{"id":3591,"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=3591"},"modified":"2018-05-12T03:03:53","modified_gmt":"2018-05-12T03:03:53","slug":"lewis-acids-and-bases-2","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-mcc-chemistryformajors-2\/chapter\/lewis-acids-and-bases-2\/","title":{"raw":"15.2 Complex Ions","rendered":"15.2 Complex Ions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\nBy the end of this module, you will be able to:\r\n<ul>\r\n \t<li>Write equations for the formation of complex ions<\/li>\r\n \t<li>Perform equilibrium calculations involving formation constants<\/li>\r\n<\/ul>\r\n<\/div>\r\n&nbsp;\r\n\r\nMany slightly soluble ionic solids dissolve when the concentration of the metal ion in solution is decreased through the formation of complex (polyatomic) ions in a Lewis acid-base reaction. For example, silver chloride dissolves in a solution of ammonia because the silver ion reacts with ammonia to form the <b>complex ion<\/b> [latex]\\text{Ag}{\\left({\\text{NH}}_{3}\\right)}_{2}{}^{+}[\/latex]. The Lewis structure of the [latex]\\text{Ag}{\\left({\\text{NH}}_{3}\\right)}_{2}{}^{+}[\/latex] ion is:\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23214007\/CNX_Chem_15_02_AgNH32_img.jpg\" alt=\"A structure is shown in brackets. The structure has a central A g atom to which N atoms are single bonded to the left and right. Each of these atoms N atom has H atoms single bonded above, below, and to the outer end of the structure. Outside the brackets is a superscripted plus.\" width=\"325\" height=\"169\" \/>\r\n\r\nThe equations for the dissolution of AgCl in a solution of NH<sub>3<\/sub> are:\r\n<p style=\"text-align: center\">[latex]\\begin{array}{rrll}{}&amp;\\text{AgCl}\\left(s\\right)&amp;\\longrightarrow&amp;{\\text{Ag}}^{\\text{+}}\\left(aq\\right)+{\\text{Cl}}^{-}\\left(aq\\right)\\\\{}&amp;{\\text{Ag}}^{\\text{+}}\\left(aq\\right)+2{\\text{NH}}_{3}\\left(aq\\right)&amp;\\longrightarrow&amp;\\text{Ag}{\\left({\\text{NH}}_{3}\\right)}_{2}{}^{\\text{+}}\\left(aq\\right)\\\\\\text{Net: }&amp;\\text{AgCl}\\left(s\\right)+2{\\text{NH}}_{3}\\left(aq\\right)&amp;\\longrightarrow &amp;\\text{Ag}{\\left({\\text{NH}}_{3}\\right)}_{2}{}^{\\text{+}}\\left(aq\\right)+{\\text{Cl}}^{-}\\left(aq\\right)\\end{array}[\/latex]<\/p>\r\nAluminum hydroxide dissolves in a solution of sodium hydroxide or another strong base because of the formation of the complex ion [latex]\\text{Al}{\\left(\\text{OH}\\right)}_{4}{}^{-}[\/latex]. The Lewis structure of the [latex]\\text{Al}{\\left(\\text{OH}\\right)}_{4}{}^{-}[\/latex] ion is:\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23214008\/CNX_Chem_15_02_AlOH4_img.jpg\" alt=\"An H atom is bonded to an O atom. The O atom has 2 dots above it and 2 dots below it. The O atom is bonded to an A l atom, which has three additional O atoms bonded to it as well. Each of these additional O atoms has 4 dots arranged around it, and is bonded to an H atom. This entire molecule is contained in brackets, to the right of which is a superscripted negative sign.\" width=\"325\" height=\"169\" \/>\r\n\r\nThe equations for the dissolution are:\r\n<p style=\"text-align: center\">[latex]\\begin{array}{rrll}{}&amp;\\text{Al}{\\left(\\text{OH}\\right)}_{3}\\left(s\\right)&amp;\\longrightarrow&amp; {\\text{Al}}^{\\text{3+}}\\left(aq\\right)+3{\\text{OH}}^{-}\\left(aq\\right)\\\\{}&amp;{\\text{Al}}^{\\text{3+}}\\left(aq\\right)+4{\\text{OH}}^{-}\\left(aq\\right)&amp;\\longrightarrow&amp;\\text{Al}{\\left(\\text{OH}\\right)}_{4}{}^{-}\\left(aq\\right)\\\\\\text{Net:}&amp;\\text{Al}{\\left(\\text{OH}\\right)}_{3}\\left(s\\right)+{\\text{OH}}^{-}\\left(aq\\right)&amp;\\longrightarrow&amp;\\text{Al}{\\left(\\text{OH}\\right)}_{4}{}^{-}\\left(aq\\right)\\end{array}[\/latex]<\/p>\r\nMercury(II) sulfide dissolves in a solution of sodium sulfide because HgS reacts with the S<sup>2\u2013<\/sup> ion:\r\n<p style=\"text-align: center\">[latex]\\begin{array}{rrll}{}&amp;\\text{HgS}\\left(s\\right)&amp;\\longrightarrow&amp;{\\text{Hg}}^{\\text{2+}}\\left(aq\\right)+{\\text{S}}^{2-}\\left(aq\\right)\\\\{}&amp;{\\text{Hg}}^{\\text{2+}}\\left(aq\\right)+2{\\text{S}}^{2-}\\left(aq\\right)&amp;\\longrightarrow&amp;{\\text{HgS}}_{2}{}^{2-}\\left(aq\\right)\\\\\\text{Net:}&amp;\\text{HgS}\\left(s\\right)+{\\text{S}}^{2-}\\left(aq\\right)&amp;\\longrightarrow &amp;{\\text{HgS}}_{2}{}^{2-}\\left(aq\\right)\\end{array}[\/latex]<\/p>\r\nA complex ion consists of a central atom, typically a transition metal cation, surrounded by ions, or molecules called <b>ligands<\/b>. These ligands can be neutral molecules like H<sub>2<\/sub>O or NH<sub>3<\/sub>, or ions such as CN<sup>\u2013<\/sup> or OH<sup>\u2013<\/sup>. Often, the ligands act as Lewis bases, donating a pair of electrons to the central atom. The ligands aggregate themselves around the central atom, creating a new ion with a charge equal to the sum of the charges and, most often, a transitional metal ion. This more complex arrangement is why the resulting ion is called a <em>complex ion<\/em>. The complex ion formed in these reactions cannot be predicted; it must be determined experimentally. The types of bonds formed in complex ions are called coordinate covalent bonds, as electrons from the ligands are being shared with the central atom. Because of this, complex ions are sometimes referred to as coordination complexes. This will be studied further in upcoming chapters.\r\n\r\nThe equilibrium constant for the reaction of the components of a complex ion to form the complex ion in solution is called a <b>formation constant (<em>K<\/em><sub>f<\/sub>)<\/b> (sometimes called a stability constant). For example, the complex ion [latex]\\text{Cu}{\\left(\\text{CN}\\right)}_{2}{}^{-}[\/latex] is shown here:\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23214009\/CNX_Chem_15_02_CuCN2-_img.jpg\" alt=\"A Cu atom is bonded to two C atoms. Each of these C atoms is triple bonded to an N atom. Each N atom has two dots on the side of it.\" width=\"325\" height=\"43\" \/>\r\n\r\nIt forms by the reaction:\r\n<p style=\"text-align: center\">[latex]{\\text{Cu}}^{\\text{+}}\\left(aq\\right)+2{\\text{CN}}^{-}\\left(aq\\right)\\rightleftharpoons \\text{Cu}{\\left(\\text{CN}\\right)}_{2}{}^{-}\\left(aq\\right)[\/latex]<\/p>\r\nAt equilibrium:\r\n<p style=\"text-align: center\">[latex]{K}_{\\text{f}}=Q=\\frac{\\left[\\text{Cu}{\\left(\\text{CN}\\right)}_{2}{}^{-}\\right]}{\\left[{\\text{Cu}}^{+}\\right]{\\left[{\\text{CN}}^{-}\\right]}^{2}}[\/latex]<\/p>\r\nThe inverse of the formation constant is the <b>dissociation constant (<em>K<\/em><sub>d<\/sub>)<\/b>, the equilibrium constant for the <em>decomposition<\/em> of a complex ion into its components in solution. We will work with dissociation constants further in the exercises for this section. <a href=\".\/chapter\/formation-constants-for-complex-ions\/\" target=\"_blank\" rel=\"noopener\">Formation Constants for Complex Ions<\/a>\u00a0and Table\u00a01\u00a0are tables of formation constants. In general, the larger the formation constant, the more stable the complex; however, as in the case of <em>K<\/em><sub>sp<\/sub> values, the stoichiometry of the compound must be considered.\r\n<table id=\"fs-idm57488576\" class=\"span-all\" summary=\"No Summary Text\">\r\n<thead>\r\n<tr valign=\"middle\">\r\n<th colspan=\"2\">Table\u00a01. Common Complex Ions by Decreasing Formulation Constants<\/th>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<th>Substance<\/th>\r\n<th><em>K<\/em><sub>f<\/sub> at 25 \u00b0C<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"middle\">\r\n<td>[latex]{\\left[\\text{Cd}{\\left(\\text{CN}\\right)}_{4}\\right]}^{2-}[\/latex]<\/td>\r\n<td>1.3 [latex]\\times [\/latex] 10<sup>7<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td>[latex]\\text{Ag}{\\left({\\text{NH}}_{3}\\right)}_{2}{}^{+}[\/latex]<\/td>\r\n<td>1.7 [latex]\\times [\/latex] 10<sup>7<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td>[latex]{\\left[{\\text{AlF}}_{6}\\right]}^{\\text{3-}}[\/latex]<\/td>\r\n<td>7 [latex]\\times [\/latex] 10<sup>19<\/sup><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nAs an example of dissolution by complex ion formation, let us consider what happens when we add aqueous ammonia to a mixture of silver chloride and water. Silver chloride dissolves slightly in water, giving a small concentration of Ag<sup>+<\/sup> ([Ag<sup>+<\/sup>] = 1.3 [latex]\\times [\/latex] 10<sup>\u20135\u00a0<\/sup><em>M<\/em>):\r\n<p style=\"text-align: center\">[latex]\\text{AgCl}\\left(s\\right)\\rightleftharpoons {\\text{Ag}}^{\\text{+}}\\left(aq\\right)+{\\text{Cl}}^{-}\\left(aq\\right)[\/latex]<\/p>\r\nHowever, if NH<sub>3<\/sub> is present in the water, the complex ion, [latex]\\text{Ag}{\\left({\\text{NH}}_{3}\\right)}_{2}{}^{+}[\/latex], can form according to the equation:\r\n<p style=\"text-align: center\">[latex]{\\text{Ag}}^{\\text{+}}\\left(aq\\right)+2{\\text{NH}}_{3}\\left(aq\\right)\\rightleftharpoons \\text{Ag}{\\left({\\text{NH}}_{3}\\right)}_{2}{}^{\\text{+}}\\left(aq\\right)[\/latex]<\/p>\r\nwith\r\n<p style=\"text-align: center\">[latex]{K}_{\\text{f}}=\\frac{\\left[\\text{Ag}{\\left({\\text{NH}}_{3}\\right)}_{2}{}^{+}\\right]}{\\left[{\\text{Ag}}^{+}\\right]{\\left[{\\text{NH}}_{3}\\right]}^{2}}=1.6\\times {10}^{7}[\/latex]<\/p>\r\nThe large size of this formation constant indicates that most of the free silver ions produced by the dissolution of AgCl combine with NH<sub>3<\/sub> to form [latex]\\text{Ag}{\\left({\\text{NH}}_{3}\\right)}_{2}{}^{+}[\/latex]. As a consequence, the concentration of silver ions, [Ag<sup>+<\/sup>], is reduced, and the reaction quotient for the dissolution of silver chloride, [Ag<sup>+<\/sup>][Cl<sup>\u2013<\/sup>], falls below the solubility product of AgCl:\r\n<p style=\"text-align: center\">[latex]Q=\\left[{\\text{Ag}}^{+}\\right]\\left[{\\text{Cl}}^{-}\\right]&lt;{K}_{\\text{sp}}[\/latex]<\/p>\r\nMore silver chloride then dissolves. If the concentration of ammonia is great enough, all of the silver chloride dissolves.\r\n<div class=\"textbox examples\">\r\n<h3>Example 1:\u00a0Dissociation of a Complex Ion<\/h3>\r\nCalculate the concentration of the silver ion in a solution that initially is 0.10 <em>M<\/em> with respect to [latex]\\text{Ag}{\\left({\\text{NH}}_{3}\\right)}_{2}{}^{+}[\/latex].\r\n[reveal-answer q=\"576838\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"576838\"]\r\n\r\nWe use the familiar path to solve this problem:\r\n\r\n<img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23214010\/CNX_Chem_15_02_Format_img.jpg\" alt=\"Four boxes are shown side by side, with three right facing arrows connecting them. The first box contains the text \u201cDetermine the direction of change.\u201d The second box contains the text \u201cDetermine x and the equilibrium concentrations.\u201d The third box contains the text \u201cSolve for x and the equilibrium concentrations.\u201d The fourth box contains the text \u201cCheck the math.\u201d\" width=\"881\" height=\"156\" \/>\r\n<ol>\r\n \t<li><em>Determine the direction of change.<\/em> The complex ion [latex]\\text{Ag}{\\left({\\text{NH}}_{3}\\right)}_{2}{}^{+}[\/latex] is in equilibrium with its components, as represented by the equation:[latex]{\\text{Ag}}^{\\text{+}}\\left(aq\\right)+2{\\text{NH}}_{3}\\left(aq\\right)\\rightleftharpoons \\text{Ag}{\\left({\\text{NH}}_{3}\\right)}_{2}{}^{\\text{+}}\\left(aq\\right)[\/latex]We write the equilibrium as a formation reaction because\u00a0<a href=\".\/chapter\/formation-constants-for-complex-ions\/\" target=\"_blank\" rel=\"noopener\">Formation Constants for Complex Ions<\/a>\u00a0lists formation constants for complex ions. Before equilibrium, the reaction quotient is larger than the equilibrium constant [<em>K<\/em><sub>f<\/sub> = 1.6 [latex]\\times [\/latex] 10<sup>7<\/sup>, and [latex]Q=\\frac{0.10}{0\\times 0}[\/latex], it is infinitely large], so the reaction shifts to the left to reach equilibrium.<\/li>\r\n \t<li><em>Determine<\/em> x<em> and equilibrium concentrations.<\/em> We let the change in concentration of Ag<sup>+<\/sup> be <em>x<\/em>. Dissociation of 1 mol of [latex]\\text{Ag}{\\left({\\text{NH}}_{3}\\right)}_{2}{}^{+}[\/latex] gives 1 mol of Ag<sup>+<\/sup> and 2 mol of NH<sub>3<\/sub>, so the change in [NH<sub>3<\/sub>] is 2<em>x<\/em> and that of [latex]\\text{Ag}{\\left({\\text{NH}}_{3}\\right)}_{2}{}^{+}[\/latex] is \u2013<em>x<\/em>. In summary:<img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23214011\/CNX_Chem_15_02_ICETable1_img.jpg\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), and Equilibrium concentration ( M ). The second column has the header, \u201cA g superscript positive sign plus 2 N H subscript 3 equilibrium sign A g ( N H subscript 3 ) subscript 2 superscript positive sign.\u201d Under the second column is a subgroup of three rows and three columns. The first column contains: 0, x, and 0 plus x. The second column contains: 0, 2 x, and 0 plus 2 x. The third column contains 0.10, negative x, and 0.10 minus x.\" width=\"842\" height=\"228\" \/><\/li>\r\n \t<li><em>Solve for x and the equilibrium concentrations.<\/em> At equilibrium:\u00a0 [latex]{K}_{\\text{f}}=\\frac{\\left[\\text{Ag}{\\left({\\text{NH}}_{3}\\right)}_{2}{}^{+}\\right]}{\\left[{\\text{Ag}}^{+}\\right]{\\left[{\\text{NH}}_{3}\\right]}^{2}}[\/latex]\r\n[latex]1.6\\times {10}^{7}\\text{=}\\frac{0.10-x}{\\left(x\\right){\\left(2x\\right)}^{2}}[\/latex]\u00a0 Both <em>Q<\/em> and <em>K<\/em><sub>f<\/sub> are much larger than 1, so let us assume that the changes in concentrations needed to reach equilibrium are small. Thus 0.10 \u2013 <em>x<\/em> is approximated as 0.10:\u00a0 [latex]1.6\\times {10}^{7}=\\frac{0.10}{\\left(x\\right){\\left(2x\\right)}^{2}}[\/latex]\r\n[latex]{x}^{3}=\\frac{0.10}{4\\left(1.6\\times {10}^{7}\\right)}=1.6\\times {10}^{-9}[\/latex]\r\n[latex]x=\\sqrt[3]{1.6\\times {10}^{-9}}=1.2\\times {10}^{-3}[\/latex]\u00a0 Because only 1.2% of the [latex]\\text{Ag}{\\left({\\text{NH}}_{3}\\right)}_{2}{}^{+}[\/latex] dissociates into Ag<sup>+<\/sup> and NH<sub>3<\/sub>, The assumption that <em>x<\/em> is small is justified.Now we determine the equilibrium concentrations: [latex]\\left[{\\text{Ag}}^{+}\\right]=0+x=1.2\\times {10}^{-3}M[\/latex]\r\n[latex]\\left[{\\text{NH}}_{3}\\right]=0+2x=2.4\\times {10}^{-3}M[\/latex]\r\n[latex]\\left[\\text{Ag}{\\left({\\text{NH}}_{3}\\right)}_{2}{}^{+}\\right]=0.10-x=0.10 - 0.0012=0.099[\/latex]\u00a0 The concentration of free silver ion in the solution is 0.0012 <em>M<\/em>.<\/li>\r\n \t<li><em>Check the work.<\/em> The value of <em>Q<\/em> calculated using the equilibrium concentrations is equal to <em>K<\/em><sub>f<\/sub> within the error associated with the significant figures in the calculation.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n<h4>Check Your Learning<\/h4>\r\nCalculate the silver ion concentration, [Ag<sup>+<\/sup>], of a solution prepared by dissolving 1.00 g of AgNO<sub>3<\/sub> and 10.0 g of KCN in sufficient water to make 1.00 L of solution. (Hint: Because <em>Q<\/em> &lt; <em>K<\/em><sub>f<\/sub>, assume the reaction goes to completion then calculate the [Ag<sup>+<\/sup>] produced by dissociation of the complex.)\r\n[reveal-answer q=\"747092\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"747092\"][latex]3\\times10^{-22}M[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Key Takeaways<\/h3>\r\nComplex ions are examples of Lewis acid-base adducts. In a complex ion, we have a central atom, often consisting of a transition metal cation, which acts as a Lewis acid, and several neutral molecules or ions surrounding them called ligands that act as Lewis bases. Complex ions form by sharing electron pairs to form coordinate covalent bonds. The equilibrium reaction that occurs when forming a complex ion has an equilibrium constant associated with it called a formation constant, <em>K<\/em><sub>f<\/sub>. This is often referred to as a stability constant, as it represents the stability of the complex ion. Formation of complex ions in solution can have a profound effect on the solubility of a transition metal compound.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Exercises<\/h3>\r\n<ol>\r\n \t<li>Under what circumstances, if any, does a sample of solid AgCl completely dissolve in pure water?<\/li>\r\n \t<li>Explain why the addition of NH<sub>3<\/sub> or HNO<sub>3<\/sub> to a saturated solution of Ag<sub>2<\/sub>CO<sub>3<\/sub> in contact with solid Ag<sub>2<\/sub>CO<sub>3<\/sub> increases the solubility of the solid.<\/li>\r\n \t<li>Calculate the cadmium ion concentration, [Cd<sup>2+<\/sup>], in a solution prepared by mixing 0.100 L of 0.0100 <em>M<\/em> Cd(NO<sub>3<\/sub>)<sub>2<\/sub> with 1.150 L of 0.100 NH<sub>3<\/sub>(<em>aq<\/em>).<\/li>\r\n \t<li>Explain why addition of NH<sub>3<\/sub> or HNO<sub>3<\/sub> to a saturated solution of Cu(OH)<sub>2<\/sub> in contact with solid Cu(OH)<sub>2<\/sub> increases the solubility of the solid.<\/li>\r\n \t<li>Sometimes equilibria for complex ions are described in terms of dissociation constants, <em>K<\/em><sub>d<\/sub>. For the complex ion [latex]{\\text{AlF}}_{6}{}^{\\text{3-}}[\/latex] the dissociation reaction is:[latex]{\\text{AlF}}_{6}^{3-}\\rightleftharpoons {\\text{Al}}^{\\text{3+}}+6{\\text{F}}^{-}[\/latex]\u00a0and\u00a0[latex]{K}_{\\text{d}}=\\frac{\\left[{\\text{Al}}^{\\text{3+}}\\right]{\\left[{\\text{F}}^{-}\\right]}^{6}}{\\left[{\\text{AlF}}_{6}^{3-}\\right]}=2\\times {10}^{-24}[\/latex]Calculate the value of the formation constant, <em>K<\/em><sub>f<\/sub>, for [latex]{\\text{AlF}}_{\\text{6}}^{\\text{3}-}[\/latex].<\/li>\r\n \t<li>Using the value of the formation constant for the complex ion [latex]\\text{Co}{\\left({\\text{NH}}_{3}\\right)}_{6}{}^{\\text{2+}}[\/latex], calculate the dissociation constant.<\/li>\r\n \t<li>Using the dissociation constant, <em>K<\/em><sub>d<\/sub> = 7.8 [latex]\\times [\/latex] 10<sup>\u201318<\/sup>, calculate the equilibrium concentrations of Cd<sup>2+<\/sup> and CN<sup>\u2013<\/sup> in a 0.250-<em>M<\/em> solution of [latex]\\text{Cd}{\\left(\\text{CN}\\right)}_{4}^{2-}[\/latex].<\/li>\r\n \t<li>Using the dissociation constant, <em>K<\/em><sub>d<\/sub> = 3.4 [latex]\\times [\/latex] 10<sup>\u201315<\/sup>, calculate the equilibrium concentrations of Zn<sup>2+<\/sup> and OH<sup>\u2013<\/sup> in a 0.0465-<em>M<\/em> solution of [latex]\\text{Zn}{\\left(\\text{OH}\\right)}_{4}^{2-}[\/latex].<\/li>\r\n \t<li>Using the dissociation constant, <em>K<\/em><sub>d<\/sub> = 2.2 [latex]\\times [\/latex] 10<sup>\u201334<\/sup>, calculate the equilibrium concentrations of Co<sup>3+<\/sup> and NH<sub>3<\/sub> in a 0.500-<em>M<\/em> solution of [latex]\\text{Co}{\\left({\\text{NH}}_{3}\\right)}_{6}^{\\text{3+}}[\/latex].<\/li>\r\n \t<li>Using the dissociation constant, <em>K<\/em><sub>d<\/sub> = 1 [latex]\\times [\/latex] 10<sup>\u201344<\/sup>, calculate the equilibrium concentrations of Fe<sup>3+<\/sup> and CN<sup>\u2013<\/sup> in a 0.333 M solution of [latex]\\text{Fe}{\\left(\\text{CN}\\right)}_{6}^{3-}[\/latex].<\/li>\r\n \t<li>Calculate the mass of potassium cyanide ion that must be added to 100 mL of solution to dissolve 2.0 [latex]\\times [\/latex] 10<sup>\u20132<\/sup> mol of silver cyanide, AgCN.<\/li>\r\n \t<li>Calculate the minimum concentration of ammonia needed in 1.0 L of solution to dissolve 3.0 [latex]\\times [\/latex] 10<sup>\u20133<\/sup> mol of silver bromide.<\/li>\r\n \t<li>A roll of 35-mm black and white photographic film contains about 0.27 g of unexposed AgBr before developing. What mass of Na<sub>2<\/sub>S<sub>2<\/sub>O<sub>3<\/sub>.5H<sub>2<\/sub>O (sodium thiosulfate pentahydrate or hypo) in 1.0 L of developer is required to dissolve the AgBr as [latex]\\text{Ag}{\\left({\\text{S}}_{2}{\\text{O}}_{3}\\right)}_{2}{}^{\\text{3-}}[\/latex] (K<sub>f<\/sub> = 4.7 [latex]\\times [\/latex] 10<sup>13<\/sup>)?<\/li>\r\n \t<li>Calculate [latex]\\left[{\\text{HgCl}}_{4}{}^{2-}\\right][\/latex] in a solution prepared by adding 0.0200 mol of NaCl to 0.250 L of a 0.100-<em>M<\/em> HgCl<sub>2<\/sub> solution.<\/li>\r\n \t<li>In a titration of cyanide ion, 28.72 mL of 0.0100 <em>M<\/em> AgNO<sub>3<\/sub> is added before precipitation begins. [The reaction of Ag<sup>+<\/sup> with CN<sup>\u2013<\/sup> goes to completion, producing the [latex]\\text{Ag}{\\left(\\text{CN}\\right)}_{2}{}^{-}[\/latex] complex.] Precipitation of solid AgCN takes place when excess Ag<sup>+<\/sup> is added to the solution, above the amount needed to complete the formation of [latex]\\text{Ag}{\\left(\\text{CN}\\right)}_{2}{}^{-}[\/latex]. How many grams of NaCN were in the original sample?<\/li>\r\n \t<li>What are the concentrations of Ag<sup>+<\/sup>, CN<sup>\u2013<\/sup>, and [latex]\\text{Ag}{\\left(\\text{CN}\\right)}_{2}{}^{-}[\/latex] in a saturated solution of AgCN?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"246861\"]Show Selected Answers[\/reveal-answer]\r\n[hidden-answer a=\"246861\"]\r\n\r\n1.\u00a0When the amount of solid is so small that a saturated solution is not produced.\r\n\r\n3.\u00a0Cadmium ions associate with ammonia molecules in solution to form the complex ion [latex]{\\left[\\text{Cd}{\\left({\\text{NH}}_{3}\\right)}_{4}\\right]}^{\\text{2+}}[\/latex], which is defined by the following equilibrium:\r\n<p style=\"text-align: center\">[latex]{\\text{Cd}}^{\\text{2+}}\\left(aq\\right)+4{\\text{NH}}_{3}\\left(aq\\right)\\longrightarrow {\\left[\\text{Cd}{\\left({\\text{NH}}_{3}\\right)}_{4}\\right]}^{\\text{2+}}\\left(aq\\right){K}_{\\text{f}}=4.0\\times {10}^{6}[\/latex]<\/p>\r\nThe formation of the complex ion requires 4 mol of NH<sub>3<\/sub> for each mol of Cd<sup>2+<\/sup>. First, calculate the initial amounts of Cd<sup>2+<\/sup> and of NH<sub>3<\/sub> available for association:\r\n<p style=\"text-align: center\">[latex]\\left[{\\text{Cd}}^{\\text{2+}}\\right]=\\frac{\\left(0.100\\text{L}\\right)\\left(0.0100\\text{mol}{\\text{L}}^{-1}\\right)}{0.250\\text{L}}=4.00\\times {10}^{-3}M[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\left[{\\text{NH}}_{3}\\right]=\\frac{\\left(0.150\\text{L}\\right)\\left(0.100\\text{mol}{\\text{L}}^{-1}\\right)}{0.250\\text{L}}=6.00\\times {10}^{-2}M[\/latex]<\/p>\r\nFor the reaction, 4.00 [latex]\\times [\/latex] 10<sup>\u20133<\/sup> mol\/L of Cd<sup>2+<\/sup> would require 4(4.00 [latex]\\times [\/latex] 10<sup>\u20133<\/sup> mol\/L) of NH<sub>3<\/sub> or a 1.6 [latex]\\times [\/latex] 10<sup>\u20132<\/sup>-<em>M<\/em> solution. Due to the large value of <em>K<\/em><sub>f<\/sub> and the substantial excess of NH<sub>3<\/sub>, it can be assumed that the reaction goes to completion with only a small amount of the complex dissociating to form the ions. After reaction, concentrations of the species in the solution are\r\n\r\n[NH<sub>3<\/sub>] = 6.00 [latex]\\times [\/latex] 10<sup>\u20132<\/sup> mol\/L \u2013 1.6 [latex]\\times [\/latex] 10<sup>\u20132<\/sup> mol L<sup>\u20131<\/sup> = 4.4 [latex]\\times [\/latex] 10<sup>\u20132<\/sup><em>M<\/em>\r\n\r\nLet <em>x<\/em> be the change in concentration of [Cd<sup>2+<\/sup>]:\r\n<table>\r\n<thead>\r\n<tr>\r\n<th><\/th>\r\n<th>[Cd(NH<sub>3<\/sub>)<sub>4<\/sub><sup>2+<\/sup>]<\/th>\r\n<th>[Cd<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>4.00\u00a0\u00d7 10<sup>\u22123<\/sup><\/td>\r\n<td>0<\/td>\r\n<td>4.4\u00a0\u00d7 10<sup>\u22122<\/sup><\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Equilibrium (<em>M<\/em>)<\/th>\r\n<td>4.00\u00a0\u00d7 10<sup>\u22123<\/sup> \u2212 <em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<td>4.4\u00a0\u00d7 10<sup>\u22122<\/sup> +\u00a04<em>x<\/em><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p style=\"text-align: center\">[latex]{K}_{\\text{f}}=4.0\\times {10}^{6}=\\frac{\\left[\\text{Cd}{\\left({\\text{NH}}_{3}\\right)}_{4}{}^{\\text{2+}}\\right]}{\\left[{\\text{Cd}}^{\\text{2+}}\\right]{\\left[{\\text{NH}}_{3}\\right]}^{4}}[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]4.00\\times {10}^{6}=\\frac{\\left(4.00\\times {10}^{-3}-x\\right)}{\\left(x\\right){\\left(4.4\\times {10}^{-2}+4x\\right)}^{4}}[\/latex]<\/p>\r\nAs <em>x<\/em> is expected to be about the same size as the number from which it is subtracted, the entire expression must be expanded and solved, in this case, by successive approximations where substitution of values for <em>x<\/em> into the equation continues until the remainder is judged small enough. This is a slightly different method than used in most problems. We have:\r\n<p style=\"text-align: center\">4.0 [latex]\\times [\/latex] 10<sup>6<\/sup><em>x<\/em> (4.4 [latex]\\times [\/latex] 10<sup>\u20132<\/sup> + 4<em>x<\/em>)<sup>4<\/sup> = 4.00 [latex]\\times [\/latex] 10<sup>\u20133<\/sup> \u2013 <em>x<\/em><\/p>\r\n<p style=\"text-align: center\">4.0 [latex]\\times [\/latex] 10<sup>6<\/sup><em>x<\/em> (3.75 [latex]\\times [\/latex] 10<sup>\u20136<\/sup> + 1.36 [latex]\\times [\/latex] 10<sup>\u20133<\/sup><em>x<\/em> + 0.186<em>x<\/em><sup>2<\/sup> + 11.264<em>x<\/em><sup>3<\/sup> +256<em>x<\/em><sup>4<\/sup>) = 4.00 [latex]\\times [\/latex] 10<sup>\u20133<\/sup><\/p>\r\n<p style=\"text-align: center\">16<em>x<\/em> + 5440<em>x<\/em><sup>2<\/sup> + 7.44 [latex]\\times [\/latex] 10<sup>5<\/sup><em>x<\/em><sup>3<\/sup> + 4.51 [latex]\\times [\/latex] 10<sup>7<\/sup><em>x<\/em><sup>4<\/sup> + 1.024 [latex]\\times [\/latex] 10<sup>9<\/sup><em>x<\/em> <sup>5<\/sup> = 4.00 [latex]\\times [\/latex] 10<sup>\u20133<\/sup><\/p>\r\nSubstitution of different values <em>x<\/em> will give a number to be compared with 4.00 [latex]\\times [\/latex] 10<sup>\u20133<\/sup>. Using 2.50 [latex]\\times [\/latex] 10<sup>\u20134<\/sup> for <em>x<\/em> gives 4.35 [latex]\\times [\/latex] 10<sup>\u20133<\/sup> compared with 4.00 [latex]\\times [\/latex] 10<sup>\u20133<\/sup>. Using 2.40 [latex]\\times [\/latex] 10<sup>\u20134<\/sup> gives 4.16 [latex]\\times [\/latex] 10<sup>\u20133<\/sup> compared with 4.00 [latex]\\times [\/latex] 10<sup>\u20133<\/sup>. Using 2.30 [latex]\\times [\/latex] 10<sup>\u20134<\/sup> gives 3.98 [latex]\\times [\/latex] 10<sup>\u20133<\/sup> compared with 4.00 [latex]\\times [\/latex] 10<sup>\u20133<\/sup>. Thus 2.30 [latex]\\times [\/latex] 10<sup>\u20134<\/sup> is close enough to the true value of <em>x<\/em> to make the difference equal to zero. If the approximation to drop 4<em>x<\/em> is compared with 4.4 [latex]\\times [\/latex] 10<sup>\u20132<\/sup>, the value of <em>x<\/em> obtained is 2.35 [latex]\\times [\/latex] 10<sup>\u20134<\/sup><em>M<\/em>.\r\n\r\n5.\u00a0For the formation reaction:\r\n\r\n[latex]\\begin{array}{l}{\\text{Al}}^{\\text{3+}}\\left(aq\\right)+6{\\text{F}}^{-}\\left(aq\\right)\\rightleftharpoons {\\text{AlF}}_{6}{}^{\\text{3-}}\\left(aq\\right)\\\\{K}_{\\text{f}}=\\frac{\\left[{\\text{AlF}}_{6}{}^{\\text{3-}}\\right]}{\\left[{\\text{Al}}^{\\text{3+}}\\right]{\\left[{\\text{F}}^{-}\\right]}^{6}}=\\frac{1}{{K}_{\\text{d}}}=\\frac{1}{2\\times {10}^{-24}}=5\\times {10}^{23}\\end{array}[\/latex]\r\n\r\n7.\r\n<table>\r\n<thead>\r\n<tr>\r\n<th><\/th>\r\n<th>[Cd(CN)<sub>4<\/sub><sup>2\u2212<\/sup>]<\/th>\r\n<th>[CN<sup>\u2212<\/sup>]<\/th>\r\n<th>[Cd<sup>2+<\/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>0.250<\/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>0.250\u00a0\u2212\u00a0<em>x<\/em><\/td>\r\n<td>4<em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p style=\"text-align: center\">[latex]{K}_{\\text{d}}=\\frac{\\left[{\\text{Cd}}^{\\text{2+}}\\right]\\left[{\\text{CN}}^{-}\\right]}{\\left[\\text{Cd}{\\left(\\text{CN}\\right)}_{4}{}^{2-}\\right]}=7.8\\times {10}^{-18}=\\frac{x{\\left(4x\\right)}^{4}}{0.250-x}[\/latex]<\/p>\r\nAssume that <em>x<\/em> is small when compared with 0.250 <em>M<\/em>.\r\n\r\n256<em>x<\/em><sup>5<\/sup> = 0.250 [latex]\\times [\/latex] 7.8 [latex]\\times [\/latex] 10<sup>\u201318<\/sup>\r\n\r\n<em>x<\/em><sup>5<\/sup> = 7.617 [latex]\\times [\/latex] 10<sup>\u201321<\/sup>\r\n\r\n<em>x<\/em> = [Cd<sup>2+<\/sup>] = 9.5 [latex]\\times [\/latex] 10<sup>\u20135<\/sup><em>M<\/em>\r\n\r\n4<em>x<\/em> = [CN<sup>\u2013<\/sup>] = 3.8 [latex]\\times [\/latex] 10<sup>\u20134<\/sup><em>M<\/em>\r\n\r\n9.\r\n<table>\r\n<thead>\r\n<tr>\r\n<th><\/th>\r\n<th>[Co(NH<sub>3<\/sub>)<sub>6<\/sub><sup>3+<\/sup>]<\/th>\r\n<th>[Co<sup>3+<\/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>0.500<\/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>0.500\u00a0\u2212\u00a0<em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<td>6<em>x<\/em><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p style=\"text-align: center\">[latex]{K}_{\\text{d}}=\\frac{\\left[{\\text{Co}}^{\\text{2+}}\\right]{\\left[{\\text{NH}}_{3}\\right]}^{6}}{\\left[\\text{Co}{\\left({\\text{NH}}_{3}\\right)}_{6}{}^{\\text{3+}}\\right]}=\\frac{x{\\left(6x\\right)}^{6}}{0.500-x}=2.2\\times {10}^{-34}[\/latex]<\/p>\r\nAssume that <em>x<\/em> is small when compared with 0.500 <em>M<\/em>.\r\n\r\n4.67 [latex]\\times [\/latex] 104<em>x<\/em><sup>7<\/sup> = 0.500 [latex]\\times [\/latex] 2.2 [latex]\\times [\/latex] 10<sup>\u201334<\/sup>\r\n\r\n<em>x<\/em><sup>7<\/sup> = 2.358 [latex]\\times [\/latex] 10<sup>\u201339<\/sup>\r\n\r\n<em>x<\/em> = [Co<sup>3+<\/sup>] = 3.0 [latex]\\times [\/latex] 10<sup>\u20136<\/sup><em>M<\/em>\r\n\r\n6<em>x<\/em> = [NH<sub>3<\/sub>] = 1.8 [latex]\\times [\/latex] 10<sup>\u20135<\/sup><em>M<\/em>\r\n\r\n11.\u00a0Because <em>K<\/em><sub>sp<\/sub> is small and <em>K<\/em><sub>f<\/sub> is large, most of the Ag<sup>+<\/sup> is used to form [latex]\\text{Ag}{\\left(\\text{CN}\\right)}_{2}{}^{-}[\/latex]; that is:\r\n<p style=\"text-align: center\">[latex]\\begin{array}{rll}\\left[\\text{Ag}^{+}\\right]&amp;\\lt&amp;\\left[\\text{Ag}\\left(\\text{CN}\\right)_{2}^{-}\\right]\\\\\\left[\\text{Ag}\\left(\\text{CN}\\right)_{2}^{-}\\right]&amp;\\approx&amp;2.0\\times{10}^{-1}M\\end{array}[\/latex]<\/p>\r\nThe CN<sup>\u2013<\/sup> from the dissolution and the added CN<sup>\u2013<\/sup> exist as CN<sup>\u2013<\/sup> and [latex]\\text{Ag}{\\left(\\text{CN}\\right)}_{2}{}^{-}[\/latex]. Let <em>x<\/em> be the change in concentration upon addition of CN<sup>\u2013<\/sup>. Its initial concentration is approximately 0.\r\n<p style=\"text-align: center\">[CN<sup>\u2013<\/sup>] + 2 [latex]\\left[\\text{Ag}{\\left(\\text{CN}\\right)}_{2}{}^{-}\\right][\/latex] = 2 [latex]\\times [\/latex] 10<sup>\u20131<\/sup> + <em>x<\/em><\/p>\r\nBecause <em>K<\/em><sub>sp<\/sub> is small and <em>K<\/em><sub>f<\/sub> is large, most of the CN<sup>\u2013<\/sup> is used to form [latex]\\left[\\text{Ag}{\\left(\\text{CN}\\right)}_{2}{}^{-}\\right][\/latex]; that is:\r\n<p style=\"text-align: center\">[latex]\\begin{array}{rll}\\left[{\\text{CN}}^{-}\\right]&amp;&lt;&amp;2\\left[\\text{Ag}{\\left(\\text{CN}\\right)}_{2}{}^{-}\\right]\\\\2\\left[\\text{Ag}{\\left(\\text{CN}\\right)}_{2}{}^{-}\\right]&amp;\\approx&amp;2.0\\times {10}^{-1}+x\\end{array}[\/latex]<\/p>\r\n2(2.0 [latex]\\times [\/latex] 10<sup>\u20131<\/sup>) \u2013 2.0 [latex]\\times [\/latex] 10<sup>\u20131<\/sup> = <em>x<\/em>\r\n\r\n2.0 [latex]\\times [\/latex] 10<sup>\u20131<\/sup><em>M<\/em> [latex]\\times [\/latex] L = mol CN<sup>\u2013<\/sup> added\r\n\r\nThe solution has a volume of 100 mL.\r\n\r\n2 [latex]\\times [\/latex] 10<sup>\u20131<\/sup> mol\/L [latex]\\times [\/latex] 0.100 L = 2 [latex]\\times [\/latex] 10<sup>\u20132<\/sup> mol\r\n\r\nmass KCN = 2.0 [latex]\\times [\/latex] 10<sup>\u20132<\/sup> mol KCN [latex]\\times [\/latex] 65.120 g\/mol = 1.3 g\r\n\r\n13.\u00a0The reaction is governed by two equilibria, both of which must be satisfied:\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\text{AgBr}\\left(s\\right)\\rightleftharpoons\\text{Ag}^{+}\\left(aq\\right)+\\text{Br}^{-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,{;}\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{sp}}=3.3\\times{10}^{-13}\\\\\\text{Ag}^{+}\\left(aq\\right)+2{\\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}\\end{array}[\/latex]<\/p>\r\nThe overall equilibrium is obtained by adding the two equations and multiplying their <em>K<\/em>s:\r\n<p style=\"text-align: center\">[latex]\\frac{\\left[\\text{Ag}{\\left({\\text{S}}_{2}{\\text{O}}_{3}\\right)}^{\\text{3-}}\\right]\\left[{\\text{Br}}^{-}\\right]}{{\\left[{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}\\right]}^{2}}=15.51[\/latex]<\/p>\r\nIf all Ag is to be dissolved, the concentration of the complex is the molar concentration of AgBr.\r\n\r\nformula mass (AgBr) = 187.772 g\/mol\r\n\r\n[latex]\\text{moles present}=\\frac{0.27\\text{g}\\text{AgBr}}{187.772\\text{g}{\\text{mol}}^{-1}}=1.438\\times {10}^{-3}\\text{mol}[\/latex]\r\n\r\nLet <em>x<\/em> be the change in concentration of [latex]{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}:[\/latex]\r\n<table>\r\n<thead>\r\n<tr>\r\n<th><\/th>\r\n<th>[Ag<sup>+<\/sup>]<\/th>\r\n<th>[S<sub>2<\/sub>O<sub>3<\/sub><sup>2\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<\/td>\r\n<td>0<\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Equilibrium (<em>M<\/em>)<\/th>\r\n<td>[latex]\\frac{1}{2}x[\/latex]<\/td>\r\n<td><em>x<\/em><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[latex]\\frac{\\left(1.438\\times {10}^{-3}\\right)\\left(1.438\\times {10}^{-3}\\right)}{{x}^{2}}=15.51[\/latex]\r\n\r\n<em>x<\/em><sup>2<\/sup> = 1.333 [latex]\\times [\/latex] 10<sup>\u20137<\/sup>\r\n\r\n[latex]x=3.65\\times {10}^{-4}M=\\left[{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}\\right][\/latex]\r\n\r\nThe formula mass of Na<sub>2<\/sub>S<sub>2<\/sub>O<sub>3<\/sub>\u20225H<sub>2<\/sub>O is 248.13 g\/mol. The total [latex]\\left[{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}\\right][\/latex] needed is:\r\n\r\n2(1.438 [latex]\\times [\/latex] 10<sup>\u20133<\/sup>) + 3.65 [latex]\\times [\/latex] 10<sup>\u20134<\/sup> = 3.241 [latex]\\times [\/latex] 10<sup>\u20133<\/sup> mol\r\n\r\ng(hypo) = 3.241 [latex]\\times [\/latex] 10<sup>\u20133<\/sup> mol [latex]\\times [\/latex] 248.13 g\/mol = 0.80 g\r\n\r\n15. The equilibrium is:[latex]{\\text{Ag}}^{\\text{+}}\\left(aq\\right)+2{\\text{CN}}^{-}\\left(aq\\right)\\rightleftharpoons \\text{Ag}{\\left(\\text{CN}\\right)}_{2}{}^{-}\\left(aq\\right){K}_{\\text{f}}=1\\times {10}^{20}[\/latex]\r\n\r\n&nbsp;\r\n\r\nThe number of moles of AgNO<sub>3<\/sub> added is:\r\n\r\n0.02872 L [latex]\\times [\/latex] 0.0100 mol\/L = 2.87 [latex]\\times [\/latex] 10<sup>\u20134<\/sup> mol\r\n\r\nThis compound reacts with CN<sup>\u2013<\/sup> to form [latex]\\text{Ag}{\\left(\\text{CN}\\right)}_{2}{}^{-}[\/latex], so there are 2.87 [latex]\\times [\/latex] 10<sup>\u20134<\/sup> mol [latex]\\text{Ag}{\\left(\\text{CN}\\right)}_{2}{}^{-}[\/latex]. This amount requires 2 [latex]\\times [\/latex] 2.87 [latex]\\times [\/latex] 10<sup>\u20134<\/sup> mol, or 5.74 [latex]\\times [\/latex] 10<sup>\u20134<\/sup> mol, of CN<sup>\u2013<\/sup>. The titration is stopped just as precipitation of AgCN begins:\r\n\r\n[latex]{\\text{AgCN}}_{2}{}^{-}\\left(aq\\right)+{\\text{Ag}}^{\\text{+}}\\left(aq\\right)\\rightleftharpoons 2\\text{AgCN}\\left(s\\right)[\/latex]\r\n\r\nso only the first equilibrium is applicable. The value of <em>K<\/em><sub>f<\/sub> is very large.\r\n\r\nmol CN<sup>\u2013<\/sup> &lt; [latex]\\left[\\text{Ag}{\\left(\\text{CN}\\right)}_{2}{}^{-}\\right][\/latex]\r\n\r\nmol NaCN = 2 mol [ [latex]\\text{Ag}{\\left(\\text{CN}\\right)}_{2}{}^{-}[\/latex] ] = 5.74 [latex]\\times [\/latex] 10<sup>\u20134<\/sup> mol\r\n\r\n[latex]\\text{mass}\\left(\\text{NaCN}\\right)=5.74\\times {10}^{-4}\\text{mol}\\times \\frac{49.007\\text{g}}{1\\text{mol}}=0.0281\\text{g}[\/latex]\r\n\r\n&nbsp;\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Glossary<\/h2>\r\n<b>complex ion:\u00a0<\/b>ion consisting of a transition metal central atom and surrounding molecules or ions called ligands\r\n\r\n<b>dissociation constant:\u00a0<\/b>(<em>K<\/em><sub>d<\/sub>) equilibrium constant for the decomposition of a complex ion into its components in solution\r\n\r\n<b>formation constant:\u00a0<\/b>(<em>K<\/em><sub>f<\/sub>) (also, stability constant) equilibrium constant for the formation of a complex ion from its components in solution\r\n\r\n&nbsp;\r\n\r\n<b>ligand:\u00a0<\/b>molecule or ion that surrounds a transition metal and forms a complex ion; ligands act as Lewis bases","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<p>By the end of this module, you will be able to:<\/p>\n<ul>\n<li>Write equations for the formation of complex ions<\/li>\n<li>Perform equilibrium calculations involving formation constants<\/li>\n<\/ul>\n<\/div>\n<p>&nbsp;<\/p>\n<p>Many slightly soluble ionic solids dissolve when the concentration of the metal ion in solution is decreased through the formation of complex (polyatomic) ions in a Lewis acid-base reaction. For example, silver chloride dissolves in a solution of ammonia because the silver ion reacts with ammonia to form the <b>complex ion<\/b> [latex]\\text{Ag}{\\left({\\text{NH}}_{3}\\right)}_{2}{}^{+}[\/latex]. The Lewis structure of the [latex]\\text{Ag}{\\left({\\text{NH}}_{3}\\right)}_{2}{}^{+}[\/latex] ion is:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23214007\/CNX_Chem_15_02_AgNH32_img.jpg\" alt=\"A structure is shown in brackets. The structure has a central A g atom to which N atoms are single bonded to the left and right. Each of these atoms N atom has H atoms single bonded above, below, and to the outer end of the structure. Outside the brackets is a superscripted plus.\" width=\"325\" height=\"169\" \/><\/p>\n<p>The equations for the dissolution of AgCl in a solution of NH<sub>3<\/sub> are:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{rrll}{}&\\text{AgCl}\\left(s\\right)&\\longrightarrow&{\\text{Ag}}^{\\text{+}}\\left(aq\\right)+{\\text{Cl}}^{-}\\left(aq\\right)\\\\{}&{\\text{Ag}}^{\\text{+}}\\left(aq\\right)+2{\\text{NH}}_{3}\\left(aq\\right)&\\longrightarrow&\\text{Ag}{\\left({\\text{NH}}_{3}\\right)}_{2}{}^{\\text{+}}\\left(aq\\right)\\\\\\text{Net: }&\\text{AgCl}\\left(s\\right)+2{\\text{NH}}_{3}\\left(aq\\right)&\\longrightarrow &\\text{Ag}{\\left({\\text{NH}}_{3}\\right)}_{2}{}^{\\text{+}}\\left(aq\\right)+{\\text{Cl}}^{-}\\left(aq\\right)\\end{array}[\/latex]<\/p>\n<p>Aluminum hydroxide dissolves in a solution of sodium hydroxide or another strong base because of the formation of the complex ion [latex]\\text{Al}{\\left(\\text{OH}\\right)}_{4}{}^{-}[\/latex]. The Lewis structure of the [latex]\\text{Al}{\\left(\\text{OH}\\right)}_{4}{}^{-}[\/latex] ion is:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23214008\/CNX_Chem_15_02_AlOH4_img.jpg\" alt=\"An H atom is bonded to an O atom. The O atom has 2 dots above it and 2 dots below it. The O atom is bonded to an A l atom, which has three additional O atoms bonded to it as well. Each of these additional O atoms has 4 dots arranged around it, and is bonded to an H atom. This entire molecule is contained in brackets, to the right of which is a superscripted negative sign.\" width=\"325\" height=\"169\" \/><\/p>\n<p>The equations for the dissolution are:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{rrll}{}&\\text{Al}{\\left(\\text{OH}\\right)}_{3}\\left(s\\right)&\\longrightarrow& {\\text{Al}}^{\\text{3+}}\\left(aq\\right)+3{\\text{OH}}^{-}\\left(aq\\right)\\\\{}&{\\text{Al}}^{\\text{3+}}\\left(aq\\right)+4{\\text{OH}}^{-}\\left(aq\\right)&\\longrightarrow&\\text{Al}{\\left(\\text{OH}\\right)}_{4}{}^{-}\\left(aq\\right)\\\\\\text{Net:}&\\text{Al}{\\left(\\text{OH}\\right)}_{3}\\left(s\\right)+{\\text{OH}}^{-}\\left(aq\\right)&\\longrightarrow&\\text{Al}{\\left(\\text{OH}\\right)}_{4}{}^{-}\\left(aq\\right)\\end{array}[\/latex]<\/p>\n<p>Mercury(II) sulfide dissolves in a solution of sodium sulfide because HgS reacts with the S<sup>2\u2013<\/sup> ion:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{rrll}{}&\\text{HgS}\\left(s\\right)&\\longrightarrow&{\\text{Hg}}^{\\text{2+}}\\left(aq\\right)+{\\text{S}}^{2-}\\left(aq\\right)\\\\{}&{\\text{Hg}}^{\\text{2+}}\\left(aq\\right)+2{\\text{S}}^{2-}\\left(aq\\right)&\\longrightarrow&{\\text{HgS}}_{2}{}^{2-}\\left(aq\\right)\\\\\\text{Net:}&\\text{HgS}\\left(s\\right)+{\\text{S}}^{2-}\\left(aq\\right)&\\longrightarrow &{\\text{HgS}}_{2}{}^{2-}\\left(aq\\right)\\end{array}[\/latex]<\/p>\n<p>A complex ion consists of a central atom, typically a transition metal cation, surrounded by ions, or molecules called <b>ligands<\/b>. These ligands can be neutral molecules like H<sub>2<\/sub>O or NH<sub>3<\/sub>, or ions such as CN<sup>\u2013<\/sup> or OH<sup>\u2013<\/sup>. Often, the ligands act as Lewis bases, donating a pair of electrons to the central atom. The ligands aggregate themselves around the central atom, creating a new ion with a charge equal to the sum of the charges and, most often, a transitional metal ion. This more complex arrangement is why the resulting ion is called a <em>complex ion<\/em>. The complex ion formed in these reactions cannot be predicted; it must be determined experimentally. The types of bonds formed in complex ions are called coordinate covalent bonds, as electrons from the ligands are being shared with the central atom. Because of this, complex ions are sometimes referred to as coordination complexes. This will be studied further in upcoming chapters.<\/p>\n<p>The equilibrium constant for the reaction of the components of a complex ion to form the complex ion in solution is called a <b>formation constant (<em>K<\/em><sub>f<\/sub>)<\/b> (sometimes called a stability constant). For example, the complex ion [latex]\\text{Cu}{\\left(\\text{CN}\\right)}_{2}{}^{-}[\/latex] is shown here:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23214009\/CNX_Chem_15_02_CuCN2-_img.jpg\" alt=\"A Cu atom is bonded to two C atoms. Each of these C atoms is triple bonded to an N atom. Each N atom has two dots on the side of it.\" width=\"325\" height=\"43\" \/><\/p>\n<p>It forms by the reaction:<\/p>\n<p style=\"text-align: center\">[latex]{\\text{Cu}}^{\\text{+}}\\left(aq\\right)+2{\\text{CN}}^{-}\\left(aq\\right)\\rightleftharpoons \\text{Cu}{\\left(\\text{CN}\\right)}_{2}{}^{-}\\left(aq\\right)[\/latex]<\/p>\n<p>At equilibrium:<\/p>\n<p style=\"text-align: center\">[latex]{K}_{\\text{f}}=Q=\\frac{\\left[\\text{Cu}{\\left(\\text{CN}\\right)}_{2}{}^{-}\\right]}{\\left[{\\text{Cu}}^{+}\\right]{\\left[{\\text{CN}}^{-}\\right]}^{2}}[\/latex]<\/p>\n<p>The inverse of the formation constant is the <b>dissociation constant (<em>K<\/em><sub>d<\/sub>)<\/b>, the equilibrium constant for the <em>decomposition<\/em> of a complex ion into its components in solution. We will work with dissociation constants further in the exercises for this section. <a href=\".\/chapter\/formation-constants-for-complex-ions\/\" target=\"_blank\" rel=\"noopener\">Formation Constants for Complex Ions<\/a>\u00a0and Table\u00a01\u00a0are tables of formation constants. In general, the larger the formation constant, the more stable the complex; however, as in the case of <em>K<\/em><sub>sp<\/sub> values, the stoichiometry of the compound must be considered.<\/p>\n<table id=\"fs-idm57488576\" class=\"span-all\" summary=\"No Summary Text\">\n<thead>\n<tr valign=\"middle\">\n<th colspan=\"2\">Table\u00a01. Common Complex Ions by Decreasing Formulation Constants<\/th>\n<\/tr>\n<tr valign=\"middle\">\n<th>Substance<\/th>\n<th><em>K<\/em><sub>f<\/sub> at 25 \u00b0C<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"middle\">\n<td>[latex]{\\left[\\text{Cd}{\\left(\\text{CN}\\right)}_{4}\\right]}^{2-}[\/latex]<\/td>\n<td>1.3 [latex]\\times[\/latex] 10<sup>7<\/sup><\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td>[latex]\\text{Ag}{\\left({\\text{NH}}_{3}\\right)}_{2}{}^{+}[\/latex]<\/td>\n<td>1.7 [latex]\\times[\/latex] 10<sup>7<\/sup><\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td>[latex]{\\left[{\\text{AlF}}_{6}\\right]}^{\\text{3-}}[\/latex]<\/td>\n<td>7 [latex]\\times[\/latex] 10<sup>19<\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>As an example of dissolution by complex ion formation, let us consider what happens when we add aqueous ammonia to a mixture of silver chloride and water. Silver chloride dissolves slightly in water, giving a small concentration of Ag<sup>+<\/sup> ([Ag<sup>+<\/sup>] = 1.3 [latex]\\times[\/latex] 10<sup>\u20135\u00a0<\/sup><em>M<\/em>):<\/p>\n<p style=\"text-align: center\">[latex]\\text{AgCl}\\left(s\\right)\\rightleftharpoons {\\text{Ag}}^{\\text{+}}\\left(aq\\right)+{\\text{Cl}}^{-}\\left(aq\\right)[\/latex]<\/p>\n<p>However, if NH<sub>3<\/sub> is present in the water, the complex ion, [latex]\\text{Ag}{\\left({\\text{NH}}_{3}\\right)}_{2}{}^{+}[\/latex], can form according to the equation:<\/p>\n<p style=\"text-align: center\">[latex]{\\text{Ag}}^{\\text{+}}\\left(aq\\right)+2{\\text{NH}}_{3}\\left(aq\\right)\\rightleftharpoons \\text{Ag}{\\left({\\text{NH}}_{3}\\right)}_{2}{}^{\\text{+}}\\left(aq\\right)[\/latex]<\/p>\n<p>with<\/p>\n<p style=\"text-align: center\">[latex]{K}_{\\text{f}}=\\frac{\\left[\\text{Ag}{\\left({\\text{NH}}_{3}\\right)}_{2}{}^{+}\\right]}{\\left[{\\text{Ag}}^{+}\\right]{\\left[{\\text{NH}}_{3}\\right]}^{2}}=1.6\\times {10}^{7}[\/latex]<\/p>\n<p>The large size of this formation constant indicates that most of the free silver ions produced by the dissolution of AgCl combine with NH<sub>3<\/sub> to form [latex]\\text{Ag}{\\left({\\text{NH}}_{3}\\right)}_{2}{}^{+}[\/latex]. As a consequence, the concentration of silver ions, [Ag<sup>+<\/sup>], is reduced, and the reaction quotient for the dissolution of silver chloride, [Ag<sup>+<\/sup>][Cl<sup>\u2013<\/sup>], falls below the solubility product of AgCl:<\/p>\n<p style=\"text-align: center\">[latex]Q=\\left[{\\text{Ag}}^{+}\\right]\\left[{\\text{Cl}}^{-}\\right]<{K}_{\\text{sp}}[\/latex]<\/p>\n<p>More silver chloride then dissolves. If the concentration of ammonia is great enough, all of the silver chloride dissolves.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 1:\u00a0Dissociation of a Complex Ion<\/h3>\n<p>Calculate the concentration of the silver ion in a solution that initially is 0.10 <em>M<\/em> with respect to [latex]\\text{Ag}{\\left({\\text{NH}}_{3}\\right)}_{2}{}^{+}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q576838\">Show Answer<\/span><\/p>\n<div id=\"q576838\" class=\"hidden-answer\" style=\"display: none\">\n<p>We use the familiar path to solve this problem:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23214010\/CNX_Chem_15_02_Format_img.jpg\" alt=\"Four boxes are shown side by side, with three right facing arrows connecting them. The first box contains the text \u201cDetermine the direction of change.\u201d The second box contains the text \u201cDetermine x and the equilibrium concentrations.\u201d The third box contains the text \u201cSolve for x and the equilibrium concentrations.\u201d The fourth box contains the text \u201cCheck the math.\u201d\" width=\"881\" height=\"156\" \/><\/p>\n<ol>\n<li><em>Determine the direction of change.<\/em> The complex ion [latex]\\text{Ag}{\\left({\\text{NH}}_{3}\\right)}_{2}{}^{+}[\/latex] is in equilibrium with its components, as represented by the equation:[latex]{\\text{Ag}}^{\\text{+}}\\left(aq\\right)+2{\\text{NH}}_{3}\\left(aq\\right)\\rightleftharpoons \\text{Ag}{\\left({\\text{NH}}_{3}\\right)}_{2}{}^{\\text{+}}\\left(aq\\right)[\/latex]We write the equilibrium as a formation reaction because\u00a0<a href=\".\/chapter\/formation-constants-for-complex-ions\/\" target=\"_blank\" rel=\"noopener\">Formation Constants for Complex Ions<\/a>\u00a0lists formation constants for complex ions. Before equilibrium, the reaction quotient is larger than the equilibrium constant [<em>K<\/em><sub>f<\/sub> = 1.6 [latex]\\times[\/latex] 10<sup>7<\/sup>, and [latex]Q=\\frac{0.10}{0\\times 0}[\/latex], it is infinitely large], so the reaction shifts to the left to reach equilibrium.<\/li>\n<li><em>Determine<\/em> x<em> and equilibrium concentrations.<\/em> We let the change in concentration of Ag<sup>+<\/sup> be <em>x<\/em>. Dissociation of 1 mol of [latex]\\text{Ag}{\\left({\\text{NH}}_{3}\\right)}_{2}{}^{+}[\/latex] gives 1 mol of Ag<sup>+<\/sup> and 2 mol of NH<sub>3<\/sub>, so the change in [NH<sub>3<\/sub>] is 2<em>x<\/em> and that of [latex]\\text{Ag}{\\left({\\text{NH}}_{3}\\right)}_{2}{}^{+}[\/latex] is \u2013<em>x<\/em>. In summary:<img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23214011\/CNX_Chem_15_02_ICETable1_img.jpg\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), and Equilibrium concentration ( M ). The second column has the header, \u201cA g superscript positive sign plus 2 N H subscript 3 equilibrium sign A g ( N H subscript 3 ) subscript 2 superscript positive sign.\u201d Under the second column is a subgroup of three rows and three columns. The first column contains: 0, x, and 0 plus x. The second column contains: 0, 2 x, and 0 plus 2 x. The third column contains 0.10, negative x, and 0.10 minus x.\" width=\"842\" height=\"228\" \/><\/li>\n<li><em>Solve for x and the equilibrium concentrations.<\/em> At equilibrium:\u00a0 [latex]{K}_{\\text{f}}=\\frac{\\left[\\text{Ag}{\\left({\\text{NH}}_{3}\\right)}_{2}{}^{+}\\right]}{\\left[{\\text{Ag}}^{+}\\right]{\\left[{\\text{NH}}_{3}\\right]}^{2}}[\/latex]<br \/>\n[latex]1.6\\times {10}^{7}\\text{=}\\frac{0.10-x}{\\left(x\\right){\\left(2x\\right)}^{2}}[\/latex]\u00a0 Both <em>Q<\/em> and <em>K<\/em><sub>f<\/sub> are much larger than 1, so let us assume that the changes in concentrations needed to reach equilibrium are small. Thus 0.10 \u2013 <em>x<\/em> is approximated as 0.10:\u00a0 [latex]1.6\\times {10}^{7}=\\frac{0.10}{\\left(x\\right){\\left(2x\\right)}^{2}}[\/latex]<br \/>\n[latex]{x}^{3}=\\frac{0.10}{4\\left(1.6\\times {10}^{7}\\right)}=1.6\\times {10}^{-9}[\/latex]<br \/>\n[latex]x=\\sqrt[3]{1.6\\times {10}^{-9}}=1.2\\times {10}^{-3}[\/latex]\u00a0 Because only 1.2% of the [latex]\\text{Ag}{\\left({\\text{NH}}_{3}\\right)}_{2}{}^{+}[\/latex] dissociates into Ag<sup>+<\/sup> and NH<sub>3<\/sub>, The assumption that <em>x<\/em> is small is justified.Now we determine the equilibrium concentrations: [latex]\\left[{\\text{Ag}}^{+}\\right]=0+x=1.2\\times {10}^{-3}M[\/latex]<br \/>\n[latex]\\left[{\\text{NH}}_{3}\\right]=0+2x=2.4\\times {10}^{-3}M[\/latex]<br \/>\n[latex]\\left[\\text{Ag}{\\left({\\text{NH}}_{3}\\right)}_{2}{}^{+}\\right]=0.10-x=0.10 - 0.0012=0.099[\/latex]\u00a0 The concentration of free silver ion in the solution is 0.0012 <em>M<\/em>.<\/li>\n<li><em>Check the work.<\/em> The value of <em>Q<\/em> calculated using the equilibrium concentrations is equal to <em>K<\/em><sub>f<\/sub> within the error associated with the significant figures in the calculation.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<h4>Check Your Learning<\/h4>\n<p>Calculate the silver ion concentration, [Ag<sup>+<\/sup>], of a solution prepared by dissolving 1.00 g of AgNO<sub>3<\/sub> and 10.0 g of KCN in sufficient water to make 1.00 L of solution. (Hint: Because <em>Q<\/em> &lt; <em>K<\/em><sub>f<\/sub>, assume the reaction goes to completion then calculate the [Ag<sup>+<\/sup>] produced by dissociation of the complex.)<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q747092\">Show Answer<\/span><\/p>\n<div id=\"q747092\" class=\"hidden-answer\" style=\"display: none\">[latex]3\\times10^{-22}M[\/latex]<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Key Takeaways<\/h3>\n<p>Complex ions are examples of Lewis acid-base adducts. In a complex ion, we have a central atom, often consisting of a transition metal cation, which acts as a Lewis acid, and several neutral molecules or ions surrounding them called ligands that act as Lewis bases. Complex ions form by sharing electron pairs to form coordinate covalent bonds. The equilibrium reaction that occurs when forming a complex ion has an equilibrium constant associated with it called a formation constant, <em>K<\/em><sub>f<\/sub>. This is often referred to as a stability constant, as it represents the stability of the complex ion. Formation of complex ions in solution can have a profound effect on the solubility of a transition metal compound.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Exercises<\/h3>\n<ol>\n<li>Under what circumstances, if any, does a sample of solid AgCl completely dissolve in pure water?<\/li>\n<li>Explain why the addition of NH<sub>3<\/sub> or HNO<sub>3<\/sub> to a saturated solution of Ag<sub>2<\/sub>CO<sub>3<\/sub> in contact with solid Ag<sub>2<\/sub>CO<sub>3<\/sub> increases the solubility of the solid.<\/li>\n<li>Calculate the cadmium ion concentration, [Cd<sup>2+<\/sup>], in a solution prepared by mixing 0.100 L of 0.0100 <em>M<\/em> Cd(NO<sub>3<\/sub>)<sub>2<\/sub> with 1.150 L of 0.100 NH<sub>3<\/sub>(<em>aq<\/em>).<\/li>\n<li>Explain why addition of NH<sub>3<\/sub> or HNO<sub>3<\/sub> to a saturated solution of Cu(OH)<sub>2<\/sub> in contact with solid Cu(OH)<sub>2<\/sub> increases the solubility of the solid.<\/li>\n<li>Sometimes equilibria for complex ions are described in terms of dissociation constants, <em>K<\/em><sub>d<\/sub>. For the complex ion [latex]{\\text{AlF}}_{6}{}^{\\text{3-}}[\/latex] the dissociation reaction is:[latex]{\\text{AlF}}_{6}^{3-}\\rightleftharpoons {\\text{Al}}^{\\text{3+}}+6{\\text{F}}^{-}[\/latex]\u00a0and\u00a0[latex]{K}_{\\text{d}}=\\frac{\\left[{\\text{Al}}^{\\text{3+}}\\right]{\\left[{\\text{F}}^{-}\\right]}^{6}}{\\left[{\\text{AlF}}_{6}^{3-}\\right]}=2\\times {10}^{-24}[\/latex]Calculate the value of the formation constant, <em>K<\/em><sub>f<\/sub>, for [latex]{\\text{AlF}}_{\\text{6}}^{\\text{3}-}[\/latex].<\/li>\n<li>Using the value of the formation constant for the complex ion [latex]\\text{Co}{\\left({\\text{NH}}_{3}\\right)}_{6}{}^{\\text{2+}}[\/latex], calculate the dissociation constant.<\/li>\n<li>Using the dissociation constant, <em>K<\/em><sub>d<\/sub> = 7.8 [latex]\\times[\/latex] 10<sup>\u201318<\/sup>, calculate the equilibrium concentrations of Cd<sup>2+<\/sup> and CN<sup>\u2013<\/sup> in a 0.250-<em>M<\/em> solution of [latex]\\text{Cd}{\\left(\\text{CN}\\right)}_{4}^{2-}[\/latex].<\/li>\n<li>Using the dissociation constant, <em>K<\/em><sub>d<\/sub> = 3.4 [latex]\\times[\/latex] 10<sup>\u201315<\/sup>, calculate the equilibrium concentrations of Zn<sup>2+<\/sup> and OH<sup>\u2013<\/sup> in a 0.0465-<em>M<\/em> solution of [latex]\\text{Zn}{\\left(\\text{OH}\\right)}_{4}^{2-}[\/latex].<\/li>\n<li>Using the dissociation constant, <em>K<\/em><sub>d<\/sub> = 2.2 [latex]\\times[\/latex] 10<sup>\u201334<\/sup>, calculate the equilibrium concentrations of Co<sup>3+<\/sup> and NH<sub>3<\/sub> in a 0.500-<em>M<\/em> solution of [latex]\\text{Co}{\\left({\\text{NH}}_{3}\\right)}_{6}^{\\text{3+}}[\/latex].<\/li>\n<li>Using the dissociation constant, <em>K<\/em><sub>d<\/sub> = 1 [latex]\\times[\/latex] 10<sup>\u201344<\/sup>, calculate the equilibrium concentrations of Fe<sup>3+<\/sup> and CN<sup>\u2013<\/sup> in a 0.333 M solution of [latex]\\text{Fe}{\\left(\\text{CN}\\right)}_{6}^{3-}[\/latex].<\/li>\n<li>Calculate the mass of potassium cyanide ion that must be added to 100 mL of solution to dissolve 2.0 [latex]\\times[\/latex] 10<sup>\u20132<\/sup> mol of silver cyanide, AgCN.<\/li>\n<li>Calculate the minimum concentration of ammonia needed in 1.0 L of solution to dissolve 3.0 [latex]\\times[\/latex] 10<sup>\u20133<\/sup> mol of silver bromide.<\/li>\n<li>A roll of 35-mm black and white photographic film contains about 0.27 g of unexposed AgBr before developing. What mass of Na<sub>2<\/sub>S<sub>2<\/sub>O<sub>3<\/sub>.5H<sub>2<\/sub>O (sodium thiosulfate pentahydrate or hypo) in 1.0 L of developer is required to dissolve the AgBr as [latex]\\text{Ag}{\\left({\\text{S}}_{2}{\\text{O}}_{3}\\right)}_{2}{}^{\\text{3-}}[\/latex] (K<sub>f<\/sub> = 4.7 [latex]\\times[\/latex] 10<sup>13<\/sup>)?<\/li>\n<li>Calculate [latex]\\left[{\\text{HgCl}}_{4}{}^{2-}\\right][\/latex] in a solution prepared by adding 0.0200 mol of NaCl to 0.250 L of a 0.100-<em>M<\/em> HgCl<sub>2<\/sub> solution.<\/li>\n<li>In a titration of cyanide ion, 28.72 mL of 0.0100 <em>M<\/em> AgNO<sub>3<\/sub> is added before precipitation begins. [The reaction of Ag<sup>+<\/sup> with CN<sup>\u2013<\/sup> goes to completion, producing the [latex]\\text{Ag}{\\left(\\text{CN}\\right)}_{2}{}^{-}[\/latex] complex.] Precipitation of solid AgCN takes place when excess Ag<sup>+<\/sup> is added to the solution, above the amount needed to complete the formation of [latex]\\text{Ag}{\\left(\\text{CN}\\right)}_{2}{}^{-}[\/latex]. How many grams of NaCN were in the original sample?<\/li>\n<li>What are the concentrations of Ag<sup>+<\/sup>, CN<sup>\u2013<\/sup>, and [latex]\\text{Ag}{\\left(\\text{CN}\\right)}_{2}{}^{-}[\/latex] in a saturated solution of AgCN?<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q246861\">Show Selected Answers<\/span><\/p>\n<div id=\"q246861\" class=\"hidden-answer\" style=\"display: none\">\n<p>1.\u00a0When the amount of solid is so small that a saturated solution is not produced.<\/p>\n<p>3.\u00a0Cadmium ions associate with ammonia molecules in solution to form the complex ion [latex]{\\left[\\text{Cd}{\\left({\\text{NH}}_{3}\\right)}_{4}\\right]}^{\\text{2+}}[\/latex], which is defined by the following equilibrium:<\/p>\n<p style=\"text-align: center\">[latex]{\\text{Cd}}^{\\text{2+}}\\left(aq\\right)+4{\\text{NH}}_{3}\\left(aq\\right)\\longrightarrow {\\left[\\text{Cd}{\\left({\\text{NH}}_{3}\\right)}_{4}\\right]}^{\\text{2+}}\\left(aq\\right){K}_{\\text{f}}=4.0\\times {10}^{6}[\/latex]<\/p>\n<p>The formation of the complex ion requires 4 mol of NH<sub>3<\/sub> for each mol of Cd<sup>2+<\/sup>. First, calculate the initial amounts of Cd<sup>2+<\/sup> and of NH<sub>3<\/sub> available for association:<\/p>\n<p style=\"text-align: center\">[latex]\\left[{\\text{Cd}}^{\\text{2+}}\\right]=\\frac{\\left(0.100\\text{L}\\right)\\left(0.0100\\text{mol}{\\text{L}}^{-1}\\right)}{0.250\\text{L}}=4.00\\times {10}^{-3}M[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\left[{\\text{NH}}_{3}\\right]=\\frac{\\left(0.150\\text{L}\\right)\\left(0.100\\text{mol}{\\text{L}}^{-1}\\right)}{0.250\\text{L}}=6.00\\times {10}^{-2}M[\/latex]<\/p>\n<p>For the reaction, 4.00 [latex]\\times[\/latex] 10<sup>\u20133<\/sup> mol\/L of Cd<sup>2+<\/sup> would require 4(4.00 [latex]\\times[\/latex] 10<sup>\u20133<\/sup> mol\/L) of NH<sub>3<\/sub> or a 1.6 [latex]\\times[\/latex] 10<sup>\u20132<\/sup>&#8211;<em>M<\/em> solution. Due to the large value of <em>K<\/em><sub>f<\/sub> and the substantial excess of NH<sub>3<\/sub>, it can be assumed that the reaction goes to completion with only a small amount of the complex dissociating to form the ions. After reaction, concentrations of the species in the solution are<\/p>\n<p>[NH<sub>3<\/sub>] = 6.00 [latex]\\times[\/latex] 10<sup>\u20132<\/sup> mol\/L \u2013 1.6 [latex]\\times[\/latex] 10<sup>\u20132<\/sup> mol L<sup>\u20131<\/sup> = 4.4 [latex]\\times[\/latex] 10<sup>\u20132<\/sup><em>M<\/em><\/p>\n<p>Let <em>x<\/em> be the change in concentration of [Cd<sup>2+<\/sup>]:<\/p>\n<table>\n<thead>\n<tr>\n<th><\/th>\n<th>[Cd(NH<sub>3<\/sub>)<sub>4<\/sub><sup>2+<\/sup>]<\/th>\n<th>[Cd<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>4.00\u00a0\u00d7 10<sup>\u22123<\/sup><\/td>\n<td>0<\/td>\n<td>4.4\u00a0\u00d7 10<sup>\u22122<\/sup><\/td>\n<\/tr>\n<tr>\n<th>Equilibrium (<em>M<\/em>)<\/th>\n<td>4.00\u00a0\u00d7 10<sup>\u22123<\/sup> \u2212 <em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<td>4.4\u00a0\u00d7 10<sup>\u22122<\/sup> +\u00a04<em>x<\/em><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: center\">[latex]{K}_{\\text{f}}=4.0\\times {10}^{6}=\\frac{\\left[\\text{Cd}{\\left({\\text{NH}}_{3}\\right)}_{4}{}^{\\text{2+}}\\right]}{\\left[{\\text{Cd}}^{\\text{2+}}\\right]{\\left[{\\text{NH}}_{3}\\right]}^{4}}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]4.00\\times {10}^{6}=\\frac{\\left(4.00\\times {10}^{-3}-x\\right)}{\\left(x\\right){\\left(4.4\\times {10}^{-2}+4x\\right)}^{4}}[\/latex]<\/p>\n<p>As <em>x<\/em> is expected to be about the same size as the number from which it is subtracted, the entire expression must be expanded and solved, in this case, by successive approximations where substitution of values for <em>x<\/em> into the equation continues until the remainder is judged small enough. This is a slightly different method than used in most problems. We have:<\/p>\n<p style=\"text-align: center\">4.0 [latex]\\times[\/latex] 10<sup>6<\/sup><em>x<\/em> (4.4 [latex]\\times[\/latex] 10<sup>\u20132<\/sup> + 4<em>x<\/em>)<sup>4<\/sup> = 4.00 [latex]\\times[\/latex] 10<sup>\u20133<\/sup> \u2013 <em>x<\/em><\/p>\n<p style=\"text-align: center\">4.0 [latex]\\times[\/latex] 10<sup>6<\/sup><em>x<\/em> (3.75 [latex]\\times[\/latex] 10<sup>\u20136<\/sup> + 1.36 [latex]\\times[\/latex] 10<sup>\u20133<\/sup><em>x<\/em> + 0.186<em>x<\/em><sup>2<\/sup> + 11.264<em>x<\/em><sup>3<\/sup> +256<em>x<\/em><sup>4<\/sup>) = 4.00 [latex]\\times[\/latex] 10<sup>\u20133<\/sup><\/p>\n<p style=\"text-align: center\">16<em>x<\/em> + 5440<em>x<\/em><sup>2<\/sup> + 7.44 [latex]\\times[\/latex] 10<sup>5<\/sup><em>x<\/em><sup>3<\/sup> + 4.51 [latex]\\times[\/latex] 10<sup>7<\/sup><em>x<\/em><sup>4<\/sup> + 1.024 [latex]\\times[\/latex] 10<sup>9<\/sup><em>x<\/em> <sup>5<\/sup> = 4.00 [latex]\\times[\/latex] 10<sup>\u20133<\/sup><\/p>\n<p>Substitution of different values <em>x<\/em> will give a number to be compared with 4.00 [latex]\\times[\/latex] 10<sup>\u20133<\/sup>. Using 2.50 [latex]\\times[\/latex] 10<sup>\u20134<\/sup> for <em>x<\/em> gives 4.35 [latex]\\times[\/latex] 10<sup>\u20133<\/sup> compared with 4.00 [latex]\\times[\/latex] 10<sup>\u20133<\/sup>. Using 2.40 [latex]\\times[\/latex] 10<sup>\u20134<\/sup> gives 4.16 [latex]\\times[\/latex] 10<sup>\u20133<\/sup> compared with 4.00 [latex]\\times[\/latex] 10<sup>\u20133<\/sup>. Using 2.30 [latex]\\times[\/latex] 10<sup>\u20134<\/sup> gives 3.98 [latex]\\times[\/latex] 10<sup>\u20133<\/sup> compared with 4.00 [latex]\\times[\/latex] 10<sup>\u20133<\/sup>. Thus 2.30 [latex]\\times[\/latex] 10<sup>\u20134<\/sup> is close enough to the true value of <em>x<\/em> to make the difference equal to zero. If the approximation to drop 4<em>x<\/em> is compared with 4.4 [latex]\\times[\/latex] 10<sup>\u20132<\/sup>, the value of <em>x<\/em> obtained is 2.35 [latex]\\times[\/latex] 10<sup>\u20134<\/sup><em>M<\/em>.<\/p>\n<p>5.\u00a0For the formation reaction:<\/p>\n<p>[latex]\\begin{array}{l}{\\text{Al}}^{\\text{3+}}\\left(aq\\right)+6{\\text{F}}^{-}\\left(aq\\right)\\rightleftharpoons {\\text{AlF}}_{6}{}^{\\text{3-}}\\left(aq\\right)\\\\{K}_{\\text{f}}=\\frac{\\left[{\\text{AlF}}_{6}{}^{\\text{3-}}\\right]}{\\left[{\\text{Al}}^{\\text{3+}}\\right]{\\left[{\\text{F}}^{-}\\right]}^{6}}=\\frac{1}{{K}_{\\text{d}}}=\\frac{1}{2\\times {10}^{-24}}=5\\times {10}^{23}\\end{array}[\/latex]<\/p>\n<p>7.<\/p>\n<table>\n<thead>\n<tr>\n<th><\/th>\n<th>[Cd(CN)<sub>4<\/sub><sup>2\u2212<\/sup>]<\/th>\n<th>[CN<sup>\u2212<\/sup>]<\/th>\n<th>[Cd<sup>2+<\/sup>]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<th>Initial concentration (<em>M<\/em>)<\/th>\n<td>0.250<\/td>\n<td>0<\/td>\n<td>0<\/td>\n<\/tr>\n<tr>\n<th>Equilibrium (<em>M<\/em>)<\/th>\n<td>0.250\u00a0\u2212\u00a0<em>x<\/em><\/td>\n<td>4<em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: center\">[latex]{K}_{\\text{d}}=\\frac{\\left[{\\text{Cd}}^{\\text{2+}}\\right]\\left[{\\text{CN}}^{-}\\right]}{\\left[\\text{Cd}{\\left(\\text{CN}\\right)}_{4}{}^{2-}\\right]}=7.8\\times {10}^{-18}=\\frac{x{\\left(4x\\right)}^{4}}{0.250-x}[\/latex]<\/p>\n<p>Assume that <em>x<\/em> is small when compared with 0.250 <em>M<\/em>.<\/p>\n<p>256<em>x<\/em><sup>5<\/sup> = 0.250 [latex]\\times[\/latex] 7.8 [latex]\\times[\/latex] 10<sup>\u201318<\/sup><\/p>\n<p><em>x<\/em><sup>5<\/sup> = 7.617 [latex]\\times[\/latex] 10<sup>\u201321<\/sup><\/p>\n<p><em>x<\/em> = [Cd<sup>2+<\/sup>] = 9.5 [latex]\\times[\/latex] 10<sup>\u20135<\/sup><em>M<\/em><\/p>\n<p>4<em>x<\/em> = [CN<sup>\u2013<\/sup>] = 3.8 [latex]\\times[\/latex] 10<sup>\u20134<\/sup><em>M<\/em><\/p>\n<p>9.<\/p>\n<table>\n<thead>\n<tr>\n<th><\/th>\n<th>[Co(NH<sub>3<\/sub>)<sub>6<\/sub><sup>3+<\/sup>]<\/th>\n<th>[Co<sup>3+<\/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>0.500<\/td>\n<td>0<\/td>\n<td>0<\/td>\n<\/tr>\n<tr>\n<th>Equilibrium (<em>M<\/em>)<\/th>\n<td>0.500\u00a0\u2212\u00a0<em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<td>6<em>x<\/em><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: center\">[latex]{K}_{\\text{d}}=\\frac{\\left[{\\text{Co}}^{\\text{2+}}\\right]{\\left[{\\text{NH}}_{3}\\right]}^{6}}{\\left[\\text{Co}{\\left({\\text{NH}}_{3}\\right)}_{6}{}^{\\text{3+}}\\right]}=\\frac{x{\\left(6x\\right)}^{6}}{0.500-x}=2.2\\times {10}^{-34}[\/latex]<\/p>\n<p>Assume that <em>x<\/em> is small when compared with 0.500 <em>M<\/em>.<\/p>\n<p>4.67 [latex]\\times[\/latex] 104<em>x<\/em><sup>7<\/sup> = 0.500 [latex]\\times[\/latex] 2.2 [latex]\\times[\/latex] 10<sup>\u201334<\/sup><\/p>\n<p><em>x<\/em><sup>7<\/sup> = 2.358 [latex]\\times[\/latex] 10<sup>\u201339<\/sup><\/p>\n<p><em>x<\/em> = [Co<sup>3+<\/sup>] = 3.0 [latex]\\times[\/latex] 10<sup>\u20136<\/sup><em>M<\/em><\/p>\n<p>6<em>x<\/em> = [NH<sub>3<\/sub>] = 1.8 [latex]\\times[\/latex] 10<sup>\u20135<\/sup><em>M<\/em><\/p>\n<p>11.\u00a0Because <em>K<\/em><sub>sp<\/sub> is small and <em>K<\/em><sub>f<\/sub> is large, most of the Ag<sup>+<\/sup> is used to form [latex]\\text{Ag}{\\left(\\text{CN}\\right)}_{2}{}^{-}[\/latex]; that is:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{rll}\\left[\\text{Ag}^{+}\\right]&\\lt&\\left[\\text{Ag}\\left(\\text{CN}\\right)_{2}^{-}\\right]\\\\\\left[\\text{Ag}\\left(\\text{CN}\\right)_{2}^{-}\\right]&\\approx&2.0\\times{10}^{-1}M\\end{array}[\/latex]<\/p>\n<p>The CN<sup>\u2013<\/sup> from the dissolution and the added CN<sup>\u2013<\/sup> exist as CN<sup>\u2013<\/sup> and [latex]\\text{Ag}{\\left(\\text{CN}\\right)}_{2}{}^{-}[\/latex]. Let <em>x<\/em> be the change in concentration upon addition of CN<sup>\u2013<\/sup>. Its initial concentration is approximately 0.<\/p>\n<p style=\"text-align: center\">[CN<sup>\u2013<\/sup>] + 2 [latex]\\left[\\text{Ag}{\\left(\\text{CN}\\right)}_{2}{}^{-}\\right][\/latex] = 2 [latex]\\times[\/latex] 10<sup>\u20131<\/sup> + <em>x<\/em><\/p>\n<p>Because <em>K<\/em><sub>sp<\/sub> is small and <em>K<\/em><sub>f<\/sub> is large, most of the CN<sup>\u2013<\/sup> is used to form [latex]\\left[\\text{Ag}{\\left(\\text{CN}\\right)}_{2}{}^{-}\\right][\/latex]; that is:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{rll}\\left[{\\text{CN}}^{-}\\right]&<&2\\left[\\text{Ag}{\\left(\\text{CN}\\right)}_{2}{}^{-}\\right]\\\\2\\left[\\text{Ag}{\\left(\\text{CN}\\right)}_{2}{}^{-}\\right]&\\approx&2.0\\times {10}^{-1}+x\\end{array}[\/latex]<\/p>\n<p>2(2.0 [latex]\\times[\/latex] 10<sup>\u20131<\/sup>) \u2013 2.0 [latex]\\times[\/latex] 10<sup>\u20131<\/sup> = <em>x<\/em><\/p>\n<p>2.0 [latex]\\times[\/latex] 10<sup>\u20131<\/sup><em>M<\/em> [latex]\\times[\/latex] L = mol CN<sup>\u2013<\/sup> added<\/p>\n<p>The solution has a volume of 100 mL.<\/p>\n<p>2 [latex]\\times[\/latex] 10<sup>\u20131<\/sup> mol\/L [latex]\\times[\/latex] 0.100 L = 2 [latex]\\times[\/latex] 10<sup>\u20132<\/sup> mol<\/p>\n<p>mass KCN = 2.0 [latex]\\times[\/latex] 10<sup>\u20132<\/sup> mol KCN [latex]\\times[\/latex] 65.120 g\/mol = 1.3 g<\/p>\n<p>13.\u00a0The reaction is governed by two equilibria, both of which must be satisfied:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\text{AgBr}\\left(s\\right)\\rightleftharpoons\\text{Ag}^{+}\\left(aq\\right)+\\text{Br}^{-}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,{;}\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{sp}}=3.3\\times{10}^{-13}\\\\\\text{Ag}^{+}\\left(aq\\right)+2{\\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}\\end{array}[\/latex]<\/p>\n<p>The overall equilibrium is obtained by adding the two equations and multiplying their <em>K<\/em>s:<\/p>\n<p style=\"text-align: center\">[latex]\\frac{\\left[\\text{Ag}{\\left({\\text{S}}_{2}{\\text{O}}_{3}\\right)}^{\\text{3-}}\\right]\\left[{\\text{Br}}^{-}\\right]}{{\\left[{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}\\right]}^{2}}=15.51[\/latex]<\/p>\n<p>If all Ag is to be dissolved, the concentration of the complex is the molar concentration of AgBr.<\/p>\n<p>formula mass (AgBr) = 187.772 g\/mol<\/p>\n<p>[latex]\\text{moles present}=\\frac{0.27\\text{g}\\text{AgBr}}{187.772\\text{g}{\\text{mol}}^{-1}}=1.438\\times {10}^{-3}\\text{mol}[\/latex]<\/p>\n<p>Let <em>x<\/em> be the change in concentration of [latex]{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}:[\/latex]<\/p>\n<table>\n<thead>\n<tr>\n<th><\/th>\n<th>[Ag<sup>+<\/sup>]<\/th>\n<th>[S<sub>2<\/sub>O<sub>3<\/sub><sup>2\u2212<\/sup>]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<th>Initial concentration (<em>M<\/em>)<\/th>\n<td>0<\/td>\n<td>0<\/td>\n<\/tr>\n<tr>\n<th>Equilibrium (<em>M<\/em>)<\/th>\n<td>[latex]\\frac{1}{2}x[\/latex]<\/td>\n<td><em>x<\/em><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>[latex]\\frac{\\left(1.438\\times {10}^{-3}\\right)\\left(1.438\\times {10}^{-3}\\right)}{{x}^{2}}=15.51[\/latex]<\/p>\n<p><em>x<\/em><sup>2<\/sup> = 1.333 [latex]\\times[\/latex] 10<sup>\u20137<\/sup><\/p>\n<p>[latex]x=3.65\\times {10}^{-4}M=\\left[{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}\\right][\/latex]<\/p>\n<p>The formula mass of Na<sub>2<\/sub>S<sub>2<\/sub>O<sub>3<\/sub>\u20225H<sub>2<\/sub>O is 248.13 g\/mol. The total [latex]\\left[{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}\\right][\/latex] needed is:<\/p>\n<p>2(1.438 [latex]\\times[\/latex] 10<sup>\u20133<\/sup>) + 3.65 [latex]\\times[\/latex] 10<sup>\u20134<\/sup> = 3.241 [latex]\\times[\/latex] 10<sup>\u20133<\/sup> mol<\/p>\n<p>g(hypo) = 3.241 [latex]\\times[\/latex] 10<sup>\u20133<\/sup> mol [latex]\\times[\/latex] 248.13 g\/mol = 0.80 g<\/p>\n<p>15. The equilibrium is:[latex]{\\text{Ag}}^{\\text{+}}\\left(aq\\right)+2{\\text{CN}}^{-}\\left(aq\\right)\\rightleftharpoons \\text{Ag}{\\left(\\text{CN}\\right)}_{2}{}^{-}\\left(aq\\right){K}_{\\text{f}}=1\\times {10}^{20}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>The number of moles of AgNO<sub>3<\/sub> added is:<\/p>\n<p>0.02872 L [latex]\\times[\/latex] 0.0100 mol\/L = 2.87 [latex]\\times[\/latex] 10<sup>\u20134<\/sup> mol<\/p>\n<p>This compound reacts with CN<sup>\u2013<\/sup> to form [latex]\\text{Ag}{\\left(\\text{CN}\\right)}_{2}{}^{-}[\/latex], so there are 2.87 [latex]\\times[\/latex] 10<sup>\u20134<\/sup> mol [latex]\\text{Ag}{\\left(\\text{CN}\\right)}_{2}{}^{-}[\/latex]. This amount requires 2 [latex]\\times[\/latex] 2.87 [latex]\\times[\/latex] 10<sup>\u20134<\/sup> mol, or 5.74 [latex]\\times[\/latex] 10<sup>\u20134<\/sup> mol, of CN<sup>\u2013<\/sup>. The titration is stopped just as precipitation of AgCN begins:<\/p>\n<p>[latex]{\\text{AgCN}}_{2}{}^{-}\\left(aq\\right)+{\\text{Ag}}^{\\text{+}}\\left(aq\\right)\\rightleftharpoons 2\\text{AgCN}\\left(s\\right)[\/latex]<\/p>\n<p>so only the first equilibrium is applicable. The value of <em>K<\/em><sub>f<\/sub> is very large.<\/p>\n<p>mol CN<sup>\u2013<\/sup> &lt; [latex]\\left[\\text{Ag}{\\left(\\text{CN}\\right)}_{2}{}^{-}\\right][\/latex]<\/p>\n<p>mol NaCN = 2 mol [ [latex]\\text{Ag}{\\left(\\text{CN}\\right)}_{2}{}^{-}[\/latex] ] = 5.74 [latex]\\times[\/latex] 10<sup>\u20134<\/sup> mol<\/p>\n<p>[latex]\\text{mass}\\left(\\text{NaCN}\\right)=5.74\\times {10}^{-4}\\text{mol}\\times \\frac{49.007\\text{g}}{1\\text{mol}}=0.0281\\text{g}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Glossary<\/h2>\n<p><b>complex ion:\u00a0<\/b>ion consisting of a transition metal central atom and surrounding molecules or ions called ligands<\/p>\n<p><b>dissociation constant:\u00a0<\/b>(<em>K<\/em><sub>d<\/sub>) equilibrium constant for the decomposition of a complex ion into its components in solution<\/p>\n<p><b>formation constant:\u00a0<\/b>(<em>K<\/em><sub>f<\/sub>) (also, stability constant) equilibrium constant for the formation of a complex ion from its components in solution<\/p>\n<p>&nbsp;<\/p>\n<p><b>ligand:\u00a0<\/b>molecule or ion that surrounds a transition metal and forms a complex ion; ligands act as Lewis bases<\/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-3591\">\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":3,"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-3591","chapter","type-chapter","status-publish","hentry"],"part":2983,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-chemistryformajors-2\/wp-json\/pressbooks\/v2\/chapters\/3591","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-chemistryformajors-2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-chemistryformajors-2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-chemistryformajors-2\/wp-json\/wp\/v2\/users\/17"}],"version-history":[{"count":17,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-chemistryformajors-2\/wp-json\/pressbooks\/v2\/chapters\/3591\/revisions"}],"predecessor-version":[{"id":6574,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-chemistryformajors-2\/wp-json\/pressbooks\/v2\/chapters\/3591\/revisions\/6574"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-chemistryformajors-2\/wp-json\/pressbooks\/v2\/parts\/2983"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-chemistryformajors-2\/wp-json\/pressbooks\/v2\/chapters\/3591\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-chemistryformajors-2\/wp-json\/wp\/v2\/media?parent=3591"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-chemistryformajors-2\/wp-json\/pressbooks\/v2\/chapter-type?post=3591"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-chemistryformajors-2\/wp-json\/wp\/v2\/contributor?post=3591"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-chemistryformajors-2\/wp-json\/wp\/v2\/license?post=3591"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}