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 ($Ca{F}_2$), 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: ${\mathrm{CaF}}_{2(\mathrm{s})}\rightleftharpoons {\mathrm{Ca}}_{(\mathrm{aq})}^{2+}+2{\mathrm{F}}_{(\mathrm{aq})}^{-}$

The equilibrium constant expression ($K_c$) for this reaction is: $K_c=\frac{[{\mathrm{Ca}}^{2+}][{\mathrm{F}}^{-}{]}^2}{[{\mathrm{CaF}}_2]}$

Since $Ca{F}_2$ 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: $A_m{B}_{n(s)}\rightleftharpoons m{A}_{(aq)}^{n+}+n{B}_{(aq)}^{m-}$

The solubility product expression is: $K_{sp}=[A^{n+}{]}^m[B^{m-}{]}^n$

9.10.2 Calculating $K_{sp}$ from Molar Solubility

The molar solubility ($s$) of a salt is the number of moles that dissolve per litre of solution to form a saturated solution. Once $s$ is known, the equilibrium ion concentrations can be determined from stoichiometry and substituted into the $K_{sp}$ expression.

Example: For $AgBr$ (1:1 salt): $[A{g}^{+}]=[B{r}^{-}]=s$, so $K_{sp}=s^2$.

Example: For $Ca{F}_2$: $[C{a}^{2+}]=s$ and $[F^{-}]=2s$, so $K_{sp}=(s)(2s{)}^2=4s^3$.

9.10.3 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 $Ca{F}_2$ 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.

9.10.4 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.

Example: Adding $NaF$ to a saturated $Ca{F}_2$ solution introduces extra $F^{-}$ ions. By Le Chatelier's Principle, the equilibrium shifts left (toward undissolved solid), reducing the solubility of $Ca{F}_2$.

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: ${\mathrm{AgBr}}_{(\mathrm{s})}\rightleftharpoons {\mathrm{Ag}}_{(\mathrm{aq})}^{+}+{\mathrm{Br}}_{(\mathrm{aq})}^{-}$

2. Determine ion concentrations: Since the solubility of $AgBr$ is $7.1\times 10^{-7}M$:

   • $[{\mathrm{Ag}}^{+}]=7.1\times 10^{-7}M$
   • $[{\mathrm{Br}}^{-}]=7.1\times 10^{-7}M$

3. Apply the $K_{sp}$ formula: $K_{sp}=[{\mathrm{Ag}}^{+}][{\mathrm{Br}}^{-}]$

4. Calculate the value: $K_{sp}=(7.1\times 10^{-7})(7.1\times 10^{-7})=(7.1\times 10^{-7}{)}^2$ ${\mathbf{K}}_{\mathbf{sp}}=5.04\times 10^{-13}$

Possible Questions/Answers

Concept Assessment Exercise 9.5

i. Write $K_{sp}$ expressions for:

• a) Iron (II) Hydroxide, $Fe(OH{)}_2$

  • Dissolution: $Fe(OH{)}_{2(s)}\rightleftharpoons F{e}_{(aq)}^{2+}+2O{H}_{(aq)}^{-}$
  • Expression: $K_{sp}=[F{e}^{2+}][O{H}^{-}{]}^2$

• b) Calcium Sulphate, $CaS{O}_4$

  • Dissolution: CaSO_{4(s)} \rightleftharpoons Ca^{2+}_{(aq)} + SO_4^{2-}_{(aq)}
  • Expression: $K_{sp}=[C{a}^{2+}][S{O}_4^{2-}]$

ii. Lead(II) Sulphate ($PbS{O}_4$) is used as a white pigment. What is the solubility of $PbS{O}_4$ if $K_{sp}=1.96\times 10^{-8}$ at $25^{\circ }C$?

• Solution:

  1. Let $s$ be the molar solubility of $PbS{O}_4$. The dissolution is: PbSO_{4(s)} \rightleftharpoons Pb^{2+}_{(aq)} + SO_4^{2-}_{(aq)} At equilibrium: $[P{b}^{2+}]=s$ and $[S{O}_4^{2-}]=s$.

  2. Write the $K_{sp}$ expression: $K_{sp}=[P{b}^{2+}][S{O}_4^{2-}]=(s)(s)=s^2$

  3. Solve for $s$: $s=\sqrt{K_{sp}}=\sqrt{1.96\times 10^{-8}}$ $\mathbf{s}=1.4\times 10^{-4}\mathbf{M}$

  The solubility of $PbS{O}_4$ is $1.4\times 10^{-4}$ mol/L at $25^{\circ }C$.