# 9.10 Solubility Product and Precipitation Reactions This section explores the equilibrium between sparingly soluble ionic compounds and their ions in solution, focusing on the concepts of solubility product and the prediction of precipitation. ### 9.10.1 Solubility Product ($K_{sp}$) When an excess of a sparingly soluble ionic compound, like Calcium Fluoride ($CaF_2$), is mixed with water, a dynamic equilibrium is established between the undissolved solid and its dissolved ions in a saturated solution. The equilibrium reaction is: $$ \mathrm{CaF}_{2(\mathrm{~s})} \rightleftharpoons \mathrm{Ca}^{2+}_{(\mathrm{aq})} + 2\mathrm{F}^{-}_{(\mathrm{aq})} $$ The equilibrium constant expression ($K_c$) for this reaction is: $$ K_c = \frac{[\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^2}{[\mathrm{CaF}_2]} $$ Since $CaF_2$ is a solid, its concentration is considered constant and is incorporated into the equilibrium constant. This new constant is called the **solubility product constant ($K_{sp}$)**. $$ K_c[\mathrm{CaF}_2] = [\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^2 $$ $$ K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^2 $$ ***Definition:*** The **solubility product constant ($K_{sp}$)** is the product of the equilibrium concentrations of the constituent ions in a saturated solution of a sparingly soluble salt, with each concentration raised to the power of its stoichiometric coefficient in the balanced dissolution equation. **General Expression** For a general sparingly soluble ionic compound $A_m B_n$, the dissolution equilibrium is: $$ A_m B_{n(s)} \rightleftharpoons m A^{n+}_{(aq)} + n B^{m-}_{(aq)} $$ The solubility product expression is: $$ K_{sp} = [A^{n+}]^m [B^{m-}]^n $$ > **Did You Know?** > Precipitates are insoluble ionic solid products that form when certain cations and anions combine in an aqueous solution. ### 9.10.2 Precipitation Reactions Precipitation is the reverse process of dissolution, where a solid forms from a solution. A **precipitation reaction** occurs when two or more solutions are mixed, yielding an insoluble solid substance. To predict whether a precipitate will form, we use the **ion product ($Q_{sp}$)**. The ion product has the same form as the $K_{sp}$ expression but uses the *initial concentrations* of the ions rather than their equilibrium concentrations. For the $CaF_2$ example, the ion product is: $$ Q_{sp} = [\text{Initial conc. of } \mathrm{Ca}^{2+}] [\text{Initial conc. of } \mathrm{F}^{-}]^2 $$ **Predicting Precipitation** By comparing the ion product ($Q_{sp}$) with the solubility product constant ($K_{sp}$), we can determine if a precipitate will form: * **If $Q_{sp} > K_{sp}$**: The solution is *supersaturated*. Precipitation will occur, and the ion concentrations will decrease until the product equals $K_{sp}$. * **If $Q_{sp} < K_{sp}$**: The solution is *unsaturated*. No precipitation will occur. * **If $Q_{sp} = K_{sp}$**: The solution is *saturated*. The system is at equilibrium, and no net change will occur. ### Common Ion Effect and Solubility The solubility of a sparingly soluble salt is decreased in the presence of a common ion. This is an application of Le Chatelier's Principle. ## Worked Examples ### Example 9.7 The solubility of $AgBr$ is $7.1 \times 10^{-7} \, M$ at $25^{\circ}C$. Calculate its $K_{sp}$. ### Solution 1. **Write the dissolution equation:** When $AgBr$ dissolves, it forms silver ($Ag^+$) and bromide ($Br^−$) ions in a 1:1 molar ratio. $$ \mathrm{AgBr}_{(\mathrm{s})} \rightleftharpoons \mathrm{Ag}^{+}_{(\mathrm{aq})} + \mathrm{Br}^{-}_{(\mathrm{aq})} $$ 2. **Determine ion concentrations:** Since the solubility of $AgBr$ is $7.1 \times 10^{-7} \, M$, the concentrations of the ions at equilibrium are: * $[\mathrm{Ag}^{+}] = 7.1 \times 10^{-7} \, M$ * $[\mathrm{Br}^{-}] = 7.1 \times 10^{-7} \, M$ 3. **Apply the $K_{sp}$ formula:** $$ K_{sp} = [\mathrm{Ag}^{+}][\mathrm{Br}^{-}] $$ 4. **Calculate the value:** $$ K_{sp} = (7.1 \times 10^{-7})(7.1 \times 10^{-7}) $$ $$ K_{sp} = 50.41 \times 10^{-14} $$ $$ \boldsymbol{K_{sp} = 5.04 \times 10^{-13}} $$ ## Possible Questions/Answers ### Concept Assessment Exercise 9.5 **i. Write $K_{sp}$ expressions for:** * **a) Iron (II) Hydroxide, $Fe(OH)_2$** * **Dissolution:** $Fe(OH)_{2(s)} \rightleftharpoons Fe^{2+}_{(aq)} + 2OH^{-}_{(aq)}$ * **Expression:** $K_{sp} = [Fe^{2+}][OH^{-}]^2$ * **b) Calcium Sulphate, $CaSO_4$** * **Dissolution:** ${CaSO_4}_{(s)} \rightleftharpoons {Ca^{2+}}_{(aq)} + {SO_4^{2-}}_{(aq)}$ * **Expression:** $K_{sp} = [Ca^{2+}][SO_{4}^{2-}]$ **ii. Lead(II) Sulphate ($PbSO_4$) is used as a white pigment. What is the solubility of $PbSO_4$ if $K_{sp} = 1.96 \times 10^{-8}$ at $25^{\circ}C$?** * **A:** 1. **Let 's' be the molar solubility of $PbSO_4$.** The dissolution is: $\mathrm{{PbSO_4}_{(s)} \rightleftharpoons {Pb^{2+}}_{(aq)} + {SO_4^{2-}}_{(aq)}}$ At equilibrium: $[Pb^{2+}] = s$ and $[SO_{4}^{2-}] = s$. 2. **Write the $K_{sp}$ expression:** $K_{sp} = [Pb^{2+}][SO_{4}^{2-}] = (s)(s) = s^2$ 3. **Solve for 's':** $s = \sqrt{K_{sp}}$ $s = \sqrt{1.96 \times 10^{-8}}$ $s = 1.4 \times 10^{-4} \, M$ The solubility of $PbSO_4$ is **$1.4 \times 10^{-4} \, M$**. **iii. Phosphate in natural water often precipitates as insoluble $Ca_3(PO_4)_2$. In the Indus river, the concentration of $Ca^{2+}$ and $PO_4^{3-}$ ions is $1.0 \times 10^{-9} \, M$ each. Will calcium phosphate precipitate? ($K_{sp} = 1.2 \times 10^{-20}$ at $25^{\circ}C$)** * **A:** No, it will not precipitate. 1. **Write the dissolution equation and ion product expression ($Q_{sp}$):** $\mathrm{{Ca_3(PO_4)_2}_{(s)} \rightleftharpoons 3{Ca^{2+}}_{(aq)} + 2{PO_4^{3-}}_{(aq)}}$ $Q_{sp} = [Ca^{2+}]^3[PO_4^{3-}]^2$ 2. **Calculate $Q_{sp}$ using initial concentrations:** Given: $[Ca^{2+}] = 1.0 \times 10^{-9} \, M$ and $[PO_4^{3-}] = 1.0 \times 10^{-9} \, M$. $Q_{sp} = (1.0 \times 10^{-9})^3 (1.0 \times 10^{-9})^2$ $Q_{sp} = (1.0 \times 10^{-27}) (1.0 \times 10^{-18})$ $Q_{sp} = 1.0 \times 10^{-45}$ 3. **Compare $Q_{sp}$ to $K_{sp}$:** $Q_{sp} = 1.0 \times 10^{-45}$ $K_{sp} = 1.2 \times 10^{-20}$ Since $1.0 \times 10^{-45} < 1.2 \times 10^{-20}$, **$Q_{sp} < K_{sp}$**. Therefore, no precipitate will form.