] is fixed by <em>K<\/em><sub>a<\/sub> of acetic acid:\r\n<p style=\"text-align: center;\">[latex]{K}_{\\text{a}}\\left({\\text{H}}_{2}{\\text{CO}}_{3}\\right)=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{HCP}}_{3}{}^{-}\\right]}{\\left[{\\text{H}}_{2}{\\text{CO}}_{3}{}^{-}\\right]}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\left[{\\text{H}}_{2}{\\text{CO}}_{3}{}^{-}\\right]=\\frac{{K}_{\\text{a}}\\left[{\\text{H}}_{2}{\\text{CO}}_{3}\\right]}{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}=\\frac{4.3\\times {10}^{-7}\\left[0.010\\right]}{\\left[1.2\\times {10}^{-5}\\right]}=3.58\\times {10}^{-4}M[\/latex]<\/p>\r\nFrom [latex]{K}_{\\text{a}}\\left({\\text{HCO}}_{3}{}^{-}\\right)[\/latex] we obtain [latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right][\/latex] :\r\n<p style=\"text-align: center;\">[latex]{K}_{a}=7\\times {10}^{-11}=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{CO}}_{3}{}^{2-}\\right]}{\\left[{\\text{HCO}}_{3}{}^{-}\\right]}=\\frac{1.2\\times {10}^{-17}\\left[{\\text{CO}}_{3}{}^{2-}\\right]}{\\left[3.58\\times {10}^{-4}\\right]}\\text{ }[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right]=\\frac{7\\times {10}^{-11}\\times 3.58\\times {10}^{-4}}{1.2\\times {10}^{-5}}=2.09\\times {10}^{-9}[\/latex]<\/p>\r\nFrom the solubility product:\r\n<p style=\"text-align: center;\"><em>K<\/em><sub>sp<\/sub> = [Cd<sup>2+<\/sup>] [latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right][\/latex] = 2.5 [latex]\\times [\/latex] 10<sup>\u201314<\/sup> = Cd<sup>2+<\/sup> (2.09 [latex]\\times [\/latex] 10<sup>\u20139<\/sup>)<\/p>\r\n<p style=\"text-align: center;\">[latex]\\left[{\\text{Cd}}^{\\text{2+}}\\right]=\\frac{2.5\\times {10}^{-14}}{2.09\\times {10}^{-9}}=1\\times {10}^{-5}M[\/latex]<\/p>\r\n22.\u00a0[latex]{\\text{CaCO}}_{3}:0.350\\cancel{\\text{g}}\\times \\frac{1\\text{mol}}{100.09\\cancel{\\text{g}}}=3.50\\times {10}^{-3}\\text{mol}[\/latex]\r\n\r\n[latex]{\\text{SrCO}}_{3}:0.350\\cancel{\\text{g}}\\times \\frac{1\\text{mol}}{147.63\\cancel{\\text{g}}}=2.37\\times {10}^{-3}\\text{mol}[\/latex]\r\n\r\n[latex]{\\text{BaCO}}_{3}:0.350\\cancel{\\text{g}}\\times \\frac{1\\text{mol}}{197.34\\cancel{\\text{g}}}=1.77\\times {10}^{-3}\\text{mol}[\/latex]\r\n<ul>\r\n \t<li>Total: 7.64 [latex]\\times [\/latex] 10<sup>\u20133<\/sup> mol<\/li>\r\n \t<li>CaCO<sub>3<\/sub>: <em>K<\/em><sub>sp<\/sub> = 4.8 [latex]\\times [\/latex] 10<sup>\u20139<\/sup><\/li>\r\n \t<li>SrCO<sub>3<\/sub>: <em>K<\/em><sub>sp<\/sub> = 9.42 [latex]\\times [\/latex] 10<sup>\u201310<\/sup><\/li>\r\n \t<li>BaCO<sub>3<\/sub>: <em>K<\/em><sub>sp<\/sub> = 8.1 [latex]\\times [\/latex] 10<sup>\u20139<\/sup><\/li>\r\n<\/ul>\r\nSolubilities are approximately equal.\r\n\r\nWhen the three solids dissolve, 7.64 [latex]\\times [\/latex] 10<sup>\u20133<\/sup> mol of metal ions and 7.64 [latex]\\times [\/latex] 10<sup>\u20133<\/sup> mol of [latex]{\\text{CO}}_{3}{}^{2-}[\/latex] are initially present. The [latex]{\\text{CO}}_{3}{}^{2-}[\/latex] reacts with CH<sub>3<\/sub>CO<sub>2<\/sub>H to form [latex]{\\text{HCO}}_{3}{}^{-}[\/latex] :\r\n<p style=\"text-align: center;\">[latex]{\\text{ CO}}_{3}{}^{2-}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\left(aq\\right)\\rightleftharpoons {\\text{HCO}}_{3}{}^{-}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}\\left(aq\\right)[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]K=\\frac{\\left[{\\text{HCO}}_{3}{}^{-}\\right]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}\\right]}{\\left[{\\text{CO}}_{3}{}^{2-}\\right]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right]}\\times \\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}=\\frac{{K}_{\\text{a}}\\left(\\text{for}{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right)}{{K}_{\\text{a}}\\left(\\text{for}{\\text{HCO}}_{3}{}^{-}\\right)}=\\frac{1.8\\times {10}^{-5}}{7\\times {10}^{-11}}=2.6\\times {10}^{5}[\/latex]<\/p>\r\nThis <em>K<\/em> value is large, so virtually all [latex]{\\text{CO}}_{3}{}^{2-}[\/latex] undergoes this reaction and approximately 7.64 [latex]\\times [\/latex] 10<sup>\u20133<\/sup> mol of [latex]{\\text{HCO}}_{3}{}^{-}[\/latex] forms. The [latex]{\\text{HCO}}_{3}{}^{-}[\/latex] reacts with CH<sub>3<\/sub>CO<sub>2<\/sub>H:\r\n<p style=\"text-align: center;\">[latex]{\\text{HCO}}_{3}{}^{-}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\left(aq\\right)\\longrightarrow {\\text{H}}_{2}{\\text{CO}}_{3}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}\\left(aq\\right)[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]K=\\frac{{K}_{\\text{a}}\\left(\\text{for}{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right)}{{K}_{\\text{a}}\\left(\\text{for}{\\text{HCO}}_{3}{}^{-}\\right)}=\\frac{1.8\\times {10}^{-5}}{4.3\\times {10}^{-7}}=42[\/latex]<\/p>\r\nThis reaction is virtually complete as well (<em>K<\/em> is large).\r\n\r\nFor each mol of [latex]{\\text{CO}}_{3}{}^{2-}[\/latex] produced, 2 mol of CH<sub>3<\/sub>CO<sub>2<\/sub>H is required for conversion to H<sub>2<\/sub>CO<sub>3<\/sub>. Thus 15.3 [latex]\\times [\/latex] 10<sup>\u20133<\/sup> mol CH<sub>3<\/sub>CO<sub>2<\/sub>H is needed:\r\n<p style=\"text-align: center;\">[latex]\\text{Volume}=\\frac{15.3\\times {10}^{-3}\\text{mol}}{1.50\\text{mol}{\\text{L}}^{-1}}=0.0102\\text{L}\\left(10.2\\text{mL}\\right)[\/latex]<\/p>\r\n24.\u00a0There are two equilibria involved: [latex]\\begin{array}{l}\\text{Mg}{\\left(\\text{OH}\\right)}_{2}\\left(s\\right)\\rightleftharpoons {\\text{Mg}}^{\\text{2+}}\\left(aq\\right)+2{\\text{OH}}^{-}\\left(aq\\right){K}_{\\text{sp}}=1.5\\times {10}^{-11}\\\\ {\\text{CN}}^{-}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons \\text{HCN}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)\\left({K}_{\\text{b}}=\\frac{{K}_{\\text{w}}}{{K}_{\\text{a}}}=\\frac{1.0\\times {10}^{-14}}{4\\times {10}^{-10}}=2.5\\times {10}^{-5}\\right)\\end{array}[\/latex] The Mg(NO<sub>3<\/sub>)<sub>2<\/sub> dissolves, and [Mg<sup>2+<\/sup>] = 0.010 <em>M<\/em>.\r\n\r\n[Mg<sup>2+<\/sup>][OH<sup>\u2013<\/sup>]<sup>2<\/sup> = <em>K<\/em><sub>sp<\/sub> = 1.5 [latex]\\times [\/latex] 10<sup>\u201311<\/sup>\r\n\r\n(0.010)[OH<sup>\u2013<\/sup>]<sup>2<\/sup> = 1.5 [latex]\\times [\/latex] 10<sup>\u201311<\/sup>\r\n\r\n[OH<sup>\u2013<\/sup>] = 3.9 [latex]\\times [\/latex] 10<sup>\u20135<\/sup><em>M<\/em>\r\n\r\nWe need to add enough CN<sup>\u2013<\/sup> to make [OH<sup>\u2013<\/sup>] = 3.9 [latex]\\times [\/latex] 10<sup>\u20135<\/sup><em>M<\/em>. Both OH<sup>\u2013<\/sup> and HCN come from CN<sup>\u2013<\/sup>, so [OH<sup>\u2013<\/sup>] = [HCN]:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\frac{\\left[\\text{HCN}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\text{CN}}^{-}\\right]}={K}_{\\text{b}}=2\\times {10}^{-5}\\\\ \\frac{{\\left(3.9\\times {10}^{-5}\\right)}^{2}}{\\left[{\\text{CN}}^{-}\\right]}=2.5\\times {10}^{-5}\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[CN<sup>\u2013<\/sup>] = 6.1 [latex]\\times [\/latex] 10<sup>\u20135<\/sup><em>M<\/em><\/p>\r\nmol NaCN = mol HCN + mol CN<sup>\u2013<\/sup> = 3.9 [latex]\\times [\/latex] 10<sup>\u20135<\/sup> + 6.1 [latex]\\times [\/latex] 10<sup>\u20135<\/sup> = 1.0 [latex]\\times [\/latex] 10<sup>\u20134<\/sup>\r\n\r\n[latex]\\text{mass}\\left(\\text{NaCN}\\right)=1.0\\times {10}^{-4}\\text{mol}\\times \\frac{49.007\\text{g}}{1\\text{mol}}=5\\times {10}^{-3}\\text{g}[\/latex]\r\n\r\n26. The answers are as follows:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li><em>K<\/em><sub>sp<\/sub> = [Mg<sup>2+<\/sup>][F<sup>\u2013<\/sup>]<sup>2<\/sup> = (1.21 [latex]\\times [\/latex] 10<sup>\u20133<\/sup>)(2 [latex]\\times [\/latex] 1.21 [latex]\\times [\/latex] 10<sup>\u20133<\/sup>)<sup>2<\/sup> = 7.09 [latex]\\times [\/latex] 10<sup>\u20139<\/sup><\/li>\r\n \t<li><em>K<\/em><sub>sp<\/sub> = [Mg<sup>2+<\/sup>][F<sup>\u2013<\/sup>]<sup>2<\/sup> = [<em>x<\/em>][0.100 + 2<em>x<\/em>]<sup>2<\/sup> = 7.09 [latex]\\times [\/latex] 10<sup>\u20139\r\n<\/sup>Assume that 2<em>x<\/em> is small when compared with 0.100 <em>M<\/em>.\r\n0.100<em>x<\/em> = 7.09 [latex]\\times [\/latex] 10<sup>\u20139\r\n<\/sup><em>x<\/em> = [MgF<sub>2<\/sub>] = 7.09 [latex]\\times [\/latex] 10<sup>\u20137<\/sup><em>M\r\n<\/em>The value 7.09 [latex]\\times [\/latex] 10<sup>\u20137<\/sup><em>M<\/em> is quite small when compared with 0.100 <em>M<\/em>, so the assumption is valid.<\/li>\r\n \t<li>Determine the concentration of Mg<sup>2+<\/sup> and F<sup>\u2013<\/sup> that will be present in the final volume. Compare the value of the ion product [Mg<sup>2+<\/sup>][F<sup>\u2013<\/sup>]<sup>2<\/sup> with <em>K<\/em><sub>sp<\/sub>. If this value is larger than <em>K<\/em><sub>sp<\/sub>, precipitation will occur.\r\n0.1000 L [latex]\\times [\/latex] 3.00 [latex]\\times [\/latex] 10<sup>\u20133<\/sup><em>M<\/em> Mg(NO<sub>3<\/sub>)<sub>2<\/sub> = 0.3000 L [latex]\\times [\/latex] <em>M<\/em> Mg(NO<sub>3<\/sub>)<sub>2\r\n<\/sub><em>M<\/em> Mg(NO<sub>3<\/sub>)<sub>2<\/sub> = 1.00 [latex]\\times [\/latex] 10<sup>\u20133<\/sup><em>M\r\n<\/em>0.2000 L [latex]\\times [\/latex] 2.00 [latex]\\times [\/latex] 10<sup>\u20133<\/sup><em>M<\/em> NaF = 0.3000 L [latex]\\times [\/latex] <em>M<\/em> NaF\r\n<em>M<\/em> NaF = 1.33 [latex]\\times [\/latex] 10<sup>\u20133<\/sup><em>M\r\n<\/em>ion product = (1.00 [latex]\\times [\/latex] 10<sup>\u20133<\/sup>)(1.33 [latex]\\times [\/latex] 10<sup>\u20133<\/sup>)<sup>2<\/sup> = 1.77 [latex]\\times [\/latex] 10<sup>\u20139\r\n<\/sup>This value is smaller than <em>K<\/em><sub>sp<\/sub>, so no precipitation will occur.<\/li>\r\n \t<li>MgF<sub>2<\/sub> is less soluble at 27 \u00b0C than at 18 \u00b0C. Because added heat acts like an added reagent, when it appears on the product side, the Le Ch\u00e2telier\u2019s principle states that the equilibrium will shift to the reactants\u2019 side to counter the stress. Consequently, less reagent will dissolve. This situation is found in our case. Therefore, the reaction is exothermic.<\/li>\r\n<\/ol>\r\n28.\u00a0Effect on amount of solid Mg(OH)<sub>2<\/sub>, [Mg<sup>2+<\/sup>], [OH<sup>\u2013<\/sup>]:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>increase, increase, decrease;<\/li>\r\n \t<li>increase, decrease, increase;<\/li>\r\n \t<li>decrease, increase, decrease;<\/li>\r\n \t<li>no effect predicted;<\/li>\r\n \t<li>increase, no effect, no effect<\/li>\r\n<\/ol>\r\n30.\u00a0[H<sub>2<\/sub>O] &gt; [Ca<sup>2+<\/sup>] &gt; [latex]\\left[{\\text{PO}}_{4}{}^{3-}\\right][\/latex] &gt; [latex]{\\text{HPO}}_{4}{}^{2-}[\/latex] = [OH<sup>\u2013<\/sup>] &gt; [latex]\\left[{\\text{HPO}}_{4}{}^{-}\\right][\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Glossary<\/h2>\r\n<b>multiple equilibrium: <\/b>system characterized by more than one state of balance between a slightly soluble ionic solid and an aqueous solution of ions working simultaneously","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>Describe examples of systems involving two (or more) simultaneous chemical equilibria<\/li>\n<li>Calculate reactant and product concentrations for multiple equilibrium systems<\/li>\n<li>Compare dissolution and weak electrolyte formation<\/li>\n<\/ul>\n<\/div>\n<p>There are times when one equilibrium reaction does not adequately describe the system being studied. Sometimes we have more than one type of equilibrium occurring at once (for example, an acid-base reaction and a precipitation reaction).<\/p>\n<p>The ocean is a unique example of a system with <b>multiple equilibria<\/b>, or multiple states of solubility equilibria working simultaneously. Carbon dioxide in the air dissolves in sea water, forming carbonic acid (H<sub>2<\/sub>CO<sub>3<\/sub>). The carbonic acid then ionizes to form hydrogen ions and bicarbonate ions [latex]\\left({\\text{HCO}}_{3}{}^{-}\\right)[\/latex], which can further ionize into more hydrogen ions and carbonate ions [latex]\\left({\\text{CO}}_{3}{}^{2-}\\right):[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rll}{}{\\text{CO}}_{2}\\left(g\\right)&\\rightleftharpoons&{\\text{CO}}_{2}\\left(aq\\right)\\\\{\\text{CO}}_{2}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}&\\rightleftharpoons&{\\text{H}}_{2}{\\text{CO}}_{3}\\left(aq\\right)\\\\{\\text{H}}_{2}{\\text{CO}}_{3}\\left(aq\\right)&\\rightleftharpoons&{\\text{H}}^{\\text{+}}\\left(aq\\right)+{\\text{HCO}}_{3}{}^{-}\\left(aq\\right)\\\\{\\text{HCO}}_{3}{}^{-}\\left(aq\\right)&\\rightleftharpoons&{\\text{H}}^{\\text{+}}\\left(aq\\right)+{\\text{CO}}_{3}{}^{2-}\\left(aq\\right)\\end{array}[\/latex]<\/p>\n<p>The excess H<sup>+<\/sup> ions make seawater more acidic. Increased ocean acidification can then have negative impacts on reef-building coral, as they cannot absorb the calcium carbonate they need to grow and maintain their skeletons (Figure\u00a01). This in turn disrupts the local biosystem that depends upon the health of the reefs for its survival. If enough local reefs are similarly affected, the disruptions to sea life can be felt globally. The world\u2019s oceans are presently in the midst of a period of intense acidification, believed to have begun in the mid-nineteenth century, and which is now accelerating at a rate faster than any change to oceanic pH in the last 20 million years.<\/p>\n<div id=\"attachment_5431\" style=\"width: 985px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5431\" class=\"size-full wp-image-5431\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09042406\/CNX_Chem_15_03_CoralReef.jpg\" alt=\"This figure contains two photographs of coral reefs. In a, a colorful reef that includes hues of purple and pink corals is shown in blue green water with fish swimming in the background. In b, grey-green mossy looking coral is shown in a blue aquatic environment. This photo does not have the colorful appearance or fish that were shown in figure a.\" width=\"975\" height=\"386\" \/><\/p>\n<p id=\"caption-attachment-5431\" class=\"wp-caption-text\">Figure\u00a01. Healthy coral reefs (a) support a dense and diverse array of sea life across the ocean food chain. But when coral are unable to adequately build and maintain their calcium carbonite skeletons because of excess ocean acidification, the unhealthy reef (b) is only capable of hosting a small fraction of the species as before, and the local food chain starts to collapse. (credit a: modification of work by NOAA Photo Library; credit b: modification of work by \u201cprilfish\u201d\/Flickr)<\/p>\n<\/div>\n<div class=\"textbox\">\n<p>Learn more about ocean acidification and how it affects other marine creatures.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ocean Acidification\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/kxPwbhFeZSw?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox\">This site has detailed information about <a href=\"http:\/\/www.teachoceanscience.net\/teaching_resources\/education_modules\/coral_reefs_and_climate_change\/how_does_climate_change_affect_coral_reefs\/\" target=\"_blank\">how ocean acidification specifically affects coral reefs<\/a>.<\/div>\n<p>Slightly soluble solids derived from weak acids generally dissolve in strong acids, unless their solubility products are extremely small. For example, we can dissolve CuCO<sub>3<\/sub>, FeS, and Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub> in HCl because their basic anions react to form weak acids (H<sub>2<\/sub>CO<sub>3<\/sub>, H<sub>2<\/sub>S, and [latex]{\\text{H}}_{2}{\\text{PO}}_{4}^{-}[\/latex] ). The resulting decrease in the concentration of the anion results in a shift of the equilibrium concentrations to the right in accordance with Le Ch\u00e2telier\u2019s principle.<\/p>\n<p>Of particular relevance to us is the dissolution of hydroxylapatite, Ca<sub>5<\/sub>(PO<sub>4<\/sub>)<sub>3<\/sub>OH, in acid. Apatites are a class of calcium phosphate minerals (Figure\u00a02); a biological form of hydroxylapatite is found as the principal mineral in the enamel of our teeth. A mixture of hydroxylapatite and water (or saliva) contains an equilibrium mixture of solid Ca<sub>5<\/sub>(PO<sub>4<\/sub>)<sub>3<\/sub>OH and dissolved Ca<sup>2+<\/sup>, [latex]{\\text{PO}}_{4}^{3-}[\/latex], and OH<sup>\u2013<\/sup> ions:<\/p>\n<p style=\"text-align: center;\">[latex]{\\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)[\/latex]<\/p>\n<div id=\"attachment_5434\" style=\"width: 660px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5434\" class=\"wp-image-5434 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09042527\/CNX_Chem_15_03_Apatite.jpg\" alt=\"This figure includes an image of two large light blue apatite crystals in a mineral conglomerate that includes white, grey, and tan crystals. The blue apatite crystals have a dull, dusty, or powdered appearance.\" width=\"650\" height=\"383\" \/><\/p>\n<p id=\"caption-attachment-5434\" class=\"wp-caption-text\">Figure\u00a02. Crystal of the mineral hydroxylapatite, Ca<sub>5<\/sub>(PO<sub>4<\/sub>)<sub>3<\/sub>OH, is shown here. Pure apatite is white, but like many other minerals, this sample is colored because of the presence of impurities.<\/p>\n<\/div>\n<p>When exposed to acid, phosphate ions react with hydronium ions to form hydrogen phosphate ions and ultimately, phosphoric acid:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{PO}}_{4}{}^{3-}\\left(aq\\right)+{\\text{H}}_{3}{\\text{O}}^{+}\\longrightarrow {\\text{H}}_{2}{\\text{PO}}_{4}{}^{2-}+{\\text{H}}_{2}\\text{O}[\/latex]<br \/>\n[latex]{\\text{PO}}_{4}{}^{2-}\\left(aq\\right)+{\\text{H}}_{3}{\\text{O}}^{+}\\longrightarrow {\\text{H}}_{2}{\\text{PO}}_{4}{}^{-}+{\\text{H}}_{2}\\text{O}[\/latex]<br \/>\n[latex]{\\text{H}}_{2}{\\text{PO}}_{4}{}^{-}+{\\text{H}}_{3}{\\text{O}}^{+}\\longrightarrow {\\text{H}}_{3}{\\text{PO}}_{4}+{\\text{H}}_{2}\\text{O}[\/latex]<\/p>\n<p>Hydroxide ion reacts to form water:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{OH}}^{-}\\left(aq\\right)+{\\text{H}}_{3}{\\text{O}}^{+}\\longrightarrow 2{\\text{H}}_{2}\\text{O}[\/latex]<\/p>\n<p>These reactions decrease the phosphate and hydroxide ion concentrations, and additional hydroxylapatite dissolves in an acidic solution in accord with Le Ch\u00e2telier\u2019s principle. Our teeth develop cavities when acid waste produced by bacteria growing on them causes the hydroxylapatite of the enamel to dissolve. Fluoride toothpastes contain sodium fluoride, NaF, or stannous fluoride [more properly named tin(II) fluoride], SnF<sub>2<\/sub>. They function by replacing the OH<sup>\u2013<\/sup> ion in hydroxylapatite with F<sup>\u2013<\/sup> ion, producing fluorapatite, Ca<sub>5<\/sub>(PO<sub>4<\/sub>)<sub>3<\/sub>F:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{NaF}+{\\text{Ca}}_{5}{\\left({\\text{PO}}_{4}\\right)}_{3}\\text{OH}\\rightleftharpoons {\\text{Ca}}_{5}{\\left({\\text{PO}}_{4}\\right)}_{3}\\text{F}+{\\text{Na}}^{+}+{\\text{OH}}^{-}[\/latex]<\/p>\n<p>The resulting Ca<sub>5<\/sub>(PO<sub>4<\/sub>)<sub>3<\/sub>F is slightly less soluble than Ca<sub>5<\/sub>(PO<sub>4<\/sub>)<sub>3<\/sub>OH, and F<sup>\u2013<\/sup> is a weaker base than OH<sup>\u2013<\/sup>. Both of these factors make the fluorapatite more resistant to attack by acids than hydroxylapatite. See the Chemistry in Everyday Life feature on the role of fluoride in preventing tooth decay for more information.<\/p>\n<div class=\"textbox shaded\">\n<h3 data-type=\"title\">Role of Fluoride in Preventing Tooth Decay<\/h3>\n<p>As we saw previously, fluoride ions help protect our teeth by reacting with hydroxylapatite to form fluorapatite, Ca<sub>5<\/sub>(PO<sub>4<\/sub>)<sub>3<\/sub>F. Since it lacks a hydroxide ion, fluorapatite is more resistant to attacks by acids in our mouths and is thus less soluble, protecting our teeth. Scientists discovered that naturally fluorinated water could be beneficial to your teeth, and so it became common practice to add fluoride to drinking water. Toothpastes and mouthwashes also contain amounts of fluoride (Figure\u00a03).<\/p>\n<div id=\"attachment_5435\" style=\"width: 660px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5435\" class=\"wp-image-5435 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09042617\/CNX_Chem_15_03_Toothpaste.jpg\" alt=\"A tube of toothpaste\" width=\"650\" height=\"269\" \/><\/p>\n<p id=\"caption-attachment-5435\" class=\"wp-caption-text\">Figure 3. Fluoride, found in many toothpastes, helps prevent tooth decay (credit: Kerry Ceszyk).<\/p>\n<\/div>\n<p>Unfortunately, excess fluoride can negate its advantages. Natural sources of drinking water in various parts of the world have varying concentrations of fluoride, and places where that concentration is high are prone to certain health risks when there is no other source of drinking water. The most serious side effect of excess fluoride is the bone disease, skeletal fluorosis. When excess fluoride is in the body, it can cause the joints to stiffen and the bones to thicken. It can severely impact mobility and can negatively affect the thyroid gland. Skeletal fluorosis is a condition that over 2.7 million people suffer from across the world. So while fluoride can protect our teeth from decay, the US Environmental Protection Agency sets a maximum level of 4 ppm\u00a0(4 mg\/L)\u00a0of fluoride in drinking water in the US. Fluoride levels in water are not regulated in all countries, so fluorosis is a problem in areas with high levels of fluoride in the groundwater.<\/p>\n<\/div>\n<p>When acid rain attacks limestone or marble, which are calcium carbonates, a reaction occurs that is similar to the acid attack on hydroxylapatite. The hydronium ion from the acid rain combines with the carbonate ion from calcium carbonates and forms the hydrogen carbonate ion, a weak acid:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{CO}}_{3}{}^{2-}\\left(aq\\right)\\longrightarrow {\\text{HCO}}_{3}{}^{-}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/p>\n<p>Calcium hydrogen carbonate, Ca(HCO<sub>3<\/sub>)<sub>2<\/sub>, is soluble, so limestone and marble objects slowly dissolve in acid rain.<\/p>\n<p>If we add calcium carbonate to a concentrated acid, hydronium ion reacts with the carbonate ion according to the equation:<\/p>\n<p style=\"text-align: center;\">[latex]2{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{CO}}_{3}{}^{2-}\\left(aq\\right)\\longrightarrow {\\text{H}}_{2}{\\text{CO}}_{3}\\left(aq\\right)+2{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/p>\n<p>(Acid rain is usually not sufficiently acidic to cause this reaction; however, laboratory acids are.) The solution may become saturated with the weak electrolyte carbonic acid, which is unstable, and carbon dioxide gas can be evolved:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{H}}_{2}{\\text{CO}}_{3}\\left(aq\\right)\\longrightarrow {\\text{CO}}_{2}\\left(g\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/p>\n<p>These reactions decrease the carbonate ion concentration, and additional calcium carbonate dissolves. If enough acid is present, the concentration of carbonate ion is reduced to such a low level that the reaction quotient for the dissolution of calcium carbonate remains less than the solubility product of calcium carbonate, even after all of the calcium carbonate has dissolved.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 1:\u00a0Prevention of Precipitation of Mg(OH)<sub>2<\/sub><\/h3>\n<p>Calculate the concentration of ammonium ion that is required to prevent the precipitation of Mg(OH)<sub>2<\/sub> in a solution with [Mg<sup>2+<\/sup>] = 0.10 <em>M<\/em> and [NH<sub>3<\/sub>] = 0.10 <em>M<\/em>.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q496110\">Show Answer<\/span><\/p>\n<div id=\"q496110\" class=\"hidden-answer\" style=\"display: none\">\n<p>Two equilibria are involved in this system:<\/p>\n<ul>\n<li>Reaction (1): [latex]\\text{Mg}{\\left(\\text{OH}\\right)}_{2}\\left(s\\right)\\rightleftharpoons {\\text{Mg}}^{\\text{2+}}\\left(aq\\right)+2{\\text{OH}}^{-}\\left(aq\\right);{K}_{\\text{sp}}=1.5\\times {10}^{-11}[\/latex]<\/li>\n<li>Reaction (2): [latex]{\\text{NH}}_{3}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{NH}}_{4}{}^{\\text{+}}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right);{K}_{\\text{sp}}=1.8\\times {10}^{-5}[\/latex]<\/li>\n<\/ul>\n<p>To prevent the formation of solid Mg(OH)<sub>2<\/sub>, we must adjust the concentration of OH<sup>\u2013<\/sup> so that the reaction quotient for Equation (1), <em>Q<\/em> = [Mg<sup>2+<\/sup>][OH<sup>\u2013<\/sup>]<sup>2<\/sup>, is less than <em>K<\/em><sub>sp<\/sub> for Mg(OH)<sub>2<\/sub>. (To simplify the calculation, we determine the concentration of OH<sup>\u2013<\/sup> when <em>Q<\/em> = <em>K<\/em><sub>sp<\/sub>.) [OH<sup>\u2013<\/sup>] can be reduced by the addition of [latex]{\\text{NH}}_{4}^{+}[\/latex], which shifts Reaction (2) to the left and reduces [OH<sup>\u2013<\/sup>].<\/p>\n<ol>\n<li><em>We determine the [OH<sup>\u2013<\/sup>] at which Q = K<sub>sp<\/sub> when [Mg<sup>2+<\/sup>] = 0.10<\/em> M<em>:<\/em><br \/>\n[latex]Q=\\left[{\\text{Mg}}^{\\text{2+}}\\right]{\\left[{\\text{OH}}^{-}\\right]}^{2}=\\left(0.10\\right){\\left[{\\text{OH}}^{-}\\right]}^{2}=1.5\\times {10}^{-11}[\/latex][latex]\\left[{\\text{OH}}^{-}\\right]=1.2\\times {10}^{-5}M[\/latex]Solid Mg(OH)<sub>2<\/sub> will not form in this solution when [OH<sup>\u2013<\/sup>] is less than 1.2 [latex]\\times[\/latex] 10<sup>\u20135<\/sup><em>M<\/em>.<\/li>\n<li><em>We calculate the<\/em> [latex]\\left[N{H}_{4}{}^{+}\\right][\/latex] <em>needed to decrease [OH<sup>\u2013<\/sup>] to 1.2 \u00d7 10<sup>\u20135<\/sup><\/em> M <em>when [NH<sub>3<\/sub>] = 0.10.<\/em>[latex]{K}_{\\text{b}}=\\frac{\\left[{\\text{NH}}_{4}{}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\text{NH}}_{3}\\right]}=\\frac{\\left[{\\text{NH}}_{4}{}^{+}\\right]\\left(1.2\\times {10}^{-5}\\right)}{0.10}=1.8\\times {10}^{-5}[\/latex][latex]\\left[{\\text{NH}}_{4}{}^{+}\\right]=0.15M[\/latex]<\/li>\n<\/ol>\n<p>When [latex]\\left[{\\text{NH}}_{4}{}^{+}\\right][\/latex] equals 0.15 <em>M<\/em>, [OH<sup>\u2013<\/sup>] will be 1.2 [latex]\\times[\/latex] 10<sup>\u20135<\/sup><em>M<\/em>. Any [latex]\\left[{\\text{NH}}_{4}{}^{+}\\right][\/latex] greater than 0.15 <em>M<\/em> will reduce [OH<sup>\u2013<\/sup>] below 1.2 [latex]\\times[\/latex] 10<sup>\u20135<\/sup><em>M<\/em> and prevent the formation of Mg(OH)<sub>2<\/sub>.<\/p>\n<\/div>\n<\/div>\n<h4 id=\"fs-idm329760\">Check Your Learning<\/h4>\n<p>Consider the two equilibria:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{ZnS}\\left(s\\right)\\rightleftharpoons {\\text{Zn}}^{\\text{2+}}\\left(aq\\right)+{\\text{S}}^{2-}\\left(aq\\right);{K}_{\\text{sp}}=1\\times {10}^{-27}[\/latex]<br \/>\n[latex]2{\\text{H}}_{2}\\text{O}\\left(l\\right)+{\\text{H}}_{2}\\text{S}\\left(aq\\right)\\rightleftharpoons 2{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{S}}^{2-}\\left(aq\\right);K=1.0\\times {10}^{\\text{-26}}[\/latex]<\/p>\n<p>and calculate the concentration of hydronium ion required to prevent the precipitation of ZnS in a solution that is 0.050 <em>M<\/em> in Zn<sup>2+<\/sup> and saturated with H<sub>2<\/sub>S (0.10 <em>M<\/em> H<sub>2<\/sub>S).<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q389689\">Show Answer<\/span><\/p>\n<div id=\"q389689\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\text{>}0.2M[\/latex]<\/p>\n<p>([S<sup>2\u2013<\/sup>] is less than 2 [latex]\\times[\/latex] 10<sup>\u201326<\/sup><em>M<\/em> and precipitation of ZnS does not occur.)<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Therefore, precise calculations of the solubility of solids from the solubility product are limited to cases in which the only significant reaction occurring when the solid dissolves is the formation of its ions.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 2:\u00a0Multiple Equilibria<\/h3>\n<p>Unexposed silver halides are removed from photographic film when they react with sodium thiosulfate (Na<sub>2<\/sub>S<sub>2<\/sub>O<sub>3<\/sub>, called hypo) to form the complex ion [latex]\\text{Ag}{\\left({\\text{S}}_{2}{\\text{O}}_{3}\\right)}_{2}{}^{3-}[\/latex] (<em>K<\/em><sub>f<\/sub> = 4.7 [latex]\\times[\/latex] 10<sup>13<\/sup>). The reaction with silver bromide is:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-5437\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09042721\/CNX_Chem_15_03_AgBr_img.jpg\" alt=\"A chemical reaction is shown using structural formulas. On the left, A g superscript plus is followed by a plus sign, the number 2, and a structure in brackets. The structure is composed of a central S atom which has O atoms single bonded above, right, and below. A second S atom is single bonded to the left. Each of these bonded atoms has 6 dots around it. Outside the brackets is a superscript 2 negative. Following a bidirectional arrow is a structure in brackets with a central A g atom. To the left and right, S atoms are single bonded to the A g atom. Each of these S atoms has four dots around it, and an S atom connected with a single bond moving out from the central A g atom, forming the ends of the structure. Each of these atoms has three O atoms attached with single bonds above, below, and at the end of the structure. Each O atom has six dots around it. Outside the brackets is a superscript 3 negative.\" width=\"1000\" height=\"176\" \/><\/p>\n<p>&nbsp;<\/p>\n<p>What mass of Na<sub>2<\/sub>S<sub>2<\/sub>O<sub>3<\/sub> is required to prepare 1.00 L of a solution that will dissolve 1.00 g of AgBr by the formation of [latex]\\text{Ag}{\\left({\\text{S}}_{2}{\\text{O}}_{3}\\right)}_{2}{}^{3-}?[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q693371\">Show Answer<\/span><\/p>\n<div id=\"q693371\" class=\"hidden-answer\" style=\"display: none\">\n<p>Two equilibria are involved when AgBr dissolves in a solution containing the [latex]{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}[\/latex] ion:<\/p>\n<ul>\n<li>Reaction (1): [latex]\\text{AgBr}\\left(s\\right)\\rightleftharpoons {\\text{Ag}}^{\\text{+}}\\left(aq\\right)+{\\text{Br}}^{-}\\left(aq\\right);{K}_{\\text{sp}}=3.3\\times {10}^{\\text{-13}}[\/latex]<\/li>\n<li>Reaction (2): [latex]{\\text{Ag}}^{\\text{+}}\\left(aq\\right)+{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}\\left(aq\\right)\\rightleftharpoons \\text{Ag}{\\left({\\text{S}}_{2}{\\text{O}}_{3}\\right)}_{2}{}^{3-}\\left(aq\\right);{K}_{\\text{f}}=4.7\\times {10}^{13}[\/latex]<\/li>\n<\/ul>\n<p>In order for 1.00 g of AgBr to dissolve, the [Ag<sup>+<\/sup>] in the solution that results must be low enough for <em>Q<\/em> for Reaction (1) to be smaller than <em>K<\/em><sub>sp<\/sub> for this reaction. We reduce [Ag<sup>+<\/sup>] by adding [latex]{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}[\/latex] and thus cause Reaction (2) to shift to the right. We need the following steps to determine what mass of Na<sub>2<\/sub>S<sub>2<\/sub>O<sub>3<\/sub> is needed to provide the necessary [latex]{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}[\/latex] .<\/p>\n<ol>\n<li><em>We calculate the [Br<sup>\u2013<\/sup>] produced by the complete dissolution of<\/em> <em>1.00 g of AgBr (5.33<\/em> [latex]\\times[\/latex] <em>10<sup>\u20133<\/sup> mol AgBr) in 1.00 L of solution:<br \/>\n<\/em>[latex]\\left[{\\text{Br}}^{-}\\right]=5.33\\times {10}^{\\text{-3}}M[\/latex]<\/li>\n<li><em>We use [Br<sup>\u2013<\/sup>] and K<sub>sp<\/sub> to determine the maximum possible concentration of Ag<sup>+<\/sup> that can be present without causing reprecipitation of AgBr:<\/em>[latex]\\left[{\\text{Ag}}^{+}\\right]=6.2\\times {10}^{-11}M[\/latex]<\/li>\n<li><em>We determine the<\/em> [latex]\\left[{S}_{2}{O}_{3}{}^{2-}\\right][\/latex] <em>required to make [Ag<sup>+<\/sup>] = 6.2<\/em> [latex]\\times[\/latex] <em>10<sup>\u201311<\/sup> M<\/em> <em>after the remaining Ag<sup>+<\/sup> ion has reacted with<\/em> [latex]{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}[\/latex] <em>according to the equation:<\/em>[latex]{\\text{Ag}}^{+}+2{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}\\rightleftharpoons \\text{Ag}{\\left({\\text{S}}_{2}{\\text{O}}_{3}\\right)}_{2}{}^{\\text{3-}}{K}_{\\text{f}}=4.7\\times {10}^{13}[\/latex]Because 5.33 [latex]\\times[\/latex] 10<sup>\u20133<\/sup> mol of AgBr dissolves:\n<p>[latex]\\left(5.33\\times {10}^{-3}\\right)-\\left(6.2\\times {10}^{\\text{-11}}\\right)=5.33\\times {10}^{-3}\\text{mol}\\text{Ag}{\\left({\\text{S}}_{2}{\\text{O}}_{3}\\right)}_{2}{}^{3-}[\/latex]<\/p>\n<p>Thus, at equilibrium: [latex]\\left[\\text{Ag}{\\left({\\text{S}}_{2}{\\text{O}}_{3}\\right)}_{2}{}^{3-}\\right][\/latex] = 5.33 [latex]\\times[\/latex] 10<sup>\u20133<\/sup><em>M<\/em>, [Ag<sup>+<\/sup>] = 6.2 [latex]\\times[\/latex] 10<sup>\u201311<\/sup><em>M<\/em>, and <em>Q<\/em> = <em>K<\/em><sub>f<\/sub> = 4.7 [latex]\\times[\/latex] 10<sup>13<\/sup>:<\/p>\n<p>[latex]{K}_{\\text{f}}=\\frac{\\left[\\text{Ag}{\\left({\\text{S}}_{2}{\\text{O}}_{3}\\right)}_{2}{}^{\\text{3-}}\\right]}{\\left[{\\text{Ag}}^{+}\\right]{\\left[{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}\\right]}^{2}}=4.7\\times {10}^{13}[\/latex]<\/p>\n<p>[latex]\\left[{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}\\right]=1.4\\times {10}^{-3}M[\/latex]<\/p>\n<p>When [latex]\\left[{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}\\right][\/latex] is 1.4 [latex]\\times[\/latex] 10<sup>\u20133<\/sup><em>M<\/em>, [Ag<sup>+<\/sup>] is 6.2 [latex]\\times[\/latex] 10<sup>\u201311<\/sup><em>M<\/em> and all AgBr remains dissolved.<\/li>\n<li><em>We determine the total number of moles of<\/em> [latex]{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}[\/latex] <em>that must be added to the solution.<\/em> This equals the amount that reacts with Ag<sup>+<\/sup> to form [latex]\\text{Ag}{\\left({\\text{S}}_{2}{\\text{O}}_{3}\\right)}_{2}{}^{3-}[\/latex] plus the amount of free [latex]{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}[\/latex] in solution at equilibrium. To form 5.33 [latex]\\times[\/latex] 10<sup>\u20133<\/sup> mol of [latex]\\text{Ag}{\\left({\\text{S}}_{2}{\\text{O}}_{3}\\right)}_{2}{}^{3-}[\/latex] requires 2 [latex]\\times[\/latex] (5.33 [latex]\\times[\/latex] 10<sup>\u20133<\/sup>) mol of [latex]{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}[\/latex]. In addition, 1.4 [latex]\\times[\/latex] 10<sup>\u20133<\/sup> mol of unreacted [latex]{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}[\/latex] is present (Step 3). Thus, the total amount of [latex]{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}[\/latex] that must be added is:[latex]2\\times \\left(5.33\\times {10}^{-3}\\text{mol}{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}\\right)+1.4\\times {10}^{-3}\\text{mol}{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}=1.21\\times {10}^{-2}\\text{mol}{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}[\/latex]<\/li>\n<li><em>We determine the mass of Na<sub>2<\/sub>S<sub>2<\/sub>O<sub>3<\/sub> required to give 1.21 [latex]\\times[\/latex] 10<sup>\u20132<\/sup> mol<\/em> [latex]{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}[\/latex] <em>using the<\/em> <em>molar mass of Na<sub>2<\/sub>S<sub>2<\/sub>O<sub>3<\/sub>:<\/em>[latex]1.21\\times {10}^{\\text{-2}}\\text{mol}{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}\\times \\frac{158.1\\text{g}{\\text{Na}}_{2}{\\text{S}}_{2}{\\text{O}}_{3}}{1\\text{mol}{\\text{Na}}_{2}{\\text{S}}_{2}{\\text{O}}_{3}}=1.9\\text{g}{\\text{Na}}_{2}{\\text{S}}_{2}{\\text{O}}_{3}[\/latex]Thus, 1.00 L of a solution prepared from 1.9 g Na<sub>2<\/sub>S<sub>2<\/sub>O<sub>3<\/sub> dissolves 1.0 g of AgBr.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<h4>Check Your Learning<\/h4>\n<p>AgCl(s), silver chloride, is well known to have a very low solubility: Ag(<em>s<\/em>) \u21cc Ag<sup>+<\/sup>(<em>aq<\/em>) + Cl<sup>\u2212<\/sup>(<em>aq<\/em>), <em>K<sub>sp<\/sub><\/em> = 1.77 \u00d7 10<sup>\u201310<\/sup>. Adding ammonia significantly increases the solubility of AgCl because a complex ion is formed: Ag<sup>+<\/sup>(<em>aq<\/em>) + 2NH<sub>3<\/sub>(<em>aq<\/em>) \u21cc Ag(NH<sub>3<\/sub>)<sub>2<\/sub><sup>+<\/sup>(<em>aq<\/em>), <em>K<sub>f<\/sub><\/em> = 1.7 \u00d7 10<sup>7<\/sup>. What mass of NH<sub>3<\/sub> is required to prepare 1.00 L of solution that will dissolve 2.00 g of AgCl by formation of Ag(NH<sub>3<\/sub>)<sub>2<\/sub><sup>+<\/sup>?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q88870\">Show Answer<\/span><\/p>\n<div id=\"q88870\" class=\"hidden-answer\" style=\"display: none\">1.00 L of a solution prepared with 4.84 g NH<sub>3<\/sub> dissolves 2.0 g of AgCl.<\/div>\n<\/div>\n<\/div>\n<h2>Dissolution versus Weak Electrolyte Formation<\/h2>\n<p>We can determine how to shift the concentration of ions in the equilibrium between a slightly soluble solid and a solution of its ions by applying Le Ch\u00e2telier\u2019s principle. For example, one way to control the concentration of manganese(II) ion, Mn<sup>2+<\/sup>, in a solution is to adjust the pH of the solution and, consequently, to manipulate the equilibrium between the slightly soluble solid manganese(II) hydroxide, manganese(II) ion, and hydroxide ion:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{Mn}{\\left(\\text{OH}\\right)}_{2}\\left(s\\right)\\rightleftharpoons {\\text{Mn}}^{\\text{2+}}\\left(aq\\right)+2{\\text{OH}}^{-}\\left(aq\\right){K}_{\\text{sp}}=\\left[{\\text{Mn}}^{\\text{2+}}\\right]{\\left[{\\text{OH}}^{-}\\right]}^{2}[\/latex]<\/p>\n<p>This could be important to a laundry because clothing washed in water that has a manganese concentration exceeding 0.1 mg per liter may be stained by the manganese. We can reduce the concentration of manganese by increasing the concentration of hydroxide ion. We could add, for example, a small amount of NaOH or some other base such as the silicates found in many laundry detergents. As the concentration of OH<sup>\u2013<\/sup> ion increases, the equilibrium responds by shifting to the left and reducing the concentration of Mn<sup>2+<\/sup> ion while increasing the amount of solid Mn(OH)<sub>2<\/sub> in the equilibrium mixture, as predicted by Le Ch\u00e2telier\u2019s principle.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 3:\u00a0Solubility Equilibrium of a Slightly Soluble Solid<\/h3>\n<p>What is the effect on the amount of solid Mg(OH)<sub>2<\/sub> that dissolves and the concentrations of Mg<sup>2+<\/sup> and OH<sup>\u2013<\/sup> when each of the following are added to a mixture of solid Mg(OH)<sub>2<\/sub> in water at equilibrium?<\/p>\n<ol>\n<li>MgCl<sub>2<\/sub><\/li>\n<li>KOH<\/li>\n<li>an acid<\/li>\n<li>NaNO<sub>3<\/sub><\/li>\n<li>Mg(OH)<sub>2<\/sub><\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q661709\">Show Answer<\/span><\/p>\n<div id=\"q661709\" class=\"hidden-answer\" style=\"display: none\">\n<p>The equilibrium among solid Mg(OH)<sub>2<\/sub> and a solution of Mg<sup>2+<\/sup> and OH<sup>\u2013<\/sup> is:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{Mg}{\\left(\\text{OH}\\right)}_{2}\\left(s\\right)\\rightleftharpoons {\\text{Mg}}^{\\text{2+}}\\left(aq\\right)+2{\\text{OH}}^{-}\\left(aq\\right)[\/latex].<\/p>\n<ol>\n<li>The reaction shifts to the left to relieve the stress produced by the additional Mg<sup>2+<\/sup> ion, in accordance with Le Ch\u00e2telier\u2019s principle. In quantitative terms, the added Mg<sup>2+<\/sup> causes the reaction quotient to be larger than the solubility product (<em>Q<\/em> &gt; <em>K<\/em><sub>sp<\/sub>), and Mg(OH)<sub>2<\/sub> forms until the reaction quotient again equals <em>K<\/em><sub>sp<\/sub>. At the new equilibrium, [OH<sup>\u2013<\/sup>] is less and [Mg<sup>2+<\/sup>] is greater than in the solution of Mg(OH)<sub>2<\/sub> in pure water. More solid Mg(OH)<sub>2<\/sub> is present.<\/li>\n<li>The reaction shifts to the left to relieve the stress of the additional OH<sup>\u2013<\/sup> ion. Mg(OH)<sub>2<\/sub> forms until the reaction quotient again equals <em>K<\/em><sub>sp<\/sub>. At the new equilibrium, [OH<sup>\u2013<\/sup>] is greater and [Mg<sup>2+<\/sup>] is less than in the solution of Mg(OH)<sub>2<\/sub> in pure water. More solid Mg(OH)<sub>2<\/sub> is present.<\/li>\n<li>The concentration of OH<sup>\u2013<\/sup> is reduced as the OH<sup>\u2013<\/sup> reacts with the acid. The reaction shifts to the right to relieve the stress of less OH<sup>\u2013<\/sup> ion. In quantitative terms, the decrease in the OH<sup>\u2013<\/sup> concentration causes the reaction quotient to be smaller than the solubility product (<em>Q<\/em> &lt; <em>K<\/em><sub>sp<\/sub>), and additional Mg(OH)<sub>2<\/sub> dissolves until the reaction quotient again equals <em>K<\/em><sub>sp<\/sub>. At the new equilibrium, [OH<sup>\u2013<\/sup>] is less and [Mg<sup>2+<\/sup>] is greater than in the solution of Mg(OH)<sub>2<\/sub> in pure water. More Mg(OH)<sub>2<\/sub> is dissolved.<\/li>\n<li>NaNO<sub>3<\/sub> contains none of the species involved in the equilibrium, so we should expect that it has no appreciable effect on the concentrations of Mg<sup>2+<\/sup> and OH<sup>\u2013<\/sup>. (As we have seen previously, dissolved salts change the activities of the ions of an electrolyte. However, the salt effect is generally small, and we shall neglect the slight errors that may result from it.)<\/li>\n<li>The addition of solid Mg(OH)<sub>2<\/sub> has no effect on the solubility of Mg(OH)<sub>2<\/sub> or on the concentration of Mg<sup>2+<\/sup> and OH<sup>\u2013<\/sup>. The concentration of Mg(OH)<sub>2<\/sub> does not appear in the equation for the reaction quotient:<br \/>\n[latex]Q=\\left[{\\text{Mg}}^{\\text{2+}}\\right]{\\left[{\\text{OH}}^{-}\\right]}^{2}[\/latex]<br \/>\nThus, changing the amount of solid magnesium hydroxide in the mixture has no effect on the value of <em>Q<\/em>, and no shift is required to restore <em>Q<\/em> to the value of the equilibrium constant.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<h4 id=\"fs-idm701008\">Check Your Learning<\/h4>\n<p>What is the effect on the amount of solid NiCO<sub>3<\/sub> that dissolves and the concentrations of Ni<sup>2+<\/sup> and [latex]{\\text{CO}}_{3}{}^{2-}[\/latex] when each of the following are added to a mixture of the slightly soluble solid NiCO<sub>3<\/sub> and water at equilibrium?<\/p>\n<ol>\n<li>Ni(NO<sub>3<\/sub>)<sub>2<\/sub><\/li>\n<li>KClO<sub>4<\/sub><\/li>\n<li>NiCO<sub>3<\/sub><\/li>\n<li>K<sub>2<\/sub>CO<sub>3<\/sub><\/li>\n<li>HNO<sub>3<\/sub> (reacts with carbonate giving [latex]{\\text{HCO}}_{3}{}^{-}[\/latex] or H<sub>2<\/sub>O and CO<sub>2<\/sub>)<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q267299\">Show Answer<\/span><\/p>\n<div id=\"q267299\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>mass of NiCO<sub>3<\/sub>(s) increases, [Ni<sup>2+<\/sup>] increases, [latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right][\/latex] decreases<\/li>\n<li>no appreciable effect<\/li>\n<li>no effect except to increase the amount of solid NiCO<sub>3<\/sub><\/li>\n<li>mass of NiCO<sub>3<\/sub>(s) increases, [Ni<sup>2+<\/sup>] decreases, [latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right][\/latex] increases<\/li>\n<li>mass of NiCO<sub>3<\/sub>(s) decreases, [Ni<sup>2+<\/sup>] increases, [latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right][\/latex] decreases<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Key Concepts and Summary<\/h3>\n<p>Several systems we encounter consist of multiple equilibria, systems where two or more equilibria processes are occurring simultaneously. Some common examples include acid rain, fluoridation, and dissolution of carbon dioxide in sea water. When looking at these systems, we need to consider each equilibrium separately and then combine the individual equilibrium constants into one solubility product or reaction quotient expression using the tools from the first equilibrium chapter. Le Ch\u00e2telier\u2019s principle also must be considered, as each reaction in a multiple equilibria system will shift toward reactants or products based on what is added to the initial reaction and how it affects each subsequent equilibrium reaction.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Exercises<\/h3>\n<ol>\n<li id=\"fs-idm301296\">A saturated solution of a slightly soluble electrolyte in contact with some of the solid electrolyte is said to be a system in equilibrium. Explain. Why is such a system called a heterogeneous equilibrium?<\/li>\n<li>Calculate the equilibrium concentration of Ni<sup>2+<\/sup> in a 1.0-<em>M<\/em> solution [Ni(NH<sub>3<\/sub>)<sub>6<\/sub>](NO<sub>3<\/sub>)<sub>2<\/sub>.<\/li>\n<li>Calculate the equilibrium concentration of Zn<sup>2+<\/sup> in a 0.30-<em>M<\/em> solution of [latex]\\text{Zn}{\\left(\\text{CN}\\right)}_{4}{}^{2-}[\/latex].<\/li>\n<li>Calculate the equilibrium concentration of Cu<sup>2+<\/sup> in a solution initially with 0.050 <em>M<\/em> Cu<sup>2+<\/sup> and 1.00 <em>M<\/em> NH<sub>3<\/sub>.<\/li>\n<li>Calculate the equilibrium concentration of Zn<sup>2+<\/sup> in a solution initially with 0.150 <em>M<\/em> Zn<sup>2+<\/sup> and 2.50 <em>M<\/em> CN<sup>\u2013<\/sup>.<\/li>\n<li>Calculate the Fe<sup>3+<\/sup> equilibrium concentration when 0.0888 mole of K<sub>3<\/sub>[Fe(CN)<sub>6<\/sub>] is added to a solution with 0.0.00010 <em>M<\/em> CN<sup>\u2013<\/sup>.<\/li>\n<li>Calculate the Co<sup>2+<\/sup> equilibrium concentration when 0.100 mole of [Co(NH<sub>3<\/sub>)<sub>6<\/sub>](NO<sub>3<\/sub>)<sub>2<\/sub> is added to a solution with 0.025 <em>M<\/em> NH<sub>3<\/sub>. Assume the volume is 1.00 L.<\/li>\n<li>The equilibrium constant for the reaction [latex]{\\text{Hg}}^{\\text{2+}}\\left(aq\\right)+2{\\text{Cl}}^{-}\\left(aq\\right)\\rightleftharpoons {\\text{HgCl}}_{2}\\left(aq\\right)[\/latex] is 1.6 [latex]\\times[\/latex] 10<sup>13<\/sup>. Is HgCl<sub>2<\/sub> a strong electrolyte or a weak electrolyte? What are the concentrations of Hg<sup>2+<\/sup> and Cl<sup>\u2013<\/sup> in a 0.015-<em>M<\/em> solution of HgCl<sub>2<\/sub>?<\/li>\n<li>Calculate the molar solubility of Sn(OH)<sub>2<\/sub> in a buffer solution containing equal concentrations of NH<sub>3<\/sub> and [latex]{\\text{NH}}_{4}{}^{+}[\/latex].<\/li>\n<li>Calculate the molar solubility of Al(OH)<sub>3<\/sub> in a buffer solution with 0.100 <em>M<\/em> NH<sub>3<\/sub> and 0.400 <em>M<\/em> [latex]{\\text{NH}}_{4}{}^{+}[\/latex].<\/li>\n<li>What is the molar solubility of CaF<sub>2<\/sub> in a 0.100-<em>M<\/em> solution of HF? <em>K<\/em><sub>a<\/sub> for HF = 7.2 [latex]\\times[\/latex] 10<sup>\u20134<\/sup>.<\/li>\n<li>What is the molar solubility of BaSO<sub>4<\/sub> in a 0.250-<em>M<\/em> solution of NaHSO<sub>4<\/sub>? <em>K<\/em><sub>a<\/sub> for [latex]{\\text{HSO}}_{4}{}^{-}[\/latex] = 1.2 [latex]\\times[\/latex] 10<sup>\u20132<\/sup>.<\/li>\n<li>What is the molar solubility of Tl(OH)<sub>3<\/sub> in a 0.10-<em>M<\/em> solution of NH<sub>3<\/sub>?<\/li>\n<li>What is the molar solubility of Pb(OH)<sub>2<\/sub> in a 0.138-<em>M<\/em> solution of CH<sub>3<\/sub>NH<sub>2<\/sub>?<\/li>\n<li>A solution of 0.075 <em>M<\/em> CoBr<sub>2<\/sub> is saturated with H<sub>2<\/sub>S ([H<sub>2<\/sub>S] = 0.10 <em>M<\/em>). What is the minimum pH at which CoS begins to precipitate?<br \/>\n[latex]\\text{CoS}\\left(s\\right)\\rightleftharpoons {\\text{Co}}^{\\text{2+}}\\left(aq\\right)+{\\text{S}}^{2-}\\left(aq\\right){K}_{\\text{sp}}=4.5\\times {10}^{-27}[\/latex][latex]{\\text{H}}_{2}\\text{S}\\left(aq\\right)+2{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons 2{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{S}}^{2-}\\left(aq\\right)K=1.0\\times {10}^{\\text{-26}}[\/latex]<\/li>\n<li>A 0.125-<em>M<\/em> solution of Mn(NO<sub>3<\/sub>)<sub>2<\/sub> is saturated with H<sub>2<\/sub>S ([H<sub>2<\/sub>S] = 0.10 <em>M<\/em>). At what pH does MnS begin to precipitate?[latex]\\text{MnS}\\left(s\\right)\\rightleftharpoons {\\text{Mn}}^{\\text{2+}}\\left(aq\\right)+{\\text{S}}^{2-}\\left(aq\\right){K}_{\\text{sp}}=4.3\\times {10}^{-22}[\/latex][latex]{\\text{H}}_{2}\\text{S}\\left(aq\\right)+2{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons 2{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{S}}^{2-}\\left(aq\\right)K=1.0\\times {10}^{-26}[\/latex]<\/li>\n<li>Calculate the molar solubility of BaF<sub>2<\/sub> in a buffer solution containing 0.20 <em>M<\/em> HF and 0.20 <em>M<\/em> NaF.<\/li>\n<li>Calculate the molar solubility of CdCO<sub>3<\/sub> in a buffer solution containing 0.115 <em>M<\/em> Na<sub>2<\/sub>CO<sub>3<\/sub> and 0.120 <em>M<\/em> NaHCO<sub>3<\/sub><\/li>\n<li>To a 0.10-<em>M<\/em> solution of Pb(NO<sub>3<\/sub>)<sub>2<\/sub> is added enough HF(<em>g<\/em>) to make [HF] = 0.10 <em>M<\/em>.\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Does PbF<sub>2<\/sub> precipitate from this solution? Show the calculations that support your conclusion.<\/li>\n<li>What is the minimum pH at which PbF<sub>2<\/sub> precipitates?<\/li>\n<\/ol>\n<\/li>\n<li>Calculate the concentration of Cd<sup>2+<\/sup> resulting from the dissolution of CdCO<sub>3<\/sub> in a solution that is 0.250 <em>M<\/em> in CH<sub>3<\/sub>CO<sub>2<\/sub>H, 0.375 <em>M<\/em> in NaCH<sub>3<\/sub>CO<sub>2<\/sub>, and 0.010 <em>M<\/em> in H<sub>2<\/sub>CO<sub>3<\/sub>.<\/li>\n<li>Both AgCl and AgI dissolve in NH<sub>3<\/sub>.\n<ol style=\"list-style-type: lower-alpha;\">\n<li>What mass of AgI dissolves in 1.0 L of 1.0 <em>M<\/em> NH<sub>3<\/sub>?<\/li>\n<li>What mass of AgCl dissolves in 1.0 L of 1.0 <em>M<\/em> NH<sub>3<\/sub>?<\/li>\n<\/ol>\n<\/li>\n<li>Calculate the volume of 1.50 <em>M<\/em> CH<sub>3<\/sub>CO<sub>2<\/sub>H required to dissolve a precipitate composed of 350 mg each of CaCO<sub>3<\/sub>, SrCO<sub>3<\/sub>, and BaCO<sub>3<\/sub>.<\/li>\n<li>Even though Ca(OH)<sub>2<\/sub> is an inexpensive base, its limited solubility restricts its use. What is the pH of a saturated solution of Ca(OH)<sub>2<\/sub>?<\/li>\n<li>What mass of NaCN must be added to 1 L of 0.010 <em>M<\/em> Mg(NO<sub>3<\/sub>)<sub>2<\/sub> in order to produce the first trace of Mg(OH)<sub>2<\/sub>?<\/li>\n<li>Magnesium hydroxide and magnesium citrate function as mild laxatives when they reach the small intestine. Why do magnesium hydroxide and magnesium citrate, two very different substances, have the same effect in your small intestine. (Hint: The contents of the small intestine are basic.)<\/li>\n<li>The following question is taken from a Chemistry Advanced Placement Examination and is used with the permission of the Educational Testing Service. Solve the following problem: [latex]{\\text{MgF}}_{2}\\left(s\\right)\\rightleftharpoons {\\text{Mg}}^{\\text{2+}}\\left(aq\\right)+2{\\text{F}}^{-}\\left(aq\\right)[\/latex]In a saturated solution of MgF<sub>2<\/sub> at 18 \u00b0C, the concentration of Mg<sup>2+<\/sup> is 1.21 [latex]\\times[\/latex] 10<sup>\u20133<\/sup><em>M<\/em>. The equilibrium is represented by the equation above.\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Write the expression for the solubility-product constant, <em>K<\/em><sub>sp<\/sub>, and calculate its value at 18 \u00b0C.<\/li>\n<li>Calculate the equilibrium concentration of Mg<sup>2+<\/sup> in 1.000 L of saturated MgF<sub>2<\/sub> solution at 18 \u00b0C to which 0.100 mol of solid KF has been added. The KF dissolves completely. Assume the volume change is negligible.<\/li>\n<li>Predict whether a precipitate of MgF<sub>2<\/sub> will form when 100.0 mL of a 3.00 [latex]\\times[\/latex] 10<sup>\u20133<\/sup>&#8211;<em>M<\/em> solution of Mg(NO<sub>3<\/sub>)<sub>2<\/sub> is mixed with 200.0 mL of a 2.00 [latex]\\times[\/latex] 10<sup>\u20133<\/sup>&#8211;<em>M<\/em> solution of NaF at 18 \u00b0C. Show the calculations to support your prediction.<\/li>\n<li>At 27 \u00b0C the concentration of Mg<sup>2+<\/sup> in a saturated solution of MgF<sub>2<\/sub> is 1.17 [latex]\\times[\/latex] 10<sup>\u20133<\/sup><em>M<\/em>. Is the dissolving of MgF<sub>2<\/sub> in water an endothermic or an exothermic process? Give an explanation to support your conclusion.<\/li>\n<\/ol>\n<\/li>\n<li>Which of the following compounds, when dissolved in a 0.01-<em>M<\/em> solution of HClO<sub>4<\/sub>, has a solubility greater than in pure water: AgBr, BaF<sub>2<\/sub>, Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>3<\/sub>, ZnS, PbI<sub>2<\/sub>? Explain your answer.<\/li>\n<li>What is the effect on the amount of solid Mg(OH)<sub>2<\/sub> that dissolves and the concentrations of Mg<sup>2+<\/sup> and OH<sup>\u2013<\/sup> when each of the following are added to a mixture of solid Mg(OH)<sub>2<\/sub> and water at equilibrium?\n<ol style=\"list-style-type: lower-alpha;\">\n<li>MgCl<sub>2<\/sub><\/li>\n<li>KOH<\/li>\n<li>HClO<sub>4<\/sub><\/li>\n<li>NaNO<sub>3<\/sub><\/li>\n<li>Mg(OH)<sub>2<\/sub><\/li>\n<\/ol>\n<\/li>\n<li>What is the effect on the amount of CaHPO<sub>4<\/sub> that dissolves and the concentrations of Ca<sup>2+<\/sup> and [latex]{\\text{HPO}}_{4}{}^{-}[\/latex] when each of the following are added to a mixture of solid CaHPO<sub>4<\/sub> and water at equilibrium?\n<ol style=\"list-style-type: lower-alpha;\">\n<li>CaCl<sub>2<\/sub><\/li>\n<li>HCl<\/li>\n<li>KClO<sub>4<\/sub><\/li>\n<li>NaOH<\/li>\n<li>CaHPO<sub>4<\/sub><\/li>\n<\/ol>\n<\/li>\n<li>Identify all chemical species present in an aqueous solution of Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub> and list these species in decreasing order of their concentrations. (Hint: Remember that the [latex]{\\text{PO}}_{4}{}^{\\text{3-}}[\/latex] ion is a weak base.)<\/li>\n<li>A volume of 50 mL of 1.8 <em>M<\/em> NH<sub>3<\/sub> is mixed with an equal volume of a solution containing 0.95 g of MgCl<sub>2<\/sub>. What mass of NH<sub>4<\/sub>Cl must be added to the resulting solution to prevent the precipitation of Mg(OH)<sub>2<\/sub>?<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q348957\">Show Selected Answers<\/span><\/p>\n<div id=\"q348957\" class=\"hidden-answer\" style=\"display: none\">\n<p>2. [latex]{\\text{Ni}}^{\\text{2+}}\\left(aq\\right)+6{\\text{NH}}_{3}\\left(aq\\right)\\rightleftharpoons {\\left[\\text{Ni}{\\left({\\text{NH}}_{3}\\right)}_{6}\\right]}^{\\text{2+}}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,{;}\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{f}}=1.8\\times {10}^{8}[\/latex]<\/p>\n<p>Let <em>x<\/em> be the change in concentration as Ni<sup>2+<\/sup> dissociates. Because the initial Ni<sup>2+<\/sup> concentration is 0, the concentration at any times is <em>x<\/em>:<\/p>\n<p style=\"text-align: center;\">[latex]1.8\\times {10}^{8}=\\frac{{\\left[\\text{Ni}{\\left({\\text{NH}}_{3}\\right)}_{6}\\right]}^{\\text{2+}}}{\\left[{\\text{Ni}}^{\\text{2+}}\\right]{\\left[{\\text{NH}}_{3}\\right]}^{6}}=\\frac{\\left(1.0-x\\right)}{x{\\left(6x\\right)}^{6}}[\/latex]<\/p>\n<p style=\"text-align: center;\">1.8 [latex]\\times[\/latex] 10<sup>8<\/sup>(46656<em>x<\/em><sup>7<\/sup>) = 1.0 \u2013 <em>x<\/em><\/p>\n<p style=\"text-align: center;\">8.40 [latex]\\times[\/latex] 10<sup>12<\/sup>(<em>x<\/em><sup>2<\/sup>) = 1.0 \u2013 <em>x<\/em><\/p>\n<p>Since <em>x<\/em> is small in comparison with 1.0, drop <em>x<\/em>:<\/p>\n<p style=\"text-align: center;\">8.40 [latex]\\times[\/latex] 10<sup>12<\/sup>(<em>x<\/em><sup>7<\/sup>) = 1.0<\/p>\n<p style=\"text-align: center;\"><em>x<\/em><sup>7<\/sup> = 1.19 [latex]\\times[\/latex] 10<sup>\u201313<\/sup><\/p>\n<p style=\"text-align: center;\"><em>x<\/em> = 0.014 <em>M<\/em><\/p>\n<p>4.\u00a0Assume that all Cu<sup>2+<\/sup> forms the complex whose concentration is 0.050 <em>M<\/em> and the remaining NH<sub>3<\/sub> has a concentration of 1.00 <em>M<\/em> \u2013 4(0.050 <em>M<\/em>) = 0.80 <em>M<\/em>. The complex dissociates:<\/p>\n<p style=\"text-align: center;\">[latex]{\\left[\\text{Cu}{\\left({\\text{NH}}_{3}\\right)}_{4}\\right]}^{\\text{2+}}\\rightleftharpoons \\left[{\\text{Cu}}^{\\text{2+}}\\right]+4\\left[{\\text{NH}}_{3}\\right][\/latex]<\/p>\n<p>Let <em>x<\/em> be the change in concentration of Cu<sup>2+<\/sup> that dissociates:<\/p>\n<table>\n<thead>\n<tr>\n<th><\/th>\n<th>[Cu(NH<sub>3<\/sub>)<sub>4<\/sub><sup>2+<\/sup>]<\/th>\n<th>[Cu<sup>2+<\/sup>]<\/th>\n<th>[NH<sub>3<\/sub>]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<th>Initial concentration (<em>M<\/em>)<\/th>\n<td>0.050<\/td>\n<td>0<\/td>\n<td>0.80<\/td>\n<\/tr>\n<tr>\n<th>Equilibrium (<em>M<\/em>)<\/th>\n<td>0.050\u00a0\u2212\u00a0<em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<td>4<em>x<\/em> + 0.80<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: center;\">[latex]\\frac{\\left[\\text{Cu}{\\left({\\text{NH}}_{3}\\right)}_{4}{}^{\\text{2+}}\\right]}{\\left[{\\text{Cu}}^{\\text{2+}}\\right]{\\left[{\\text{NH}}_{3}\\right]}^{4}}=1.2\\times {10}^{12}=\\frac{0.050-x}{x{\\left(4x+0.80\\right)}^{4}}[\/latex]<\/p>\n<p>Assume that 4<em>x<\/em> is small when compared with 0.80 and that <em>x<\/em> is small when compared with 0.050:<\/p>\n<p style=\"text-align: center;\">(0.80)<sup>4<\/sup> [latex]\\times[\/latex] 1.2 [latex]\\times[\/latex] 10<sup>12<\/sup><em>x<\/em> = 0.050<\/p>\n<p style=\"text-align: center;\"><em>x<\/em> = 1.0 [latex]\\times[\/latex] 10<sup>\u201313<\/sup><em>M<\/em><\/p>\n<p>6.\u00a0Set up a table listing initial and equilibrium concentrations for the reaction:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{Fe}}^{\\text{3+}}+6{\\text{CN}}^{-}\\rightleftharpoons {\\left(\\text{Fe}{\\left(\\text{CN}\\right)}_{6}\\right]}^{3-}{K}_{\\text{f}}=1\\times {10}^{44}[\/latex]<\/p>\n<p>Let <em>x<\/em> be the concentration of Fe<sup>3+<\/sup> that dissociates when 0.0888 mol dissolves in 1.00 L of 0.00010 <em>M<\/em> CN<sup>\u2013<\/sup>. Assume no volume change upon dissolution:<\/p>\n<table>\n<thead>\n<tr>\n<th><\/th>\n<th>[Fe(CN)<sub>6<\/sub><sup>3\u2212<\/sup>]<\/th>\n<th>[Fe<sup>3+<\/sup>]<\/th>\n<th>[CN<sup>\u2212<\/sup>]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<th>Initial concentration (<em>M<\/em>)<\/th>\n<td>0.0888<\/td>\n<td>0<\/td>\n<td>0.00010<\/td>\n<\/tr>\n<tr>\n<th>Equilibrium (<em>M<\/em>)<\/th>\n<td>0.0888\u00a0\u2212\u00a0<em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<td>0.00010\u00a0\u2212 6<em>x<\/em><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: center;\">[latex]\\frac{\\left[\\text{Fe}{\\left(\\text{CN}\\right)}_{6}{}^{3-}\\right]}{\\left[{\\text{Fe}}^{\\text{3+}}\\right]{\\left[{\\text{CN}}^{-}\\right]}^{6}}=\\frac{0.0888-x}{x{\\left(0.000010 - 6x\\right)}^{6}}=1\\times {10}^{44}[\/latex]<\/p>\n<p>Assume that <em>x<\/em> is small when compared with the terms from which it is subtracted:<\/p>\n<p style=\"text-align: center;\">0.0888 = (0.00010)<sup>6<\/sup>(<em>x<\/em>)(1 [latex]\\times[\/latex] 10<sup>44<\/sup>)<\/p>\n<p style=\"text-align: center;\">[latex]x=\\frac{0.0888}{1\\times {10}^{26}}=9\\times {10}^{-22}M[\/latex]<\/p>\n<p>8.\u00a0Let <em>x<\/em> be the change in the number of moles of Hg<sup>2+<\/sup> that form per liter:<\/p>\n<table>\n<thead>\n<tr>\n<th><\/th>\n<th>[HgCl<sub>2<\/sub>]<\/th>\n<th>[Hg<sup>2+<\/sup>]<\/th>\n<th>[Cl<sup>\u2212<\/sup>]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<th>Initial concentration (<em>M<\/em>)<\/th>\n<td>0.015<\/td>\n<td>0<\/td>\n<td>0<\/td>\n<\/tr>\n<tr>\n<th>Equilibrium (<em>M<\/em>)<\/th>\n<td>0.015\u00a0\u2212\u00a0<em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<td>2<em>x<\/em><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\frac{\\left[{\\text{HgCl}}_{2}\\right]}{\\left[{\\text{Hg}}^{\\text{2+}}\\right]{\\left[{\\text{Cl}}^{-}\\right]}^{2}}=K=1.6\\times {10}^{13}\\\\ \\frac{0.015-x}{\\left(x\\right){\\left(2x\\right)}^{2}}\\approx \\frac{0.015}{4{x}^{3}}=1.6\\times {10}^{13}\\end{array}[\/latex]<\/p>\n<p style=\"text-align: center;\"><em>x<\/em><sup>3<\/sup> = 2.3 [latex]\\times[\/latex] 10<sup>\u201316<\/sup><\/p>\n<p style=\"text-align: center;\"><em>x<\/em> = 6.2 [latex]\\times[\/latex] 10<sup>\u20136<\/sup><em>M<\/em> = [Hg<sup>2+<\/sup>]<\/p>\n<p style=\"text-align: center;\">2x = 1.2 [latex]\\times[\/latex] 10<sup>\u20135<\/sup>] <em>M<\/em> = [Cl<sup>\u2013<\/sup>]<\/p>\n<p>The substance is a weak electrolyte because very little of the initial 0.015 <em>M<\/em> HgCl<sub>2<\/sub> dissolved.<\/p>\n<p>10.\u00a0[latex]{K}_{\\text{b}}=\\frac{\\left[{\\text{NH}}_{4}{}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\text{NH}}_{3}\\right]}=\\frac{\\left(0.400\\right)\\left[{\\text{OH}}^{-}\\right]}{\\left(0.100\\right)}=1.8\\times {10}^{-5}[\/latex]<\/p>\n<p>[latex]\\left[{\\text{OH}}^{-}\\right]=\\frac{\\left(0.100\\right)\\left(1.8\\times {10}^{-5}\\right)}{0.0400}\\text{=}4.5\\times {10}^{-5}[\/latex]<\/p>\n<p><em>K<\/em><sub>sp<\/sub> = [Al<sup>3+<\/sup>][OH<sup>\u2013<\/sup>]<sup>3<\/sup> = [Al<sup>3+<\/sup>](4.5 [latex]\\times[\/latex] 10<sup>\u20135<\/sup>)<sup>3<\/sup> = 1.9 [latex]\\times[\/latex] 10<sup>\u201333<\/sup><\/p>\n<p>[Al<sup>3+<\/sup>] = 2.1 [latex]\\times[\/latex] 10<sup>\u201320<\/sup> (molar solubility)<\/p>\n<p>12.\u00a0Find the amount of [latex]{\\text{SO}}_{4}{}^{2-}[\/latex] present from <em>K<\/em><sub>a<\/sub> for the equilibrium:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{HSO}}_{4}{}^{-}+{\\text{H}}_{2}\\text{O}\\longrightarrow {\\text{H}}_{3}{\\text{O}}^{+}+{\\text{SO}}_{4}{}^{2-}[\/latex]<\/p>\n<p>Let <em>x<\/em> be the change in [latex]\\left[{\\text{SO}}_{4}{}^{2-}\\right][\/latex] :<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{\\text{a}}=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{SO}}_{4}{}^{2-}\\right]}{\\left[{\\text{HSO}}_{4}{}^{-}\\right]}=\\frac{{x}^{2}}{0.250-x}=1.2\\times {10}^{-2}[\/latex]<\/p>\n<p>Because <em>K<\/em><sub>a<\/sub> is too large to disregard <em>x<\/em> in the expression 0.250 \u2013 <em>x<\/em>, we must solve the quadratic equation:<\/p>\n<p style=\"text-align: center;\"><em>x<\/em><sup>2<\/sup> + 1.2 [latex]\\times[\/latex] 10<sup>\u20132<\/sup><em>x<\/em> \u2013 0.250(1.2 [latex]\\times[\/latex] 10<sup>\u20132<\/sup>) = 0<\/p>\n<p style=\"text-align: center;\">[latex]x=\\frac{-1.2\\times {10}^{-2}\\pm \\sqrt{{\\left(1.2\\times {10}^{-2}\\right)}^{2}+4\\left(3.0\\times {10}^{\\text{-3}}\\right)}}{2}=\\frac{-1.2\\times {10}^{-2}\\pm 0.11}{2}=0.049M[\/latex]<\/p>\n<p style=\"text-align: center;\"><em>K<\/em><sub>sp<\/sub> = [Ba<sup>2+<\/sup>] [latex]\\left[{\\text{SO}}_{4}{}^{2-}\\right][\/latex] = [Ba<sup>2+<\/sup>](0.049) = 1.08 [latex]\\times[\/latex] 10<sup>\u201310<\/sup><\/p>\n<p style=\"text-align: center;\">[Ba<sup>2+<\/sup>] = 2.2 [latex]\\times[\/latex] 10<sup>\u20139<\/sup> (molar solubility)<\/p>\n<p>14.\u00a0[latex]{\\text{CH}}_{3}{\\text{NH}}_{2}+{\\text{H}}_{2}\\text{O}\\rightleftharpoons {\\text{CH}}_{3}{\\text{NH}}_{3}{}^{+}+{\\text{OH}}^{-}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{\\text{b}}=\\frac{\\left[{\\text{CH}}_{3}{\\text{NH}}_{3}{}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\text{CH}}_{3}{\\text{NH}}_{2}\\right]}=\\frac{\\left(x\\right)\\left(x\\right)}{0.138-x}=4.4\\times {10}^{-4}[\/latex]<\/p>\n<p>Solve the quadratic equation using the quadratic formula:<\/p>\n<p style=\"text-align: center;\"><em>x<\/em><sup>2<\/sup> + 4.4 [latex]\\times[\/latex] 10<sup>\u20134<\/sup><em>x<\/em> \u2013 0.138(4.4 [latex]\\times[\/latex] 10<sup>\u20134<\/sup>) = 0<\/p>\n<p style=\"text-align: center;\">[latex]x=\\frac{-4.4\\times {10}^{-4}\\pm \\sqrt{{\\left(4.4\\times {10}^{-4}\\right)}^{2}+4\\left(6.07\\times {10}^{\\text{-5}}\\right)}}{2}=\\frac{-4.4\\times {10}^{\\text{-4}}\\pm \\sqrt{2.43\\times {10}^{-4}}}{2}=\\frac{0.0152}{2}=7.6\\times {10}^{-3}M[\/latex]<\/p>\n<p style=\"text-align: center;\"><em>K<\/em><sub>sp<\/sub> = [Pb<sup>2+<\/sup>][OH<sup>\u2013<\/sup>]<sup>2<\/sup> = [Pb<sup>2+<\/sup>](7.6 [latex]\\times[\/latex] 10<sup>\u20133<\/sup>)<sup>2<\/sup> = 2.8 [latex]\\times[\/latex] 10<sup>\u201316<\/sup><\/p>\n<p style=\"text-align: center;\">[Pb<sup>2+<\/sup>] = 4.8 [latex]\\times[\/latex] 10<sup>\u201312<\/sup> (molar solubility)<\/p>\n<p>16.\u00a0Two equilibria are in competition for the ions and must be considered simultaneously. Precipitation of MnS will occur when the concentration of S<sup>2\u2013<\/sup> in conjunction with 0.125 <em>M<\/em> Mn<sup>2+<\/sup> exceeds the <em>K<\/em><sub>sp<\/sub> of MnS. The [S<sup>2\u2013<\/sup>] must come from the ionization of H<sub>2<\/sub>S as defined by the equilibrium:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{H}}_{2}\\text{S}\\left(aq\\right)+2{\\text{H}}_{2}\\text{O}\\left(l\\right)\\longrightarrow 2{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{S}}^{2-}\\left(aq\\right)[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}^{2}\\left[{\\text{S}}^{2-}\\right]}{\\left[{\\text{H}}_{2}\\text{S}\\right]}={K}_{1}{K}_{2}\\left({\\text{H}}_{2}\\text{S}\\right)=1.0\\times {10}^{\\text{-26}}[\/latex]<\/p>\n<p>As a saturated solution of H<sub>2<\/sub>S is 0.10 <em>M<\/em>, this later expression becomes:<\/p>\n<p style=\"text-align: center;\">[latex]{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}^{2}\\left[{\\text{S}}^{2-}\\right]=1.0\\times {10}^{-27}[\/latex]<\/p>\n<p>From the equilibrium of MnS, the minimum concentration of S<sup>2\u2013<\/sup> required to cause precipitation is calculated as:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{MnS}\\left(s\\right)\\longrightarrow {\\text{Mn}}^{\\text{2+}}\\left(aq\\right)+{\\text{S}}^{2-}\\left(aq\\right)[\/latex]<\/p>\n<p style=\"text-align: center;\"><em>K<\/em><sub>sp<\/sub> = [Mn<sup>2+<\/sup>][S<sup>2\u2013<\/sup>] = 4.3 [latex]\\times[\/latex] 10<sup>\u201322<\/sup><\/p>\n<p style=\"text-align: center;\">[latex]\\left[{\\text{S}}^{2-}\\right]=\\frac{4.3\\times {10}^{-22}}{0.125}=3.44\\times {10}^{-21}[\/latex]<\/p>\n<p>This amount of S<sup>2\u2013<\/sup> will exist in solution at a pH defined by the H<sub>2<\/sub>S equilibrium:<\/p>\n<ul>\n<li>[latex]{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}^{2}\\left(3.44\\times {10}^{-21}\\right)=1.0\\times {10}^{-27}[\/latex]<\/li>\n<li>[latex]{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}^{2}=2.91\\times {10}^{-7}[\/latex]<\/li>\n<li>[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]=5.39\\times {10}^{-4}M[\/latex]<\/li>\n<li>[latex]\\text{pH}=-\\text{log}\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]=3.27[\/latex]<\/li>\n<\/ul>\n<p>18.\u00a0Three equilibria are involved:<\/p>\n<ul>\n<li>[latex]{\\text{H}}_{2}{\\text{CO}}_{3}+{\\text{H}}_{2}\\text{O}\\longrightarrow {\\text{H}}_{3}{\\text{O}}^{+}+{\\text{HCO}}_{3}{}^{-}{K}_{{\\text{a}}_{1}}=4.3\\times {10}^{-7}[\/latex]<\/li>\n<li>[latex]{\\text{HCO}}_{3}{}^{-}+{\\text{H}}_{2}\\text{O}\\longrightarrow {\\text{H}}_{3}{\\text{O}}^{+}+{\\text{CO}}_{3}{}^{2-}{K}_{{\\text{a}}_{2}}=7\\times {10}^{-11}[\/latex]<\/li>\n<li>[latex]{\\text{CdCO}}_{3}\\longrightarrow {\\text{Cd}}^{\\text{2+}}+{\\text{CO}}_{3}{}^{2-}{K}_{\\text{sp}}=2.5\\times {10}^{-14}[\/latex]<\/li>\n<\/ul>\n<p>First, find the pH of the buffer from the Henderson-Hasselbach equation. Then find [H<sub>3<\/sub>O<sup>+<\/sup>]:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{pH}=\\text{p}{K}_{\\text{a}}+\\mathrm{log}\\frac{\\left[{\\text{CO}}_{3}{}^{2-}\\right]}{\\left[{\\text{HCO}}_{3}{}^{-}\\right]}[\/latex]<\/p>\n<p style=\"text-align: center;\">10.155 + log [latex]\\frac{0.115}{0.120}[\/latex] = 10.155 \u2013 0.018 = 10.137<\/p>\n<p>In this case, several more significant figures are carried than justified so that the value of the log ratio is meaningful:<\/p>\n<p style=\"text-align: center;\">[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]=7.3\\times {10}^{-11}[\/latex]<\/p>\n<p>Now, find the concentration of [latex]{\\text{CO}}_{3}{}^{2-}[\/latex] present in the buffer solution. Next, using [latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right][\/latex] and <em>K<\/em><sub>sp<\/sub>, calculate the concentration of Cd<sup>2+<\/sup>. This latter value represents the molar solubility. From [latex]{K}_{{a}_{1}}[\/latex] determine [H<sub>2<\/sub>CO<sub>3<\/sub>]:<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{{\\text{a}}_{1}}=4.3\\times {10}^{-7}=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{HCO}}_{3}{}^{-}\\right]}{\\left[{\\text{H}}_{2}{\\text{CO}}_{3}\\right]}=\\frac{\\left(7.3\\times {10}^{-11}\\right)\\left(0.120\\right)}{\\left[{\\text{H}}_{2}{\\text{CO}}_{3}\\right]}[\/latex]<\/p>\n<p style=\"text-align: center;\">[H<sub>2<\/sub>CO<sub>3<\/sub>] = 2.05 [latex]\\times[\/latex] 10<sup>\u20135<\/sup><\/p>\n<p>From [latex]{K}_{{\\text{a}}_{1}}{K}_{{\\text{a}}_{2}}[\/latex] , find the concentration of [latex]{\\text{CO}}_{3}{}^{2-}[\/latex] :<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rll}{}{K}_{{\\text{a}}_{1}}{K}_{{\\text{a}}_{2}}&=&\\left(4.3\\times 10 - 7\\right)\\left(7\\times 10 - 11\\right)=\\frac{\\left[{\\text{HCO}}_{3}{}^{-}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}{\\left[{\\text{H}}_{2}{\\text{CO}}_{3}\\right]}\\times \\frac{\\left[{\\text{CO}}_{3}{}^{2-}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}{\\left[{\\text{H}}_{2}{\\text{CO}}_{3}\\right]}=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{CO}}_{3}{}^{2-}\\right]}{\\left[{\\text{H}}_{2}{\\text{CO}}_{3}\\right]}=\\frac{{\\left(7.3\\times {10}^{-11}\\right)}^{2}\\times \\left[{\\text{CO}}_{3}{}^{2-}\\right]}{2.04\\times {10}^{-5}}\\\\& =&3\\times {10}^{-17}\\end{array}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right][\/latex] = 0.115 <em>M<\/em><\/p>\n<p style=\"text-align: center;\"><em>K<\/em><sub>sp<\/sub> = [Cd<sup>2+<\/sup>] [latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right][\/latex] = [Cd<sup>2+<\/sup>](0.115) = 3 [latex]\\times[\/latex] 10<sup>\u201313<\/sup><\/p>\n<p style=\"text-align: center;\">[latex]\\left[{\\text{Cd}}^{\\text{2+}}\\right]=\\frac{3\\times {10}^{-13}}{0.115}=3\\times {10}^{-12}M[\/latex]<\/p>\n<p>20.\u00a0For <em>K<\/em><sub>sp<\/sub>, (CdCO<sub>3<\/sub>) = [Cd<sup>2+<\/sup>] [latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right][\/latex] = 2.5 [latex]\\times[\/latex] 10<sup>\u201314<\/sup>, the amount of [latex]{\\text{CO}}_{3}{}^{2-}[\/latex] is governed by <em>K<\/em><sub>a<\/sub> of H<sub>2<\/sub>CO<sub>3<\/sub>, 4.3 [latex]\\times[\/latex] 10<sup>\u20137<\/sup>, and <em>K<\/em><sub>a<\/sub> of [latex]{\\text{HCO}}_{3}{}^{-}[\/latex] , 7 [latex]\\times[\/latex] 10<sup>\u201311<\/sup>, and [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] by <em>K<\/em><sub>a<\/sub> of acetic acid. First, calculate the [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] from <em>K<\/em><sub>a<\/sub> of acetic acid (HOAc):<\/p>\n<p style=\"text-align: center;\">[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]=\\frac{{K}_{\\text{a}}\\left[\\text{HOAc}\\right]}{\\left[{\\text{OAc}}^{-}\\right]}=\\frac{1.8\\times {10}^{-5}\\left(0.250\\right)}{\\left(0.375\\right)}=1.2\\times {10}^{-5}M[\/latex]<\/p>\n<p>From this and <em>K<\/em><sub>a<\/sub> for H<sub>2<\/sub>CO<sub>3<\/sub>, calculate [latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right][\/latex] present. As [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] is fixed by <em>K<\/em><sub>a<\/sub> of acetic acid:<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{\\text{a}}\\left({\\text{H}}_{2}{\\text{CO}}_{3}\\right)=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{HCP}}_{3}{}^{-}\\right]}{\\left[{\\text{H}}_{2}{\\text{CO}}_{3}{}^{-}\\right]}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\left[{\\text{H}}_{2}{\\text{CO}}_{3}{}^{-}\\right]=\\frac{{K}_{\\text{a}}\\left[{\\text{H}}_{2}{\\text{CO}}_{3}\\right]}{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}=\\frac{4.3\\times {10}^{-7}\\left[0.010\\right]}{\\left[1.2\\times {10}^{-5}\\right]}=3.58\\times {10}^{-4}M[\/latex]<\/p>\n<p>From [latex]{K}_{\\text{a}}\\left({\\text{HCO}}_{3}{}^{-}\\right)[\/latex] we obtain [latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right][\/latex] :<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{a}=7\\times {10}^{-11}=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{CO}}_{3}{}^{2-}\\right]}{\\left[{\\text{HCO}}_{3}{}^{-}\\right]}=\\frac{1.2\\times {10}^{-17}\\left[{\\text{CO}}_{3}{}^{2-}\\right]}{\\left[3.58\\times {10}^{-4}\\right]}\\text{ }[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right]=\\frac{7\\times {10}^{-11}\\times 3.58\\times {10}^{-4}}{1.2\\times {10}^{-5}}=2.09\\times {10}^{-9}[\/latex]<\/p>\n<p>From the solubility product:<\/p>\n<p style=\"text-align: center;\"><em>K<\/em><sub>sp<\/sub> = [Cd<sup>2+<\/sup>] [latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right][\/latex] = 2.5 [latex]\\times[\/latex] 10<sup>\u201314<\/sup> = Cd<sup>2+<\/sup> (2.09 [latex]\\times[\/latex] 10<sup>\u20139<\/sup>)<\/p>\n<p style=\"text-align: center;\">[latex]\\left[{\\text{Cd}}^{\\text{2+}}\\right]=\\frac{2.5\\times {10}^{-14}}{2.09\\times {10}^{-9}}=1\\times {10}^{-5}M[\/latex]<\/p>\n<p>22.\u00a0[latex]{\\text{CaCO}}_{3}:0.350\\cancel{\\text{g}}\\times \\frac{1\\text{mol}}{100.09\\cancel{\\text{g}}}=3.50\\times {10}^{-3}\\text{mol}[\/latex]<\/p>\n<p>[latex]{\\text{SrCO}}_{3}:0.350\\cancel{\\text{g}}\\times \\frac{1\\text{mol}}{147.63\\cancel{\\text{g}}}=2.37\\times {10}^{-3}\\text{mol}[\/latex]<\/p>\n<p>[latex]{\\text{BaCO}}_{3}:0.350\\cancel{\\text{g}}\\times \\frac{1\\text{mol}}{197.34\\cancel{\\text{g}}}=1.77\\times {10}^{-3}\\text{mol}[\/latex]<\/p>\n<ul>\n<li>Total: 7.64 [latex]\\times[\/latex] 10<sup>\u20133<\/sup> mol<\/li>\n<li>CaCO<sub>3<\/sub>: <em>K<\/em><sub>sp<\/sub> = 4.8 [latex]\\times[\/latex] 10<sup>\u20139<\/sup><\/li>\n<li>SrCO<sub>3<\/sub>: <em>K<\/em><sub>sp<\/sub> = 9.42 [latex]\\times[\/latex] 10<sup>\u201310<\/sup><\/li>\n<li>BaCO<sub>3<\/sub>: <em>K<\/em><sub>sp<\/sub> = 8.1 [latex]\\times[\/latex] 10<sup>\u20139<\/sup><\/li>\n<\/ul>\n<p>Solubilities are approximately equal.<\/p>\n<p>When the three solids dissolve, 7.64 [latex]\\times[\/latex] 10<sup>\u20133<\/sup> mol of metal ions and 7.64 [latex]\\times[\/latex] 10<sup>\u20133<\/sup> mol of [latex]{\\text{CO}}_{3}{}^{2-}[\/latex] are initially present. The [latex]{\\text{CO}}_{3}{}^{2-}[\/latex] reacts with CH<sub>3<\/sub>CO<sub>2<\/sub>H to form [latex]{\\text{HCO}}_{3}{}^{-}[\/latex] :<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{ CO}}_{3}{}^{2-}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\left(aq\\right)\\rightleftharpoons {\\text{HCO}}_{3}{}^{-}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}\\left(aq\\right)[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]K=\\frac{\\left[{\\text{HCO}}_{3}{}^{-}\\right]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}\\right]}{\\left[{\\text{CO}}_{3}{}^{2-}\\right]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right]}\\times \\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}=\\frac{{K}_{\\text{a}}\\left(\\text{for}{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right)}{{K}_{\\text{a}}\\left(\\text{for}{\\text{HCO}}_{3}{}^{-}\\right)}=\\frac{1.8\\times {10}^{-5}}{7\\times {10}^{-11}}=2.6\\times {10}^{5}[\/latex]<\/p>\n<p>This <em>K<\/em> value is large, so virtually all [latex]{\\text{CO}}_{3}{}^{2-}[\/latex] undergoes this reaction and approximately 7.64 [latex]\\times[\/latex] 10<sup>\u20133<\/sup> mol of [latex]{\\text{HCO}}_{3}{}^{-}[\/latex] forms. The [latex]{\\text{HCO}}_{3}{}^{-}[\/latex] reacts with CH<sub>3<\/sub>CO<sub>2<\/sub>H:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{HCO}}_{3}{}^{-}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\left(aq\\right)\\longrightarrow {\\text{H}}_{2}{\\text{CO}}_{3}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}\\left(aq\\right)[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]K=\\frac{{K}_{\\text{a}}\\left(\\text{for}{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right)}{{K}_{\\text{a}}\\left(\\text{for}{\\text{HCO}}_{3}{}^{-}\\right)}=\\frac{1.8\\times {10}^{-5}}{4.3\\times {10}^{-7}}=42[\/latex]<\/p>\n<p>This reaction is virtually complete as well (<em>K<\/em> is large).<\/p>\n<p>For each mol of [latex]{\\text{CO}}_{3}{}^{2-}[\/latex] produced, 2 mol of CH<sub>3<\/sub>CO<sub>2<\/sub>H is required for conversion to H<sub>2<\/sub>CO<sub>3<\/sub>. Thus 15.3 [latex]\\times[\/latex] 10<sup>\u20133<\/sup> mol CH<sub>3<\/sub>CO<sub>2<\/sub>H is needed:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{Volume}=\\frac{15.3\\times {10}^{-3}\\text{mol}}{1.50\\text{mol}{\\text{L}}^{-1}}=0.0102\\text{L}\\left(10.2\\text{mL}\\right)[\/latex]<\/p>\n<p>24.\u00a0There are two equilibria involved: [latex]\\begin{array}{l}\\text{Mg}{\\left(\\text{OH}\\right)}_{2}\\left(s\\right)\\rightleftharpoons {\\text{Mg}}^{\\text{2+}}\\left(aq\\right)+2{\\text{OH}}^{-}\\left(aq\\right){K}_{\\text{sp}}=1.5\\times {10}^{-11}\\\\ {\\text{CN}}^{-}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons \\text{HCN}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)\\left({K}_{\\text{b}}=\\frac{{K}_{\\text{w}}}{{K}_{\\text{a}}}=\\frac{1.0\\times {10}^{-14}}{4\\times {10}^{-10}}=2.5\\times {10}^{-5}\\right)\\end{array}[\/latex] The Mg(NO<sub>3<\/sub>)<sub>2<\/sub> dissolves, and [Mg<sup>2+<\/sup>] = 0.010 <em>M<\/em>.<\/p>\n<p>[Mg<sup>2+<\/sup>][OH<sup>\u2013<\/sup>]<sup>2<\/sup> = <em>K<\/em><sub>sp<\/sub> = 1.5 [latex]\\times[\/latex] 10<sup>\u201311<\/sup><\/p>\n<p>(0.010)[OH<sup>\u2013<\/sup>]<sup>2<\/sup> = 1.5 [latex]\\times[\/latex] 10<sup>\u201311<\/sup><\/p>\n<p>[OH<sup>\u2013<\/sup>] = 3.9 [latex]\\times[\/latex] 10<sup>\u20135<\/sup><em>M<\/em><\/p>\n<p>We need to add enough CN<sup>\u2013<\/sup> to make [OH<sup>\u2013<\/sup>] = 3.9 [latex]\\times[\/latex] 10<sup>\u20135<\/sup><em>M<\/em>. Both OH<sup>\u2013<\/sup> and HCN come from CN<sup>\u2013<\/sup>, so [OH<sup>\u2013<\/sup>] = [HCN]:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\frac{\\left[\\text{HCN}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\text{CN}}^{-}\\right]}={K}_{\\text{b}}=2\\times {10}^{-5}\\\\ \\frac{{\\left(3.9\\times {10}^{-5}\\right)}^{2}}{\\left[{\\text{CN}}^{-}\\right]}=2.5\\times {10}^{-5}\\end{array}[\/latex]<\/p>\n<p style=\"text-align: center;\">[CN<sup>\u2013<\/sup>] = 6.1 [latex]\\times[\/latex] 10<sup>\u20135<\/sup><em>M<\/em><\/p>\n<p>mol NaCN = mol HCN + mol CN<sup>\u2013<\/sup> = 3.9 [latex]\\times[\/latex] 10<sup>\u20135<\/sup> + 6.1 [latex]\\times[\/latex] 10<sup>\u20135<\/sup> = 1.0 [latex]\\times[\/latex] 10<sup>\u20134<\/sup><\/p>\n<p>[latex]\\text{mass}\\left(\\text{NaCN}\\right)=1.0\\times {10}^{-4}\\text{mol}\\times \\frac{49.007\\text{g}}{1\\text{mol}}=5\\times {10}^{-3}\\text{g}[\/latex]<\/p>\n<p>26. The answers are as follows:<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li><em>K<\/em><sub>sp<\/sub> = [Mg<sup>2+<\/sup>][F<sup>\u2013<\/sup>]<sup>2<\/sup> = (1.21 [latex]\\times[\/latex] 10<sup>\u20133<\/sup>)(2 [latex]\\times[\/latex] 1.21 [latex]\\times[\/latex] 10<sup>\u20133<\/sup>)<sup>2<\/sup> = 7.09 [latex]\\times[\/latex] 10<sup>\u20139<\/sup><\/li>\n<li><em>K<\/em><sub>sp<\/sub> = [Mg<sup>2+<\/sup>][F<sup>\u2013<\/sup>]<sup>2<\/sup> = [<em>x<\/em>][0.100 + 2<em>x<\/em>]<sup>2<\/sup> = 7.09 [latex]\\times[\/latex] 10<sup>\u20139<br \/>\n<\/sup>Assume that 2<em>x<\/em> is small when compared with 0.100 <em>M<\/em>.<br \/>\n0.100<em>x<\/em> = 7.09 [latex]\\times[\/latex] 10<sup>\u20139<br \/>\n<\/sup><em>x<\/em> = [MgF<sub>2<\/sub>] = 7.09 [latex]\\times[\/latex] 10<sup>\u20137<\/sup><em>M<br \/>\n<\/em>The value 7.09 [latex]\\times[\/latex] 10<sup>\u20137<\/sup><em>M<\/em> is quite small when compared with 0.100 <em>M<\/em>, so the assumption is valid.<\/li>\n<li>Determine the concentration of Mg<sup>2+<\/sup> and F<sup>\u2013<\/sup> that will be present in the final volume. Compare the value of the ion product [Mg<sup>2+<\/sup>][F<sup>\u2013<\/sup>]<sup>2<\/sup> with <em>K<\/em><sub>sp<\/sub>. If this value is larger than <em>K<\/em><sub>sp<\/sub>, precipitation will occur.<br \/>\n0.1000 L [latex]\\times[\/latex] 3.00 [latex]\\times[\/latex] 10<sup>\u20133<\/sup><em>M<\/em> Mg(NO<sub>3<\/sub>)<sub>2<\/sub> = 0.3000 L [latex]\\times[\/latex] <em>M<\/em> Mg(NO<sub>3<\/sub>)<sub>2<br \/>\n<\/sub><em>M<\/em> Mg(NO<sub>3<\/sub>)<sub>2<\/sub> = 1.00 [latex]\\times[\/latex] 10<sup>\u20133<\/sup><em>M<br \/>\n<\/em>0.2000 L [latex]\\times[\/latex] 2.00 [latex]\\times[\/latex] 10<sup>\u20133<\/sup><em>M<\/em> NaF = 0.3000 L [latex]\\times[\/latex] <em>M<\/em> NaF<br \/>\n<em>M<\/em> NaF = 1.33 [latex]\\times[\/latex] 10<sup>\u20133<\/sup><em>M<br \/>\n<\/em>ion product = (1.00 [latex]\\times[\/latex] 10<sup>\u20133<\/sup>)(1.33 [latex]\\times[\/latex] 10<sup>\u20133<\/sup>)<sup>2<\/sup> = 1.77 [latex]\\times[\/latex] 10<sup>\u20139<br \/>\n<\/sup>This value is smaller than <em>K<\/em><sub>sp<\/sub>, so no precipitation will occur.<\/li>\n<li>MgF<sub>2<\/sub> is less soluble at 27 \u00b0C than at 18 \u00b0C. Because added heat acts like an added reagent, when it appears on the product side, the Le Ch\u00e2telier\u2019s principle states that the equilibrium will shift to the reactants\u2019 side to counter the stress. Consequently, less reagent will dissolve. This situation is found in our case. Therefore, the reaction is exothermic.<\/li>\n<\/ol>\n<p>28.\u00a0Effect on amount of solid Mg(OH)<sub>2<\/sub>, [Mg<sup>2+<\/sup>], [OH<sup>\u2013<\/sup>]:<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>increase, increase, decrease;<\/li>\n<li>increase, decrease, increase;<\/li>\n<li>decrease, increase, decrease;<\/li>\n<li>no effect predicted;<\/li>\n<li>increase, no effect, no effect<\/li>\n<\/ol>\n<p>30.\u00a0[H<sub>2<\/sub>O] &gt; [Ca<sup>2+<\/sup>] &gt; [latex]\\left[{\\text{PO}}_{4}{}^{3-}\\right][\/latex] &gt; [latex]{\\text{HPO}}_{4}{}^{2-}[\/latex] = [OH<sup>\u2013<\/sup>] &gt; [latex]\\left[{\\text{HPO}}_{4}{}^{-}\\right][\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Glossary<\/h2>\n<p><b>multiple equilibrium: <\/b>system characterized by more than one state of balance between a slightly soluble ionic solid and an aqueous solution of ions working simultaneously<\/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-3605\">\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":26,"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-3605","chapter","type-chapter","status-publish","hentry"],"part":2983,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapters\/3605","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\/3605\/revisions"}],"predecessor-version":[{"id":5870,"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapters\/3605\/revisions\/5870"}],"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\/3605\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/wp\/v2\/media?parent=3605"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapter-type?post=3605"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/wp\/v2\/contributor?post=3605"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/wp\/v2\/license?post=3605"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}