6. Solubility & the Solubility Product
Even "insoluble" salts dissolve a little. The solubility product $K_{sp}$ quantifies exactly how little — and predicts whether a precipitate will form when solutions mix.
The Ksp Expression
For $\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)$: $$K_{sp} = [\text{Ag}^+][\text{Cl}^-]$$ (Solids never appear in the expression.)
Example: AgCl, $K_{sp} = 1.8 \times 10^{-10}$. Let $s$ = molar solubility:
$$s^2 = 1.8 \times 10^{-10} \;\Rightarrow\; s = 1.34 \times 10^{-5}\ \text{M}$$
The Common Ion Effect
Adding an ion already in the equilibrium shifts it left (Le Châtelier) and crushes solubility.
Example: AgCl in 0.10 M NaCl: $[\text{Cl}^-] \approx 0.10$ M, so $s = \frac{K_{sp}}{0.10} = 1.8 \times 10^{-9}$ M — about 7,000 times less soluble than in pure water.
Predicting Precipitation: Q vs Ksp
Calculate $Q$ with the current concentrations. If $Q > K_{sp}$, a precipitate forms; if $Q < K_{sp}$, it does not.
Example: Mix to give $[\text{Pb}^{2+}] = 1.0 \times 10^{-3}$ M and $[\text{I}^-] = 2.0 \times 10^{-3}$ M. For PbI2 ($K_{sp} = 9.8 \times 10^{-9}$):
$$Q = (1.0 \times 10^{-3})(2.0 \times 10^{-3})^2 = 4.0 \times 10^{-9} < K_{sp}$$
No precipitate forms.
Practice Problems
- Find the molar solubility of BaSO4 (Ksp = 1.1 × 10−10).
- CaF2 has a molar solubility of 2.1 × 10−4 M. Calculate its Ksp.
- Will AgCl precipitate if [Ag+] = [Cl−] = 1.0 × 10−6 M? (Ksp = 1.8 × 10−10)
- How does adding NaF affect the solubility of CaF2?
Answers: 1. 1.05 × 10−5 M 2. Ksp = 4s3 = 3.7 × 10−11 3. No — Q = 1.0 × 10−12 < Ksp 4. It decreases — the common ion F− shifts the equilibrium left