Writing Equations and Solubility Products

Write the ionic equation for the dissolution and the solubility product expression for each of the following slightly soluble ionic compounds:

(a) AgI, silver iodide, a solid with antiseptic properties

(b) CaCO₃, calcium carbonate, the active ingredient in many over-the-counter chewable antacids

(c) Mg(OH)₂, magnesium hydroxide, the active ingredient in Milk of Magnesia

(d) Mg(NH₄)PO₄, magnesium ammonium phosphate, an essentially insoluble substance used in tests for magnesium

(e) Ca₅(PO₄)₃OH, the mineral apatite, a source of phosphate for fertilizers

SOLUTION

(a) [latex]\text{AgI}(s)\;{\rightleftharpoons}\;\text{Ag}^{+}(aq)\;+\;\text{I}^{-}(aq)\;\;\;\;\;\;\;K_{\text{sp}} = [\text{Ag}^{+}][\text{I}^{-}][/latex]

(b) [latex]\text{CaCO}_3(s)\;{\rightleftharpoons}\;\text{Ca}^{2+}(aq)\;+\;\text{CO}_3^{2-}(aq) \;\;\;\;\;\;\;K_{\text{sp}} = [\text{Ca}^{2+}][\text{CO}_3^{2-}][/latex]

(c) [latex]\text{Mg(OH)}_2(s)\;{\rightleftharpoons}\;\text{Mg}^{2+}(aq)\;+\;2\text{ OH}^{-}(aq)\;\;\;\;\;\;\;K_{\text{sp}} = [\text{Mg}^{2+}][\text{OH}^{-}]^2[/latex]

(d) [latex]\text{Mg(NH}_4)\text{PO}_4(s)\;{\rightleftharpoons}\;\text{Mg}^{2+}(aq)\;+\;\text{NH}_4^{+}(aq)\;+\;\text{PO}_4^{3-}(aq)\;\;\;\;\;\;\;K_{\text{sp}} = [\text{Mg}^{2+}][\text{NH}_4^{+}][\text{PO}_4^{3-}][/latex]

(e) [latex]\text{Ca}_5(\text{PO}_4)_3\text{OH}(s)\;{\rightleftharpoons}\;5\text{ Ca}^{2+}(aq)\;+\;3\text{ PO}_4^{\;\;3-}(aq) \;+\;\text{OH}^{-}(aq)\;\;\;\;\;\;\;K_{\text{sp}} = [\text{Ca}^{2+}]^5[\text{PO}_4^{3-}]^3[\text{OH}^{-}][/latex]

Calculation of K_sp from Equilibrium Concentrations

Fluorite, CaF₂, is a slightly soluble solid that dissolves according to the equation

[latex]\text{CaF}_2(s)\;{\rightleftharpoons}\;\text{Ca}^{2+}(aq)\;+\;2\text{ F}^{-}(aq)[/latex]

The concentration of Ca²⁺ in a saturated solution of CaF₂ is 2.15 × 10^–4 M; therefore, that of F^– is 4.30 × 10^–4 M, that is, twice the concentration of Ca²⁺. What is the solubility product of fluorite?

SOLUTION

First, write out the K_sp expression, then substitute in concentrations and solve for K_sp

[latex]\text{CaF}_2(s)\;{\rightleftharpoons}\;\text{Ca}^{2+}(aq)\;+\;2\text{ F}^{-}(aq)[/latex]

A saturated solution is a solution at equilibrium with the solid. Thus

[latex]K_{\text{sp}} = [\text{Ca}^{2+}][\text{F}^{-}]^2 = (2.15\;\times\;10^{-4})(4.30\;\times\;10^{-4})^2 = \boxed{3.98\;\times\;10^{-11}}[/latex]

As with other equilibrium constants, we do not include units with K_sp.

Determination of Molar Solubility from K_sp

The K_sp of copper (I) bromide, CuBr, is 6.3 × 10^–9. Calculate the molar solubility of copper (I) bromide.

SOLUTION

The dissolution equation and solubility product expression are

CuBr(s) ⇌ Cu⁺(aq) + Br⁻(aq) K_sp = [Cu⁺][Br⁻]

Following the ICE approach to this calculation yields the table

Substituting the equilibrium concentration terms into the solubility product expression and solving for x yields

[latex]K_{\text{sp}}=[\text{Cu}^+][\text{Br}^−] \\ 6.3×10^{−9}=(x)(x)=x^2 \\ x=\sqrt{6.3×10^{−9}} =\boxed{7.9×10^{−5}\text{ M}}[/latex]

Since the dissolution stoichiometry shows one mole of copper (I) ion and one mole of bromide ion are produced for each mole of CuBr dissolved, the molar solubility of CuBr is 7.9 × 10^–5 M.

Determination of Molar Solubility from K_sp

The K_sp of calcium hydroxide, Ca(OH)₂, is 1.3 × 10^–6. Calculate the molar solubility of calcium hydroxide.

SOLUTION

The dissolution equation and solubility product expression are

Ca(OH)₂(s) ⇌ Ca²⁺(aq) + 2 OH⁻(aq)                       K_sp = [Ca²⁺][OH⁻]²

The ICE table for this system is

Substituting terms for the equilibrium concentrations into the solubility product expression and solving for x gives

[latex]K_{\text{sp}}=[\text{Ca}^{2+}][\text{OH}^−]^2 \\ 1.3×10^{−6}​=(x)(2x)^2=(x)(4x^2)=4x^3 \\ x=\sqrt[3]{3.3×10^{−7}}=\boxed{{6.9×10^{−3}}\text{ M}}[/latex]

As defined in the ICE table, x is the molarity of calcium ion in the saturated solution. The dissolution stoichiometry shows a 1:1 relation between moles of calcium ion in solution and moles of compound dissolved, and so, the molar solubility of Ca(OH)₂ is 6.9 × 10^–3 M.

Precipitation of Mg(OH)₂

The first step in the preparation of magnesium metal is the precipitation of Mg(OH)₂ from sea water by the addition of lime, Ca(OH)₂, a readily available inexpensive source of OH^–

[latex]\text{Mg(OH)}_2(s)\;{\rightleftharpoons}\;\text{Mg}^{2+}(aq)\;+\;2\text{ OH}^{-}(aq)\;\;\;\;\;\;\;K_{\text{sp}} = 8.9\times 10^{-12}[/latex]

The concentration of Mg²⁺(aq) in sea water is 0.0537 M. Will Mg(OH)₂ precipitate when enough Ca(OH)₂ is added to give a [OH^–] of 0.0010 M?

SOLUTION

Calculation of the reaction quotient under these conditions is shown here

[latex]Q = [\text{Mg}^{2+}][\text{OH}^{-}]^2 = (0.0537)(0.0010)^2 = 5.4\times 10^{-8}[/latex]

Because Q is greater than K_sp (Q = 5.4 × 10^–8 is larger than K_sp = 8.9 × 10^–12), the reverse reaction will proceed, precipitating magnesium hydroxide until the dissolved ion concentrations have been sufficiently lowered, so that Q_sp = K_sp.

Precipitation of AgCl

Does silver chloride precipitate when equal volumes of a 2.0 × 10^–4-M solution of AgNO₃ and a 2.0 × 10^–4-M solution of NaCl are mixed?

SOLUTION

The equation for the equilibrium between solid silver chloride, silver ion, and chloride ion

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

The solubility product is 1.6 × 10^–10.

AgCl will precipitate if the reaction quotient calculated from the concentrations in the mixture of AgNO₃ and NaCl is greater than K_sp. Because the volume doubles when equal volumes of AgNO₃ and NaCl solutions are mixed, each concentration is reduced to half its initial value

[latex]\frac{1}{2}(2.0\;\times\;10^{-4})\;\text{ M} = 1.0\;\times\;10^{-4}\;\text{ M}[/latex]

The reaction quotient, Q, is greater than K_sp for AgCl, so a supersaturated solution is formed.

[latex]Q = [\text{Ag}^{+}][\text{Cl}^{-}] = (1.0\;\times\;10^{-4})(1.0\;\times\;10^{-4}) = 1.0\;\times\;10^{-8}\;>\;K_{\text{sp}}[/latex]

AgCl will precipitate from the mixture until the dissolution equilibrium is established, with Q equal to K_sp.

Concentrations Following Precipitation

Clothing washed in water that has a manganese (Mn²⁺) concentration exceeding 0.1 mg/L (1.8 × 10^–6 M) may be stained by the manganese upon oxidation, but the amount of Mn²⁺ in the water can be decreased by adding a base to precipitate Mn(OH)₂. What pH is required to keep [Mn²⁺] equal to 1.8 × 10^–6 M?

SOLUTION

The dissolution of Mn(OH)₂ is described by the equation

[latex]\text{Mn(OH)}_2(s)\;{\rightleftharpoons}\;\text{Mn}^{2+}(aq)\;+\;2\text{ OH}^{-}(aq)\;\;\;\;\;\;\;K_{\text{sp}} = 2.0\;\times\;10^{-13}[/latex]

We need to calculate the concentration of OH^– when the concentration of Mn²⁺ is 1.8 × 10^–6 M. From that, we calculate the pH. At equilibrium

[latex]K_{\text{sp}} = [\text{Mn}^{2+}][\text{OH}^{-}]^2[/latex]

or

[latex](1.8\;\times\;10^{-6})[\text{OH}^{-}]^2 = 2.0\;\times\;10^{-13}[/latex]

so

[latex][\text{OH}^{-}] = 3.3\;\times\;10^{-4}\;\text{ M}[/latex]

Now we calculate the pH from the pOH

[latex]\text{pOH} = -\log [\text{OH}^{-}] = -\log (3.3\;\times\;10^{-4}) = 3.48[/latex]

[latex]\text{pH} = 14.00\;-\;\text{pOH} = 14.00\;-\;3.80 = \boxed{10.52}[/latex]

If the person doing laundry adds a base, such as the sodium silicate (Na₄SiO₄) in some detergents, to the wash water until the pH is raised to 10.52, the manganese ion will be reduced to a concentration of 1.8 × 10^–6 M; at that concentration or less, the ion will not stain clothing.

Common Ion Effect

Calculate the molar solubility of cadmium sulfide (CdS) in a 0.010-M solution of cadmium bromide (CdBr₂). The K_sp of CdS is 1.0 × 10^–28.

SOLUTION

The first thing you should notice is that the cadmium sulfide is dissolved in a solution that contains cadmium ions. We need to use an ICE table to set up this problem and include the CdBr₂ concentration as a contributor of cadmium ions

[latex]\text{CdS}(s)\;{\rightleftharpoons}\;\text{Cd}^{2+}(aq)\;+\;\text{S}^{2-}(aq)[/latex]

[latex]K_{\text{sp}} = [\text{Cd}^{2+}][\text{S}^{2-}] = 1.0\;\times\;10^{-28}[/latex]

[latex](0.010\;+\;x)(x) = 1.0\;\times\;10^{-28}[/latex]

[latex]x^2\;+\;0.010x\;-\;1.0\;\times\;10^{-28} = 0[/latex]

We can solve this equation using the quadratic formula, but we can also make an assumption to make this calculation much simpler. Since the K_sp value is so small compared with the cadmium concentration, we can assume that the change between the initial concentration and the equilibrium concentration is negligible, so that 0.010 + x ≈ 0.010. Going back to our K_sp expression, we would now get

[latex]K_{\text{sp}} = [\text{Cd}^{2+}][\text{S}^{2-}] = 1.0\;\times\;10^{-28}[/latex]

[latex](0.010)(x) = 1.0\;\times\;10^{-28}[/latex]

[latex]x = \boxed{1.0\;\times\;10^{-26} \text{ M}}[/latex]

Check that the relative magnitude of x is, in fact, small enough to use the approximation

[latex]\begin{align*} \frac{x}{[\text{Cd}^{2+}]_{\text{initial}}}&=\frac{1.0\times10^{-26} \text{ M}}{0.010 \text{ M}} \\ &=1.0\times10^{-24} = 1.0\times10^{-22}\% < 5\% \end{align*}[/latex]

Therefore, the molar solubility of CdS in this solution is 1.0 × 10^–26 M.

Effects of Common Ions and pH on Solubility

What is the effect on the amount of solid Mg(OH)₂ and the concentrations of Mg²⁺ and OH^– when each of the following are added to a saturated solution of Mg(OH)₂?

(a) MgCl₂

(b) KOH

(c) NaNO₃

(d) Mg(OH)₂

(e) HCl

SOLUTION

The solubility equilibrium is

Mg(OH)₂(s) ⇌ Mg²⁺(aq) + 2 OH⁻(aq)

(a) Adding a common ion, Mg²⁺, will increase the concentration of this ion and shift the solubility equilibrium to the left, decreasing the concentration of hydroxide ion and increasing the amount of undissolved magnesium hydroxide.

(b) Adding a common ion, OH^–, will increase the concentration of this ion and shift the solubility equilibrium to the left, decreasing the concentration of magnesium ion and increasing the amount of undissolved magnesium hydroxide.

(c) The added compound does not contain a common ion, and no effect on the magnesium hydroxide solubility equilibrium is expected.

(d) Adding more solid magnesium hydroxide will increase the amount of undissolved compound in the mixture. The solution is already saturated, though, so the concentrations of dissolved magnesium and hydroxide ions will remain the same.

Q = [Mg²⁺][OH⁻]²

Thus, changing the amount of solid magnesium hydroxide in the mixture has no effect on the value of Q, and no shift is required to restore Q to the value of the equilibrium constant.

(e) Adding the acid HCl will consume the hydroxide ions produced from Mg(OH)₂ and increase the solubility. The concentration of magnesium ion will increase, but the concentration of hydroxide ion will decrease.

Effect of pH on Solubility

How does lowering the pH affect the solubility of the following ionic compounds?

(a) Ni₃(PO₄)₂

(b) AgCl

SOLUTION

(a) The equation for the dissolution of Ni₃(PO₄)₂ is

Ni₃(PO₄)₂(s) ⇌ 3 Ni²⁺(aq) + 2 PO₄³⁻(aq)

Lowering the pH by adding acid will cause an acid-base reaction between the acid and the basic PO₄³⁻. The consumption of PO₄³⁻ in acid-base reactions will shift the equilibrium toward the right and increase the solubility of Ni₃(PO₄)₂.

(b) The equation for the dissolution of AgCl is

AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)

Although Cl⁻ is the conjugate base of the strong acid HCl, Cl⁻ has negligible basic properties in aqueous solution. Lowering the pH does not affect the equilibrium. This is a trick question!