Chapter 16: Equilibria of Other Reaction Classes

16.3 Coupled Equilibria

Learning Outcomes

• Describe examples of systems involving two (or more) coupled chemical equilibria
• Calculate reactant and product concentrations for coupled equilibrium systems

As discussed in preceding chapters on equilibrium, coupled equilibria involve two or more separate chemical reactions that share one or more reactants or products. This section of this chapter will address solubility equilibria coupled with acid-base and complex-formation reactions.

An environmentally relevant example illustrating the coupling of solubility and acid-base equilibria is the impact of ocean acidification on the health of the ocean’s coral reefs. These reefs are built upon skeletons of sparingly soluble calcium carbonate excreted by colonies of corals (small marine invertebrates).

Carbon dioxide in the air dissolves in sea water, forming carbonic acid $\ce{(H2CO3)}$. The carbonic acid then ionizes to form hydrogen ions and bicarbonate ions $\ce{(HCO3-)}$, which can further ionize into more hydrogen ions and carbonate ions $\ce{(CO3^2-)}:$

$\begin{array}{rll}{}{\ce{CO}}_{2}\left(g\right)&\rightleftharpoons&{\ce{CO}}_{2}\left(aq\right)\\{\ce{CO}}_{2}\left(aq\right)+{\ce{H}}_{2}\ce{O}&\rightleftharpoons&{\ce{H}}_{2}{\ce{CO}}_{3}\left(aq\right)\\{\ce{H}}_{2}{\ce{CO}}_{3}\left(aq\right)&\rightleftharpoons&{\ce{H}}^{\ce{+}}\left(aq\right)+{\ce{HCO}}_{3}{}^{-}\left(aq\right)\\{\ce{HCO}}_{3}{}^{-}\left(aq\right)&\rightleftharpoons&{\ce{H}}^{\ce{+}}\left(aq\right)+{\ce{CO}}_{3}{}^{2-}\left(aq\right)\end{array}$

The excess $\ce{H+}$ ions make seawater more acidic. Increased ocean acidification can then have negative impacts on reef-building coral, as they cannot absorb the calcium carbonate they need to grow and maintain their skeletons (Figure 16.3.1). This in turn disrupts the local biosystem that depends upon the health of the reefs for its survival. If enough local reefs are similarly affected, the disruptions to sea life can be felt globally. The world’s oceans are presently in the midst of a period of intense acidification, believed to have begun in the mid-nineteenth century, and which is now accelerating at a rate faster than any change to oceanic pH in the last 20 million years.

The dramatic increase in solubility with increasing acidity described above for calcium carbonate is typical of salts containing basic anions (e.g., carbonate, fluoride, hydroxide, sulfide). Another familiar example is the formation of dental cavities in tooth enamel. The major mineral component of enamel is calcium hydroxyapatite (Figure 16.3.2), a sparingly soluble ionic compound whose dissolution equilibrium is

$\ce{Ca5(PO4)3OH}(s)\longrightarrow \ce{5Ca^2+}(aq)+\ce{3PO4^3-}(aq)+\ce{OH-}(aq)$

This compound dissolved to yield two different basic ions: triprotic phosphate ions

$\ce{PO4^3-}(aq)+\ce{H3O+}\longrightarrow \ce{H2PO4^2-}+\ce{H2O}$
$\ce{PO4^2-}(aq)+\ce{H3O+}\longrightarrow \ce{H2PO4-}+\ce{H2O}$
$\ce{H2PO4-}+\ce{H3O+}\longrightarrow \ce{H3PO4}+\ce{H2O}$

and monoprotic hydroxide ions:

$\ce{OH-}(aq)+\ce{H3O+}\longrightarrow \ce{2H2O}$

Of the two basic productions, the hydroxide is, of course, by far the stronger base (it’s the strongest base that can exist in aqueous solution), and so it is the dominant factor providing the compound an acid-dependent solubility. Dental cavities form when the acid waste of bacteria growing on the surface of teeth hastens the dissolution of tooth enamel by reacting completely with the strong base hydroxide, shifting the hydroxyapatite solubility equilibrium to the right. Some toothpastes and mouth rinses contain added $\ce{NaF}$ or $\ce{SnF2}$ that make enamel more acid resistant by replacing the strong base hydroxide with the weak base fluoride:

$\ce{NaF}+\ce{Ca5(PO4)3OH}\rightleftharpoons \ce{Ca5(PO4)3F}+\ce{Na+}+\ce{OH-}$

The weak base fluoride ion reacts only partially with the bacterial acid waste, resulting in a less extensive shift in the solubility equilibrium and an increased resistance to acid dissolution. See the Chemistry in Everyday Life feature on the role of fluoride in preventing tooth decay for more information.

Role of Fluoride in Preventing Tooth Decay

As we saw previously, fluoride ions help protect our teeth by reacting with hydroxylapatite to form fluorapatite, $\ce{Ca5(PO4)2F}$. Since it lacks a hydroxide ion, fluorapatite is more resistant to attacks by acids in our mouths and is thus less soluble, protecting our teeth. Scientists discovered that naturally fluorinated water could be beneficial to your teeth, and so it became common practice to add fluoride to drinking water. Toothpastes and mouthwashes also contain amounts of fluoride (Figure 16.3.3).

Unfortunately, excess fluoride can negate its advantages. Natural sources of drinking water in various parts of the world have varying concentrations of fluoride, and places where that concentration is high are prone to certain health risks when there is no other source of drinking water. The most serious side effect of excess fluoride is the bone disease, skeletal fluorosis. When excess fluoride is in the body, it can cause the joints to stiffen and the bones to thicken. It can severely impact mobility and can negatively affect the thyroid gland. Skeletal fluorosis is a condition that over 2.7 million people suffer from across the world. So while fluoride can protect our teeth from decay, the US Environmental Protection Agency sets a maximum level of 4 ppm (4 mg/L) of fluoride in drinking water in the US. Fluoride levels in water are not regulated in all countries, so fluorosis is a problem in areas with high levels of fluoride in the groundwater.

The solubility of ionic compounds may also be increased when dissolution is coupled to the formation of a complex ion. For example, aluminum hydroxide dissolves in a solution of sodium hydroxide or another strong base because of the formation of the complex ion $\ce{Al(OH)4-}$.

The equations for the dissolution of aluminum hydroxide, the formation of the complex ion, and the combined (net) equation are shown below. As indicated by the relatively large value of K for the net reaction, coupling complex formation with dissolution drastically increases the solubility of $\ce{Al(OH)3}$.

$\ce{Al(OH)3} (s) \rightleftharpoons \ce{Al^3+} (aq) + \ce{3OH-} (aq)$ $K_{sp} = 2 \times 10^{-32}$

$\ce{Al^3+} (aq) + \ce{4OH-} (aq) \rightleftharpoons \ce{Al(OH)4-} (aq)$ $K_{f} = 1.1 \times 10^{33}$

$\text{Net: } \ce{Al(OH)3} (s) + \ce{OH-} (aq) \rightleftharpoons \ce{Al(OH)4-} (aq)$ $K = K_{sp} K_f = 22$

Example 16.3.1: Increased solubility in acidic solutions

Compute and compare the molar solublities for aluminum hydroxide, $\ce{Al(OH)3}$, dissolved in (a) pure water and (b) a buffer containing 0.100 M acetic acid and 0.100 M sodium acetate.

Solution

Show Solution
1. The molar solubility of aluminum hydroxide in water is computed considering the dissolution equilibrium only as demonstrated in several previous examples:
2. $\ce{Al(OH)3} (s) \rightleftharpoons \ce{Al^3+} (aq) + \ce{3OH-} (aq)$ $K_{sp} = 2 \times 10^{-32}$
3. $\text{molar solubility in water} = \ce{[Al^3+]} = (2\times 10^{-32} / 27)^{1/4} = 5\times 10^{-9} M$
4. The concentration of hydroxide ion of the buffered solution is conveniently calculated by the Henderson-Hasselbalch equation:
5. $\text{pH} = \text{pKa} +\text{log}\ce{[CH3COO-]} / \ce{[CH3COOH]}$
6. $\text{pH} = 4.74 +\text{log} (0.100/0.100) = 4.74$
7. At this pH, the concentration of hydroxide ion is
8. $\text{pOH} = 14.00-4.74=9.26$
9. $\ce{[OH-]} = 10^{-9.26} = 5.5\times 10^{-10}$
10. The solubility of $\ce{Al(OH)3}$ in this buffer is then calculated from its solubility product expressions:
11. $K_{sp} = \ce{[Al^3+]}\ce{[OH-]^3}$
12. $\text{molar solubility in buffer} = \ce{[Al^3+]} = K_{sp}/\ce{[OH-]^3} = (2\times 10^{-32}) / (5\times 10^{-10})^3 = 1.2\times 10^{-4} M$
13. Compared to pure water, the solubility of aluminum hydroxide in this mildly acidic buffer is approximately ten million times greater (though still relatively low).

Example 16.3.2: Multiple Equilibria

Unexposed silver halides are removed from photographic film when they react with sodium thiosulfate ($\ce{Na2S2O3}$, called hypo) to form the complex ion $\ce{Ag(S2O3)2^3-}$ (Kf = 4.7 $\times$ 1013).

What mass of $\ce{Na2S2O3}$ is required to prepare 1.00 L of a solution that will dissolve 1.00 g of $\ce{AgBr}$ by the formation of $\ce{Ag(S2O3)2^3-}?$

Show Solution

Two equilibria are involved when AgBr dissolves in a solution containing the $\ce{S2O3^2-}$ ion:

• Reaction 1 (dissolution): $\ce{AgBr}(s)\rightleftharpoons \ce{Ag+}(aq)+\ce{Br-}(aq)\qquad{K}_{\text{sp}}=5.0\times {10}^{\text{-13}}$
• Reaction 2 (complexation): $\ce{Ag+}(aq)+\ce{S2O3^2-}(aq)\rightleftharpoons \ce{Ag(S2O3)2^3-}(aq)\qquad{K}_{\text{f}}=4.7\times {10}^{13}$

First, calculate the concentration of bromide that will result when the 1.00 g of $\ce{AgBr}$ is completely dissolved via the cited complexation reaction:

$1.00\text{ g } \ce{AgBr }/(187.77\text{ g/mol})(1\text{ mol } \ce{Br-}/1\text{ mol } \ce{AgBr})=0.00532\text{ mol } \ce{Br-}$

$0.00532\text{ mol } \ce{Br-}/1.00\text{ L}=0.00532 M\ce{ Br-}$

Next, use this bromide molarity and the solubility product for silver bromide to calculate the silver ion molarity in the solution:

$[\ce{Ag+}]=K_{\text{sp}}/ [\ce{Br-}]=5.0\times{10}^{-13}/0.00532=9.4\times {10}^{-11}M$

Based on the stoichiometry of the complex ion formation, the concentration of complex ion produced is

$0.00532-9.4\times{10}^{-11}=0.00521M$

Use the silver ion and complex ion concentrations and the formation constant for the complex ion to compute the concentration of thiosulfate ion.

$\left[{\text{S}}_{2}{\text{O}}_{3}{}^{2-}\right]^{2}=\left[\text{Ag}{\left({\text{S}}_{2}{\text{O}}_{3}\right)}_{2}{}^{\text{3-}}\right]/\left[\text{Ag}^{+}\right]K_{\text{f}}=0.00521/(9.6\times{10}^{-11})(4.7\times{10}^{13})=1.15\times{10}^{-6}$

$[\ce{S2O3^2-}]=1.1\times{10}^{-3}M$

Finally, use this molar concentration to derive the required mass of sodium thiosulfate:

$(1.1\times {10}^{\text{-3}}\text{ mol}\ce{ S2O3^2-}/\text{L})\times(1\text{ mol}\ce{ Na2S2O3}/1\text{ mol}{\text{ S}}_{2}{\text{O}}_{3}{}^{2-})\times(158.1\text{ g}\ce{ Na2S2O3}/\text{mol})=1.7\text{ g}$

Thus, 1.00 L of a solution prepared from 1.7 g $\ce{Na2S2O3}$ dissolves 1.0 g of $\ce{AgBr}$.

Key Concepts and Summary

Systems involving two or more chemical equilibria that share one or more reactant or product are called coupled equilibria. Common examples of coupled equilibria include the increased solubility of some compounds in acidic solutions (coupled dissolution and neutralization equilibria) and in solutions containing ligands (coupled dissolution and complex formation). The equilibrium tools from other chapters may be applied to describe and perform calculations on these systems.

Try It

1. Calculate the equilibrium concentration of $\ce{Ni^2+}$ in a 1.0-M solution $\ce{[Ni(NH3)6](NO3)2}$.
2. Calculate the equilibrium concentration of $\ce{Cu^2+}$ in a solution initially with 0.050 M $\ce{Cu^2+}$ and 1.00 M $\ce{NH3}$.
3. Calculate the $\ce{Fe^3+}$ equilibrium concentration when 0.0888 mole of $\ce{K3[Fe(CN)6]}$ is added to a solution with 0.0.00010 M $\ce{CN-}$.
4. The equilibrium constant for the reaction $\ce{Hg^2+}(aq)+\ce{2Cl-}\left(aq\right)\rightleftharpoons \ce{HgCl2}(aq)$ is 1.6 $\times$ 1013. Is $\ce{HgCl2}$ a strong electrolyte or a weak electrolyte? What are the concentrations of $\ce{Hg^2+}$ and $\ce{Cl-}$ in a 0.015-M solution of $\ce{HgCl2}$?
5. Calculate the molar solubility of $\ce{Al(OH)3}$ in a buffer solution with 0.100 M NH3 and 0.400 M $\ce{NH4+}$.
Show Selected Solutions

1. $\ce{Ni^2+}(aq)+6{\ce{NH}}_{3}\left(aq\right)\rightleftharpoons \ce{[Ni(NH3)6]^2+}(aq)\,\,\,\,\,\,\,{;}\,\,\,\,\,\,\,{K}_{\text{f}}=1.8\times {10}^{8}$

Let x be the change in concentration as $\ce{Ni^2+}$ dissociates. Because the initial $\ce{Ni^2+}$ concentration is 0, the concentration at any times is x:

$1.8\times {10}^{8}=\frac{{\left[\ce{Ni}{\left({\ce{NH}}_{3}\right)}_{6}\right]}^{\ce{2+}}}{\left[{\ce{Ni}}^{\ce{2+}}\right]{\left[{\ce{NH}}_{3}\right]}^{6}}=\frac{\left(1.0-x\right)}{x{\left(6x\right)}^{6}}$

1.8 $\times$ 108(46656x7) = 1.0 – x

8.40 $\times$ 1012(x2) = 1.0 – x

Since x is small in comparison with 1.0, drop x:

8.40 $\times$ 1012(x7) = 1.0

x7 = 1.19 $\times$ 10–13

x = 0.014 M

2. Assume that all $\ce{Cu^2+}$ forms the complex whose concentration is 0.050 M and the remaining $\ce{NH3}$ has a concentration of 1.00 M – 4(0.050 M) = 0.80 M. The complex dissociates:

$\ce{[Cu(NH3)4]^2+}\rightleftharpoons \ce{[Cu^2+]}+\ce{4NH3}]$

Let x be the change in concentration of $\ce{Cu^2+}$ that dissociates:

[Cu(NH3)42+] [Cu2+] [NH3]
Initial concentration (M) 0.050 0 0.80
Equilibrium (M) 0.050 − x x 4x + 0.80

$\frac{\left[\text{Cu}{\left({\text{NH}}_{3}\right)}_{4}{}^{\text{2+}}\right]}{\left[{\text{Cu}}^{\text{2+}}\right]{\left[{\text{NH}}_{3}\right]}^{4}}=1.2\times {10}^{12}=\frac{0.050-x}{x{\left(4x+0.80\right)}^{4}}$

Assume that 4x is small when compared with 0.80 and that x is small when compared with 0.050:

(0.80)4 $\times$ 1.2 $\times$ 1012x = 0.050

x = 1.0 $\times$ 10–13M

3. Set up a table listing initial and equilibrium concentrations for the reaction:

${\text{Fe}}^{\text{3+}}+6{\text{CN}}^{-}\rightleftharpoons {\left(\text{Fe}{\left(\text{CN}\right)}_{6}\right]}^{3-}{K}_{\text{f}}=1\times {10}^{44}$

Let x be the concentration of $\ce{Fe^3+}$ that dissociates when 0.0888 mol dissolves in 1.00 L of 0.00010 M $\ce{CN-}$. Assume no volume change upon dissolution:

$\ce{[Fe(CN)6^3-]}$ $\ce{[Fe^3+]}$ $\ce{[CN-]}$
Initial concentration (M) 0.0888 0 0.00010
Equilibrium (M) 0.0888 − x x 0.00010 − 6x

$\frac{\left[\text{Fe}{\left(\text{CN}\right)}_{6}{}^{3-}\right]}{\left[{\text{Fe}}^{\text{3+}}\right]{\left[{\text{CN}}^{-}\right]}^{6}}=\frac{0.0888-x}{x{\left(0.000010 - 6x\right)}^{6}}=1\times {10}^{44}$

Assume that x is small when compared with the terms from which it is subtracted:

0.0888 = (0.00010)6(x)(1 $\times$ 1044)

$x=\frac{0.0888}{1\times {10}^{26}}=9\times {10}^{-22}M$

4. Let x be the change in the number of moles of $\ce{Hg^2+}$ that form per liter:

$\ce{[HgCl2]}$ $\ce{[Hg^2+]}$ $\ce{[Cl-]}$
Initial concentration (M) 0.015 0 0
Equilibrium (M) 0.015 − x x 2x

$\begin{array}{l}\frac{\left[{\text{HgCl}}_{2}\right]}{\left[{\text{Hg}}^{\text{2+}}\right]{\left[{\text{Cl}}^{-}\right]}^{2}}=K=1.6\times {10}^{13}\\ \frac{0.015-x}{\left(x\right){\left(2x\right)}^{2}}\approx \frac{0.015}{4{x}^{3}}=1.6\times {10}^{13}\end{array}$

x3 = 2.3 $\times$ 10–16

x = 6.2 $\times$ 10–6M = $\ce{[Hg^2+]}$

2x = 1.2 $\times$ 10–5] M = $\ce{[Cl-]}$

The substance is a weak electrolyte because very little of the initial 0.015 M $\ce{HgCl2}$ dissolved.

5. ${K}_{\text{b}}=\frac{\left[{\text{NH}}_{4}{}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{NH}}_{3}\right]}=\frac{\left(0.400\right)\left[{\text{OH}}^{-}\right]}{\left(0.100\right)}=1.8\times {10}^{-5}$

$\left[{\text{OH}}^{-}\right]=\frac{\left(0.100\right)\left(1.8\times {10}^{-5}\right)}{0.0400}\text{=}4.5\times {10}^{-5}$

Ksp = [Al3+][OH]3 = [Al3+](4.5 $\times$ 10–5)3 = 1.9 $\times$ 10–33

[Al3+] = 2.1 $\times$ 10–20 (molar solubility)

Glossary

coupled equilibrium: system characterized by more than one state of balance between a slightly soluble ionic solid and an aqueous solution of ions working simultaneously