15.2 Complex Ions

Learning Objectives

By the end of this module, you will be able to:

Write equations for the formation of complex ions
Perform equilibrium calculations involving formation constants

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 complex ion [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:

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.

The equations for the dissolution of AgCl in a solution of NH₃ are:

[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]

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:

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.

The equations for the dissolution are:

[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]

Mercury(II) sulfide dissolves in a solution of sodium sulfide because HgS reacts with the S^2– ion:

[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]

A complex ion consists of a central atom, typically a transition metal cation, surrounded by ions, or molecules called ligands. These ligands can be neutral molecules like H₂O or NH₃, or ions such as CN^– or OH^–. 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 complex ion. 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.

The equilibrium constant for the reaction of the components of a complex ion to form the complex ion in solution is called a formation constant (K_f) (sometimes called a stability constant). For example, the complex ion [latex]\text{Cu}{\left(\text{CN}\right)}_{2}{}^{-}[/latex] is shown here:

It forms by the reaction:

[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]

At equilibrium:

[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]

The inverse of the formation constant is the dissociation constant (K_d), the equilibrium constant for the decomposition of a complex ion into its components in solution. We will work with dissociation constants further in the exercises for this section. Formation Constants for Complex Ions and Table 1 are tables of formation constants. In general, the larger the formation constant, the more stable the complex; however, as in the case of K_sp values, the stoichiometry of the compound must be considered.

Table 1. Common Complex Ions by Decreasing Formulation Constants
Substance	K_f at 25 °C
[latex]{\left[\text{Cd}{\left(\text{CN}\right)}_{4}\right]}^{2-}[/latex]	1.3 [latex]\times [/latex] 10⁷
[latex]\text{Ag}{\left({\text{NH}}_{3}\right)}_{2}{}^{+}[/latex]	1.7 [latex]\times [/latex] 10⁷
[latex]{\left[{\text{AlF}}_{6}\right]}^{\text{3-}}[/latex]	7 [latex]\times [/latex] 10¹⁹

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⁺ ([Ag⁺] = 1.3 [latex]\times [/latex] 10^–5M):

[latex]\text{AgCl}\left(s\right)\rightleftharpoons {\text{Ag}}^{\text{+}}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)[/latex]

However, if NH₃ is present in the water, the complex ion, [latex]\text{Ag}{\left({\text{NH}}_{3}\right)}_{2}{}^{+}[/latex], can form according to 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]

with

[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]

The large size of this formation constant indicates that most of the free silver ions produced by the dissolution of AgCl combine with NH₃ to form [latex]\text{Ag}{\left({\text{NH}}_{3}\right)}_{2}{}^{+}[/latex]. As a consequence, the concentration of silver ions, [Ag⁺], is reduced, and the reaction quotient for the dissolution of silver chloride, [Ag⁺][Cl^–], falls below the solubility product of AgCl:

[latex]Q=\left[{\text{Ag}}^{+}\right]\left[{\text{Cl}}^{-}\right]<{K}_{\text{sp}}[/latex]

More silver chloride then dissolves. If the concentration of ammonia is great enough, all of the silver chloride dissolves.

Example 1: Dissociation of a Complex Ion

Calculate the concentration of the silver ion in a solution that initially is 0.10 M with respect to [latex]\text{Ag}{\left({\text{NH}}_{3}\right)}_{2}{}^{+}[/latex].

Show Answer

We use the familiar path to solve this problem:

Four boxes are shown side by side, with three right facing arrows connecting them. The first box contains the text “Determine the direction of change.” The second box contains the text “Determine x and the equilibrium concentrations.” The third box contains the text “Solve for x and the equilibrium concentrations.” The fourth box contains the text “Check the math.”

Determine the direction of change. 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 Formation Constants for Complex Ions lists formation constants for complex ions. Before equilibrium, the reaction quotient is larger than the equilibrium constant [K_f = 1.6 [latex]\times [/latex] 10⁷, and [latex]Q=\frac{0.10}{0\times 0}[/latex], it is infinitely large], so the reaction shifts to the left to reach equilibrium.
Determine x and equilibrium concentrations. We let the change in concentration of Ag⁺ be x. Dissociation of 1 mol of [latex]\text{Ag}{\left({\text{NH}}_{3}\right)}_{2}{}^{+}[/latex] gives 1 mol of Ag⁺ and 2 mol of NH₃, so the change in [NH₃] is 2x and that of [latex]\text{Ag}{\left({\text{NH}}_{3}\right)}_{2}{}^{+}[/latex] is –x. In summary:
Solve for x and the equilibrium concentrations. At equilibrium: [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]
[latex]1.6\times {10}^{7}\text{=}\frac{0.10-x}{\left(x\right){\left(2x\right)}^{2}}[/latex] Both Q and K_f are much larger than 1, so let us assume that the changes in concentrations needed to reach equilibrium are small. Thus 0.10 – x is approximated as 0.10: [latex]1.6\times {10}^{7}=\frac{0.10}{\left(x\right){\left(2x\right)}^{2}}[/latex]
[latex]{x}^{3}=\frac{0.10}{4\left(1.6\times {10}^{7}\right)}=1.6\times {10}^{-9}[/latex]
[latex]x=\sqrt[3]{1.6\times {10}^{-9}}=1.2\times {10}^{-3}[/latex] Because only 1.2% of the [latex]\text{Ag}{\left({\text{NH}}_{3}\right)}_{2}{}^{+}[/latex] dissociates into Ag⁺ and NH₃, The assumption that x is small is justified.Now we determine the equilibrium concentrations: [latex]\left[{\text{Ag}}^{+}\right]=0+x=1.2\times {10}^{-3}M[/latex]
[latex]\left[{\text{NH}}_{3}\right]=0+2x=2.4\times {10}^{-3}M[/latex]
[latex]\left[\text{Ag}{\left({\text{NH}}_{3}\right)}_{2}{}^{+}\right]=0.10-x=0.10 - 0.0012=0.099[/latex] The concentration of free silver ion in the solution is 0.0012 M.
Check the work. The value of Q calculated using the equilibrium concentrations is equal to K_f within the error associated with the significant figures in the calculation.

Check Your Learning

Calculate the silver ion concentration, [Ag⁺], of a solution prepared by dissolving 1.00 g of AgNO₃ and 10.0 g of KCN in sufficient water to make 1.00 L of solution. (Hint: Because Q < K_f, assume the reaction goes to completion then calculate the [Ag⁺] produced by dissociation of the complex.)

Show Answer

Key Takeaways

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, K_f. 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.

Exercises

Under what circumstances, if any, does a sample of solid AgCl completely dissolve in pure water?
Explain why the addition of NH₃ or HNO₃ to a saturated solution of Ag₂CO₃ in contact with solid Ag₂CO₃ increases the solubility of the solid.
Calculate the cadmium ion concentration, [Cd²⁺], in a solution prepared by mixing 0.100 L of 0.0100 M Cd(NO₃)₂ with 1.150 L of 0.100 NH₃(aq).
Explain why addition of NH₃ or HNO₃ to a saturated solution of Cu(OH)₂ in contact with solid Cu(OH)₂ increases the solubility of the solid.
Sometimes equilibria for complex ions are described in terms of dissociation constants, K_d. 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] and [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, K_f, for [latex]{\text{AlF}}_{\text{6}}^{\text{3}-}[/latex].
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.
Using the dissociation constant, K_d = 7.8 [latex]\times [/latex] 10^–18, calculate the equilibrium concentrations of Cd²⁺ and CN^– in a 0.250-M solution of [latex]\text{Cd}{\left(\text{CN}\right)}_{4}^{2-}[/latex].
Using the dissociation constant, K_d = 3.4 [latex]\times [/latex] 10^–15, calculate the equilibrium concentrations of Zn²⁺ and OH^– in a 0.0465-M solution of [latex]\text{Zn}{\left(\text{OH}\right)}_{4}^{2-}[/latex].
Using the dissociation constant, K_d = 2.2 [latex]\times [/latex] 10^–34, calculate the equilibrium concentrations of Co³⁺ and NH₃ in a 0.500-M solution of [latex]\text{Co}{\left({\text{NH}}_{3}\right)}_{6}^{\text{3+}}[/latex].
Using the dissociation constant, K_d = 1 [latex]\times [/latex] 10^–44, calculate the equilibrium concentrations of Fe³⁺ and CN^– in a 0.333 M solution of [latex]\text{Fe}{\left(\text{CN}\right)}_{6}^{3-}[/latex].
Calculate the mass of potassium cyanide ion that must be added to 100 mL of solution to dissolve 2.0 [latex]\times [/latex] 10^–2 mol of silver cyanide, AgCN.
Calculate the minimum concentration of ammonia needed in 1.0 L of solution to dissolve 3.0 [latex]\times [/latex] 10^–3 mol of silver bromide.
A roll of 35-mm black and white photographic film contains about 0.27 g of unexposed AgBr before developing. What mass of Na₂S₂O₃.5H₂O (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_f = 4.7 [latex]\times [/latex] 10¹³)?
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-M HgCl₂ solution.
In a titration of cyanide ion, 28.72 mL of 0.0100 M AgNO₃ is added before precipitation begins. [The reaction of Ag⁺ with CN^– goes to completion, producing the [latex]\text{Ag}{\left(\text{CN}\right)}_{2}{}^{-}[/latex] complex.] Precipitation of solid AgCN takes place when excess Ag⁺ 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?
What are the concentrations of Ag⁺, CN^–, and [latex]\text{Ag}{\left(\text{CN}\right)}_{2}{}^{-}[/latex] in a saturated solution of AgCN?

Show Selected Answers

1. When the amount of solid is so small that a saturated solution is not produced.

3. Cadmium 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:

[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]

The formation of the complex ion requires 4 mol of NH₃ for each mol of Cd²⁺. First, calculate the initial amounts of Cd²⁺ and of NH₃ available for association:

[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]

[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]

For the reaction, 4.00 [latex]\times [/latex] 10^–3 mol/L of Cd²⁺ would require 4(4.00 [latex]\times [/latex] 10^–3 mol/L) of NH₃ or a 1.6 [latex]\times [/latex] 10^–2–M solution. Due to the large value of K_f and the substantial excess of NH₃, 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

[NH₃] = 6.00 [latex]\times [/latex] 10^–2 mol/L – 1.6 [latex]\times [/latex] 10^–2 mol L^–1 = 4.4 [latex]\times [/latex] 10^–2M

Let x be the change in concentration of [Cd²⁺]:

	[Cd(NH₃)₄²⁺]	[Cd²⁺]	[NH₃]
Initial concentration (M)	4.00 × 10⁻³	0	4.4 × 10⁻²
Equilibrium (M)	4.00 × 10⁻³ − x	x	4.4 × 10⁻² + 4x

[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]

[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]

As x 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 x into the equation continues until the remainder is judged small enough. This is a slightly different method than used in most problems. We have:

4.0 [latex]\times [/latex] 10⁶x (4.4 [latex]\times [/latex] 10^–2 + 4x)⁴ = 4.00 [latex]\times [/latex] 10^–3 – x

4.0 [latex]\times [/latex] 10⁶x (3.75 [latex]\times [/latex] 10^–6 + 1.36 [latex]\times [/latex] 10^–3x + 0.186x² + 11.264x³ +256x⁴) = 4.00 [latex]\times [/latex] 10^–3

16x + 5440x² + 7.44 [latex]\times [/latex] 10⁵x³ + 4.51 [latex]\times [/latex] 10⁷x⁴ + 1.024 [latex]\times [/latex] 10⁹x ⁵ = 4.00 [latex]\times [/latex] 10^–3

Substitution of different values x will give a number to be compared with 4.00 [latex]\times [/latex] 10^–3. Using 2.50 [latex]\times [/latex] 10^–4 for x gives 4.35 [latex]\times [/latex] 10^–3 compared with 4.00 [latex]\times [/latex] 10^–3. Using 2.40 [latex]\times [/latex] 10^–4 gives 4.16 [latex]\times [/latex] 10^–3 compared with 4.00 [latex]\times [/latex] 10^–3. Using 2.30 [latex]\times [/latex] 10^–4 gives 3.98 [latex]\times [/latex] 10^–3 compared with 4.00 [latex]\times [/latex] 10^–3. Thus 2.30 [latex]\times [/latex] 10^–4 is close enough to the true value of x to make the difference equal to zero. If the approximation to drop 4x is compared with 4.4 [latex]\times [/latex] 10^–2, the value of x obtained is 2.35 [latex]\times [/latex] 10^–4M.

5. For the formation reaction:

[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]

	[Cd(CN)₄²⁻]	[CN⁻]	[Cd²⁺]
Initial concentration (M)	0.250	0	0
Equilibrium (M)	0.250 − x	4x	x

[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]

Assume that x is small when compared with 0.250 M.

256x⁵ = 0.250 [latex]\times [/latex] 7.8 [latex]\times [/latex] 10^–18

x⁵ = 7.617 [latex]\times [/latex] 10^–21

x = [Cd²⁺] = 9.5 [latex]\times [/latex] 10^–5M

4x = [CN^–] = 3.8 [latex]\times [/latex] 10^–4M

	[Co(NH₃)₆³⁺]	[Co³⁺]	[NH₃]
Initial concentration (M)	0.500	0	0
Equilibrium (M)	0.500 − x	x	6x

[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]

Assume that x is small when compared with 0.500 M.

4.67 [latex]\times [/latex] 104x⁷ = 0.500 [latex]\times [/latex] 2.2 [latex]\times [/latex] 10^–34

x⁷ = 2.358 [latex]\times [/latex] 10^–39

x = [Co³⁺] = 3.0 [latex]\times [/latex] 10^–6M

6x = [NH₃] = 1.8 [latex]\times [/latex] 10^–5M

11. Because K_sp is small and K_f is large, most of the Ag⁺ is used to form [latex]\text{Ag}{\left(\text{CN}\right)}_{2}{}^{-}[/latex]; that is:

[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]

The CN^– from the dissolution and the added CN^– exist as CN^– and [latex]\text{Ag}{\left(\text{CN}\right)}_{2}{}^{-}[/latex]. Let x be the change in concentration upon addition of CN^–. Its initial concentration is approximately 0.

[CN^–] + 2 [latex]\left[\text{Ag}{\left(\text{CN}\right)}_{2}{}^{-}\right][/latex] = 2 [latex]\times [/latex] 10^–1 + x

Because K_sp is small and K_f is large, most of the CN^– is used to form [latex]\left[\text{Ag}{\left(\text{CN}\right)}_{2}{}^{-}\right][/latex]; that is:

[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]

2(2.0 [latex]\times [/latex] 10^–1) – 2.0 [latex]\times [/latex] 10^–1 = x

2.0 [latex]\times [/latex] 10^–1M [latex]\times [/latex] L = mol CN^– added

The solution has a volume of 100 mL.

2 [latex]\times [/latex] 10^–1 mol/L [latex]\times [/latex] 0.100 L = 2 [latex]\times [/latex] 10^–2 mol

mass KCN = 2.0 [latex]\times [/latex] 10^–2 mol KCN [latex]\times [/latex] 65.120 g/mol = 1.3 g

13. The reaction is governed by two equilibria, both of which must be satisfied:

[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]

The overall equilibrium is obtained by adding the two equations and multiplying their Ks:

[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]

If all Ag is to be dissolved, the concentration of the complex is the molar concentration of AgBr.

formula mass (AgBr) = 187.772 g/mol

[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]

Let x be the change in concentration of [latex]{\text{S}}_{2}{\text{O}}_{3}{}^{2-}:[/latex]

	[Ag⁺]	[S₂O₃²⁻]
Initial concentration (M)	0	0
Equilibrium (M)	[latex]\frac{1}{2}x[/latex]	x

[latex]\frac{\left(1.438\times {10}^{-3}\right)\left(1.438\times {10}^{-3}\right)}{{x}^{2}}=15.51[/latex]

x² = 1.333 [latex]\times [/latex] 10^–7

[latex]x=3.65\times {10}^{-4}M=\left[{\text{S}}_{2}{\text{O}}_{3}{}^{2-}\right][/latex]

The formula mass of Na₂S₂O₃•5H₂O is 248.13 g/mol. The total [latex]\left[{\text{S}}_{2}{\text{O}}_{3}{}^{2-}\right][/latex] needed is:

2(1.438 [latex]\times [/latex] 10^–3) + 3.65 [latex]\times [/latex] 10^–4 = 3.241 [latex]\times [/latex] 10^–3 mol

g(hypo) = 3.241 [latex]\times [/latex] 10^–3 mol [latex]\times [/latex] 248.13 g/mol = 0.80 g

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]

The number of moles of AgNO₃ added is:

0.02872 L [latex]\times [/latex] 0.0100 mol/L = 2.87 [latex]\times [/latex] 10^–4 mol

This compound reacts with CN^– to form [latex]\text{Ag}{\left(\text{CN}\right)}_{2}{}^{-}[/latex], so there are 2.87 [latex]\times [/latex] 10^–4 mol [latex]\text{Ag}{\left(\text{CN}\right)}_{2}{}^{-}[/latex]. This amount requires 2 [latex]\times [/latex] 2.87 [latex]\times [/latex] 10^–4 mol, or 5.74 [latex]\times [/latex] 10^–4 mol, of CN^–. The titration is stopped just as precipitation of AgCN begins:

[latex]{\text{AgCN}}_{2}{}^{-}\left(aq\right)+{\text{Ag}}^{\text{+}}\left(aq\right)\rightleftharpoons 2\text{AgCN}\left(s\right)[/latex]

so only the first equilibrium is applicable. The value of K_f is very large.

mol CN^– < [latex]\left[\text{Ag}{\left(\text{CN}\right)}_{2}{}^{-}\right][/latex]

mol NaCN = 2 mol [ [latex]\text{Ag}{\left(\text{CN}\right)}_{2}{}^{-}[/latex] ] = 5.74 [latex]\times [/latex] 10^–4 mol

[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]

Glossary

complex ion: ion consisting of a transition metal central atom and surrounding molecules or ions called ligands

dissociation constant: (K_d) equilibrium constant for the decomposition of a complex ion into its components in solution

formation constant: (K_f) (also, stability constant) equilibrium constant for the formation of a complex ion from its components in solution

ligand: molecule or ion that surrounds a transition metal and forms a complex ion; ligands act as Lewis bases

Chapter 15: Aqueous Equilibria