# Multiple Equilibria

## Multiple Equilibria

There are times when one equilibrium reaction does not adequately describe the system being studied. Sometimes we have more than one type of equilibrium occurring at once (for example, an acid-base reaction and a precipitation reaction).

The ocean is a unique example of a system with multiple equilibria, or multiple states of solubility equilibria working simultaneously. Carbon dioxide in the air dissolves in sea water, forming carbonic acid (H2CO3). The carbonic acid then ionizes to form hydrogen ions and bicarbonate ions $$({\text{HCO}}_{3}{}^{\text{−}}),$$ which can further ionize into more hydrogen ions and carbonate ions $$({\text{CO}}_{3}{}^{\text{2−}}):$$

$${\text{CO}}_{2}(g)⇌{\text{CO}}_{2}(aq)$$

$${\text{CO}}_{2}(aq)+{\text{H}}_{2}\text{O}⇌{\text{H}}_{2}{\text{CO}}_{3}(aq)$$

$${\text{H}}_{2}{\text{CO}}_{3}(aq)⇌{\text{H}}^{\text{+}}(aq)+{\text{HCO}}_{3}{}^{\text{−}}(aq)$$

$${\text{HCO}}_{3}{}^{\text{−}}(aq)⇌{\text{H}}^{\text{+}}(aq)+{\text{CO}}_{3}{}^{\text{2−}}(aq)$$

The excess 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 (see the figure below). 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.

Healthy coral reefs (a) support a dense and diverse array of sea life across the ocean food chain. But when coral are unable to adequately build and maintain their calcium carbonite skeletons because of excess ocean acidification, the unhealthy reef (b) is only capable of hosting a small fraction of the species as before, and the local food chain starts to collapse. (credit a: modification of work by NOAA Photo Library; credit b: modification of work by “prilfish”/Flickr)

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Slightly soluble solids derived from weak acids generally dissolve in strong acids, unless their solubility products are extremely small. For example, we can dissolve CuCO3, FeS, and Ca3(PO4)2 in HCl because their basic anions react to form weak acids (H2CO3, H2S, and $${\text{H}}_{2}{\text{PO}}_{4}{}^{\text{−}}).$$ The resulting decrease in the concentration of the anion results in a shift of the equilibrium concentrations to the right in accordance with Le Châtelier’s principle.

Of particular relevance to us is the dissolution of hydroxylapatite, Ca5(PO4)3OH, in acid. Apatites are a class of calcium phosphate minerals (see the figure below); a biological form of hydroxylapatite is found as the principal mineral in the enamel of our teeth. A mixture of hydroxylapatite and water (or saliva) contains an equilibrium mixture of solid Ca5(PO4)3OH and dissolved Ca2+, $${\text{PO}}_{4}{}^{\text{3−}},$$ and OH ions:

$${\text{Ca}}_{\text{5}}({\text{P}{\text{O}}_{4})}_{3}\text{OH}(s)\phantom{\rule{0.2em}{0ex}}⇌\phantom{\rule{0.2em}{0ex}}5{\text{Ca}}^{\text{2+}}(aq)+3{\text{PO}}_{4}{}^{\text{3−}}(aq)+{\text{OH}}^{\text{−}}(aq)$$

Crystal of the mineral hydroxylapatite, Ca5(PO4)3OH, is shown here. Pure apatite is white, but like many other minerals, this sample is colored because of the presence of impurities.

When exposed to acid, phosphate ions react with hydronium ions to form hydrogen phosphate ions and ultimately, phosphoric acid:

$${\text{PO}}_{4}{}^{\text{3−}}(\mathrm{aq})+{\text{H}}_{3}{\text{O}}^{+}(\mathrm{aq})\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}{\text{PO}}_{4}{}^{\text{2−}}(\mathrm{aq})+{\text{H}}_{2}\text{O}(l)$$

$${\text{H}}_{2}{\text{PO}}_{4}{}^{\text{2−}}(aq)+{\text{H}}_{3}{\text{O}}^{+}(\mathrm{aq})\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}{\text{PO}}_{4}{}^{\text{−}}(\mathrm{aq})+{\text{H}}_{2}\text{O}(l)$$

$${\text{H}}_{2}{\text{PO}}_{4}{}^{\text{−}}(\mathrm{aq})+{\text{H}}_{3}{\text{O}}^{+}(\mathrm{aq})\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{H}}_{3}{\text{PO}}_{4}(\mathrm{aq})+{\text{H}}_{2}\text{O}(l)$$

Hydroxide ion reacts to form water:

$${\text{OH}}^{\text{−}}(aq)+{\text{H}}_{3}{\text{O}}^{+}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}2{\text{H}}_{2}\text{O}$$

These reactions decrease the phosphate and hydroxide ion concentrations, and additional hydroxylapatite dissolves in an acidic solution in accord with Le Châtelier’s principle. Our teeth develop cavities when acid waste produced by bacteria growing on them causes the hydroxylapatite of the enamel to dissolve. Fluoride toothpastes contain sodium fluoride, NaF, or stannous fluoride [more properly named tin(II) fluoride], SnF2. They function by replacing the OH ion in hydroxylapatite with F ion, producing fluorapatite, Ca5(PO4)3F:

$$\text{NaF}+{\text{Ca}}_{\text{5}}({\text{P}{\text{O}}_{4})}_{3}\text{OH}\phantom{\rule{0.2em}{0ex}}⇌\phantom{\rule{0.2em}{0ex}}{\text{Ca}}_{\text{5}}({\text{P}{\text{O}}_{4})}_{3}\text{F}+{\text{Na}}^{+}+{\text{OH}}^{\text{−}}$$

The resulting Ca5(PO4)3F is slightly less soluble than Ca5(PO4)3OH, and F is a weaker base than OH. Both of these factors make the fluorapatite more resistant to attack by acids than hydroxylapatite. See the next lesson on the role of fluoride in preventing tooth decay for more information.

When acid rain attacks limestone or marble, which are calcium carbonates, a reaction occurs that is similar to the acid attack on hydroxylapatite. The hydronium ion from the acid rain combines with the carbonate ion from calcium carbonates and forms the hydrogen carbonate ion, a weak acid:

$${\text{H}}_{3}{\text{O}}^{\text{+}}(aq)+{\text{CO}}_{3}{}^{\text{2−}}(aq)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{HCO}}_{3}{}^{\text{−}}(aq)+{\text{H}}_{2}\text{O}(l)$$

Calcium hydrogen carbonate, Ca(HCO3)2, is soluble, so limestone and marble objects slowly dissolve in acid rain.

If we add calcium carbonate to a concentrated acid, hydronium ion reacts with the carbonate ion according to the equation:

$$2{\text{H}}_{3}{\text{O}}^{\text{+}}(aq)+{\text{CO}}_{3}{}^{\text{2−}}(aq)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}{\text{CO}}_{3}(aq)+2{\text{H}}_{2}\text{O}(l)$$

(Acid rain is usually not sufficiently acidic to cause this reaction; however, laboratory acids are.) The solution may become saturated with the weak electrolyte carbonic acid, which is unstable, and carbon dioxide gas can be evolved:

$${\text{H}}_{2}{\text{CO}}_{3}(aq)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}(g)+{\text{H}}_{2}\text{O}(l)$$

These reactions decrease the carbonate ion concentration, and additional calcium carbonate dissolves. If enough acid is present, the concentration of carbonate ion is reduced to such a low level that the reaction quotient for the dissolution of calcium carbonate remains less than the solubility product of calcium carbonate, even after all of the calcium carbonate has dissolved.

## Example

### Prevention of Precipitation of Mg(OH)2

Calculate the concentration of ammonium ion that is required to prevent the precipitation of Mg(OH)2 in a solution with [Mg2+] = 0.10 M and [NH3] = 0.10 M.

### Solution

Two equilibria are involved in this system:

Reaction (1): $$\text{Mg}{(\text{OH})}_{2}(s)⇌{\text{Mg}}^{\text{2+}}(aq)+2{\text{OH}}^{\text{−}}(aq);\phantom{\rule{4em}{0ex}}{K}_{\text{sp}}=8.9\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−12}}$$

Reaction (2): $${\text{NH}}_{3}(aq)+{\text{H}}_{2}\text{O}(l)⇌{\text{NH}}_{4}{}^{\text{+}}(aq)+{\text{OH}}^{\text{−}}(aq)\phantom{\rule{4em}{0ex}}{K}_{\text{b}}=1.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−5}}$$

To prevent the formation of solid Mg(OH)2, we must adjust the concentration of OH so that the reaction quotient for Equation (1), Q = [Mg2+][OH]2, is less than Ksp for Mg(OH)2. (To simplify the calculation, we determine the concentration of OH when Q = Ksp.) [OH] can be reduced by the addition of $${\text{NH}}_{4}{}^{+},$$ which shifts Reaction (2) to the left and reduces [OH].

1. We determine the [OH] at which Q = Ksp when [Mg2+] = 0.10M:

$$Q=[{\text{Mg}}^{\text{2+}}]{[{\text{OH}}^{\text{−}}]}^{2}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}(0.10){[{\text{OH}}^{\text{−}}]}^{2}=8.9\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−12}}$$

$$[{\text{OH}}^{\text{−}}]\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}9.4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−6}}\phantom{\rule{0.2em}{0ex}}M$$

Solid Mg(OH)2 will not form in this solution when [OH] is less than 9.4 $$×$$ 10–6M.

2. We calculate the$$\mathit{[}N{H}_{4}{}^{+}\mathit{]}$$needed to decrease [OH] to 9.4 $$×$$ 10–6Mwhen [NH3] = 0.10.

$${K}_{\text{b}}=\cfrac{[{\text{NH}}_{4}{}^{+}][{\text{OH}}^{\text{−}}]}{[{\text{NH}}_{3}]}=\cfrac{[{\text{NH}}_{4}{}^{+}](9.4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−6}})}{0.10}\phantom{\rule{0.2em}{0ex}}=1.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−5}}$$

$$[{\text{NH}}_{4}{}^{+}]\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}0.19\phantom{\rule{0.2em}{0ex}}M$$

When $$[{\text{NH}}_{4}{}^{+}]$$ equals 0.19 M, [OH] will be 9.4 $$×$$ 10–6M. Any $$[{\text{NH}}_{4}{}^{+}]$$ greater than 0.19 M will reduce [OH] below 9.4 $$×$$ 10–6M and prevent the formation of Mg(OH)2.

Therefore, precise calculations of the solubility of solids from the solubility product are limited to cases in which the only significant reaction occurring when the solid dissolves is the formation of its ions.

## Example

### Multiple Equilibria

Unexposed silver halides are removed from photographic film when they react with sodium thiosulfate (Na2S2O3, called hypo) to form the complex ion $$\text{Ag}{({\text{S}}_{2}{\text{O}}_{3})}_{2}{}^{\text{3−}}$$ (Kf = 4.7 $$×$$ 1013). The reaction with silver bromide is:

What mass of Na2S2O3 is required to prepare 1.00 L of a solution that will dissolve 1.00 g of AgBr by the formation of $$\text{Ag}{({\text{S}}_{2}{\text{O}}_{3})}_{2}{}^{\text{3−}}?$$

### Solution

Two equilibria are involved when AgBr dissolves in a solution containing the $${\text{S}}_{2}{\text{O}}_{3}{}^{\text{2−}}$$ ion:

Reaction (1): $$\text{AgBr}(s)⇌{\text{Ag}}^{\text{+}}(aq)+{\text{Br}}^{\text{−}}(aq)\phantom{\rule{4em}{0ex}}{K}_{\text{sp}}=5.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−13}}$$

Reaction (2): $${\text{Ag}}^{\text{+}}(aq)+{\text{S}}_{2}{\text{O}}_{3}{}^{\text{2−}}(aq)⇌\text{Ag}{({\text{S}}_{2}{\text{O}}_{3})}_{2}{}^{\text{3−}}(aq)\phantom{\rule{4em}{0ex}}{K}_{\text{f}}=4.7\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{13}$$

In order for 1.00 g of AgBr to dissolve, the [Ag+] in the solution that results must be low enough for Q for Reaction (1) to be smaller than Ksp for this reaction. We reduce [Ag+] by adding $${\text{S}}_{2}{\text{O}}_{3}{}^{\text{2−}}$$ and thus cause Reaction (2) to shift to the right. We need the following steps to determine what mass of Na2S2O3 is needed to provide the necessary $${\text{S}}_{2}{\text{O}}_{3}{}^{\text{2−}}.$$

1. We calculate the [Br] produced by the complete dissolution of1.00 g of AgBr (5.33$$×$$10–3 mol AgBr) in 1.00 L of solution:

$$[{\text{Br}}^{\text{−}}]\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}5.33\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−3}}\phantom{\rule{0.2em}{0ex}}M$$

2. We use [Br] and Ksp to determine the maximum possible concentration of Ag+ that can be present without causing reprecipitation of AgBr:

$$[{\text{Ag}}^{+}]\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}9.4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−11}}\phantom{\rule{0.2em}{0ex}}M$$

3. We determine the$$\mathit{[}{S}_{2}{O}_{3}{}^{\mathit{2-}}\mathit{]}$$required to make [Ag+] = 9.4$$×$$10–11 Mafter the remaining Ag+ ion has reacted with$${S}_{2}{O}_{3}{}^{\mathit{2-}}$$according to the equation:

$${\text{Ag}}^{+}+2{\text{S}}_{2}{\text{O}}_{3}{}^{\text{2−}}⇌\text{Ag}{({\text{S}}_{2}{\text{O}}_{3})}_{2}{}^{\text{3−}}\phantom{\rule{4em}{0ex}}{K}_{\text{f}}=4.7\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{13}$$

Because 5.33 $$×$$ 10–3 mol of AgBr dissolves:

$$(5.33\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−3}})\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}(9.4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−11}})\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}5.33\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−3}}\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}\text{Ag}{({\text{S}}_{2}{\text{O}}_{3})}_{2}{}^{\text{3−}}$$

Thus, at equilibrium: $$[\text{Ag}{({\text{S}}_{2}{\text{O}}_{3})}_{2}{}^{\text{3−}}]$$ = 5.33 $$×$$ 10–3M, [Ag+] = 9.4 $$×$$ 10–11M, and Q = Kf = 4.7 $$×$$ 1013:

$${K}_{\text{f}}=\phantom{\rule{0.2em}{0ex}}\cfrac{[\text{Ag}{({\text{S}}_{2}{\text{O}}_{3})}_{2}{}^{\text{3−}}]}{[{\text{Ag}}^{+}][{\text{S}}_{2}{\text{O}}_{3}{}^{\text{2−}}{]}^{2}}\phantom{\rule{0.2em}{0ex}}=4.7\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{13}$$

$$[{\text{S}}_{2}{\text{O}}_{3}{}^{\text{2−}}]\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}1.1\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−3}}\phantom{\rule{0.2em}{0ex}}M$$

When $$[{\text{S}}_{2}{\text{O}}_{3}{}^{\text{2−}}]$$ is 1.1 $$×$$ 10–3M, [Ag+] is 9.4 $$×$$ 10–11M and all AgBr remains dissolved.

4. We determine the total number of moles of$${S}_{2}{O}_{3}{}^{\mathit{2-}}$$that must be added to the solution. This equals the amount that reacts with Ag+ to form $$\text{Ag}{({\text{S}}_{2}{\text{O}}_{3})}_{2}{}^{\text{3−}}$$ plus the amount of free $${\text{S}}_{2}{\text{O}}_{3}{}^{\text{2−}}$$ in solution at equilibrium. To form 5.33 $$×$$ 10–3 mol of $$\text{Ag}{({\text{S}}_{2}{\text{O}}_{3})}_{2}{}^{\text{3−}}$$ requires 2 $$×$$ (5.33 $$×$$ 10–3) mol of $${\text{S}}_{2}{\text{O}}_{3}{}^{\text{2−}}.$$ In addition, 1.1 $$×$$ 10–3 mol of unreacted $${\text{S}}_{2}{\text{O}}_{3}{}^{\text{2−}}$$ is present (Step 3). Thus, the total amount of $${\text{S}}_{2}{\text{O}}_{3}{}^{\text{2−}}$$ that must be added is:

$$2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}(5.33\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−3}}\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{S}}_{2}{\text{O}}_{3}{}^{\text{2−}})+1.1\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−3}}\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{S}}_{2}{\text{O}}_{3}{}^{\text{2−}}=1.18\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−2}}\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{S}}_{2}{\text{O}}_{3}{}^{\text{2−}}$$

5. We determine the mass of Na2S2O3 required to give 1.18 $$×$$ 10–2 mol$${S}_{2}{O}_{3}{}^{\mathit{2-}}$$using themolar mass of Na2S2O3:

$$1.18\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−2}}\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{S}}_{2}{\text{O}}_{3}{}^{\text{2−}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\cfrac{158.1\phantom{\rule{0.2em}{0ex}}\text{g}\phantom{\rule{0.2em}{0ex}}{\text{Na}}_{2}{\text{S}}_{2}{\text{O}}_{3}}{1\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{Na}}_{2}{\text{S}}_{2}{\text{O}}_{3}}\phantom{\rule{0.2em}{0ex}}=1.9\phantom{\rule{0.2em}{0ex}}\text{g}\phantom{\rule{0.2em}{0ex}}{\text{Na}}_{2}{\text{S}}_{2}{\text{O}}_{3}$$

Thus, 1.00 L of a solution prepared from 1.9 g Na2S2O3 dissolves 1.0 g of AgBr.