Solubility Product: MDCAT Chemistry notes
Solubility Product MDCAT notes: Ksp expressions, relation between Ksp and molar solubility, ionic product and precipitation, and worked examples.
Solubility equilibrium
A saturated solution is one in which the dissolved solute is in equilibrium with undissolved solute. For a sparingly soluble salt such as AgCl:
$$\mathrm{AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)}$$
Solubility product, Ksp
The solubility product is the product of the molar concentrations of the ions in a saturated solution of a sparingly soluble salt, each raised to the power of its coefficient in the balanced equation, at a given temperature.
$$K_{sp} = [\mathrm{Ag^+}][\mathrm{Cl^-}]$$
- The solid does not appear in the expression; its concentration is constant.
- For $\mathrm{A_xB_y \rightleftharpoons xA^{y+} + yB^{x-}}$: $K_{sp} = [\mathrm{A^{y+}}]^x[\mathrm{B^{x-}}]^y$
- Like every equilibrium constant, $K_{sp}$ depends only on temperature, not on concentration, pressure or volume.
- Example value: $K_{sp}$ of $\mathrm{PbSO_4}$ at 25 °C is about $1.6\times10^{-8}\ \mathrm{mol^2\,dm^{-6}}$.
Ksp and molar solubility (s)
| Salt type | Example | Ion concentrations | $K_{sp}$ |
|---|---|---|---|
| AB | AgCl, $\mathrm{PbSO_4}$, PbS | $s$, $s$ | $s^2$ |
| $\mathrm{AB_2}$ or $\mathrm{A_2B}$ | $\mathrm{Ca(OH)_2}$, $\mathrm{CaF_2}$, $\mathrm{Ag_2CrO_4}$ | $s$, $2s$ | $4s^3$ |
| $\mathrm{AB_3}$ | $\mathrm{Fe(OH)_3}$ | $s$, $3s$ | $27s^4$ |
| $\mathrm{A_2B_3}$ | $\mathrm{Bi_2S_3}$ | $2s$, $3s$ | $108s^5$ |
Worked example 1
The solubility of a 1:1 salt MX is $1.0\times10^{-5}\ \mathrm{mol\,dm^{-3}}$. Then $K_{sp} = s^2 = 1.0\times10^{-10}\ \mathrm{mol^2\,dm^{-6}}$.
Worked example 2
A salt $\mathrm{MX_2}$ has $K_{sp} = 3.2\times10^{-11}$. Then $4s^3 = 3.2\times10^{-11}$, so $s^3 = 8\times10^{-12}$ and $s = 2\times10^{-4}\ \mathrm{mol\,dm^{-3}}$. The concentration of $\mathrm{X^-}$ is $2s = 4\times10^{-4}\ \mathrm{mol\,dm^{-3}}$.
Worked example 3
For a 1:1 salt with $K_{sp} = 9\times10^{-20}$, each ion concentration is $s = \sqrt{9\times10^{-20}} = 3\times10^{-10}\ \mathrm{mol\,dm^{-3}}$.
Comparing solubilities
A smaller $K_{sp}$ means a less soluble salt, but only when the salts are of the same type (for example, both 1:1). For salts of different types, calculate $s$ from $K_{sp}$ before comparing, because a 1:2 salt with a smaller $K_{sp}$ can still be more soluble than a 1:1 salt.
Ionic product and precipitation
The ionic product (Q) is calculated the same way as $K_{sp}$ but with the actual concentrations present at any moment.
- Q < $K_{sp}$: unsaturated; no precipitate, more solid can dissolve.
- Q = $K_{sp}$: saturated; equilibrium.
- Q > $K_{sp}$: supersaturated; precipitation occurs until Q falls to $K_{sp}$.
Key formulas
- 1:1 salt: $K_{sp} = s^2$, $s = \sqrt{K_{sp}}$
- 1:2 salt: $K_{sp} = 4s^3$, $s = \sqrt[3]{K_{sp}/4}$
- General $\mathrm{A_xB_y}$: $K_{sp} = x^x y^y s^{x+y}$
Common MDCAT traps
- Precipitation needs the ionic product to exceed $K_{sp}$, not just equal it.
- Do not include the solid salt in the $K_{sp}$ expression.
- For $\mathrm{Ca(OH)_2}$ use $4s^3$, not $s^2$.
- If a question gives a value called "solubility" with units of $\mathrm{mol^2\,dm^{-6}}$ or a very tiny number, it is really $K_{sp}$; take the square root for a 1:1 salt.
Quick revision
- $K_{sp}$ applies to saturated solutions of sparingly soluble salts.
- $K_{sp}$ changes only with temperature.
- Q > $K_{sp}$ means a precipitate forms.
- $K_{sp}$ of $\mathrm{PbSO_4}$ is about $1.6\times10^{-8}$ at 25 °C.