Write solubility product expressions, calculate Ksp and solubility, and predict precipitation by comparing the ionic product with Ksp

What this dot point is asking

When an ionic solid such as silver chloride is placed in water, a tiny amount dissolves until the solution is saturated. At that point a dynamic equilibrium exists between the undissolved solid and the dissolved ions:

For silver chloride the expression is $K_{sp} = [\text{Ag}^+][\text{Cl}^-]$. For a salt with a more complex formula such as calcium fluoride, $\text{CaF}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\text{F}^-(aq)$, each ion is raised to the power of its coefficient:

Linking Ksp to molar solubility

The molar solubility is the number of moles of the salt that dissolve per litre to give a saturated solution. You can convert between $K_{sp}$ and solubility using the stoichiometry of the dissolving equation. If the molar solubility of $\text{CaF}_2$ is $s$, then $[\text{Ca}^{2+}] = s$ and $[\text{F}^-] = 2s$, so $K_{sp} = s(2s)^2 = 4s^3$.

The common ion effect

Adding a soluble salt that shares an ion with the equilibrium reduces solubility. If sodium chloride is added to a saturated silver chloride solution, the extra chloride ions shift the equilibrium

to the left, so more solid precipitates and the solubility of AgCl falls. This is just Le Chatelier's principle applied to a solubility equilibrium.

Predicting precipitation with Q

When two solutions are mixed, you decide whether a precipitate forms by calculating the ionic product $Q$ (the same form as $K_{sp}$ but using the actual mixed concentrations) and comparing it with $K_{sp}$.

• If $Q > K_{sp}$, the solution is supersaturated and a precipitate forms until $Q = K_{sp}$.
• If $Q = K_{sp}$, the solution is exactly saturated; no change.
• If $Q < K_{sp}$, the solution is unsaturated and no precipitate forms.

Remember to account for dilution: when two solutions are combined, the volume increases, so each ion concentration is diluted before you substitute into $Q$.

Comparing solubilities across different formulas

A subtle trap is comparing two salts by their $K_{sp}$ values alone. You can only compare $K_{sp}$ directly when the salts have the same ion ratio. For two 1:1 salts (such as $\text{AgCl}$ and $\text{AgBr}$) the one with the larger $K_{sp}$ is more soluble. But comparing a 1:1 salt with a 1:2 salt requires converting each $K_{sp}$ back to a molar solubility $s$ first, because the relationship differs ($s = \sqrt{K_{sp}}$ for the 1:1 salt, $s = \sqrt[3]{K_{sp}/4}$ for the 1:2 salt). A salt with a smaller $K_{sp}$ can actually be more soluble in $\text{mol L}^{-1}$ if it releases more ions per formula unit, which is a favourite higher-order WACE question.

Why this matters

Solubility equilibria explain qualitative analysis (selective precipitation of ions), the formation of stalactites and kidney stones, water hardness and scale, and the use of barium sulfate as a contrast agent in medical imaging because its very small $K_{sp}$ keeps toxic barium ions out of solution.