The Common-Ion Effect

Le Chatelier’s principle applies to dissolving too. If the water already contains one of a salt's ions, the solubility equilibrium is pushed back toward the solid, and less of the salt dissolves. This is the Q-and-K argument once more.

Key idea

In a solution that already holds an ion at concentration $c$, set up $K_{sp}$ with that ion at $c$ plus what the salt adds. When $c$ is much larger than what the salt adds, use $c$ alone: for a 1:1 salt, $s = K_{sp}/c$.

Why it works

Adding $\mathrm{Cl^-}$ to saturated AgCl raises $Q = [\mathrm{Ag^+}][\mathrm{Cl^-}]$ above $K_{sp}$, so AgCl precipitates until $Q = K_{sp}$ again. In a solution that starts with $0.10\ \text{M}$ $\mathrm{Cl^-}$, $[\mathrm{Ag^+}]$ can reach only $K_{sp}/0.10$ before the product hits the limit. The salt's own contribution to $[\mathrm{Cl^-}]$ is then negligible.

The molar solubility of AgCl (exact, from $s(s + c) = K_{sp}$) against the chloride already present, both in units of $10^{-5}\ \text{M}$. Even a little common ion cuts the solubility sharply.

Worked examples

Worked example: AgCl in salt water

Find the molar solubility of AgCl ($K_{sp} = 1.8 \times 10^{-10}$) in $0.10\ \text{M}$ NaCl.

$[\mathrm{Ag^+}] = s$, $[\mathrm{Cl^-}] = 0.10 + s \approx 0.10$.
$s(0.10) = 1.8 \times 10^{-10}$, so $s = 1.8 \times 10^{-9}\ \text{M}$.
Check: $s$ is tiny next to 0.10. In pure water $s = 1.3 \times 10^{-5}\ \text{M}$, about 7000 times more.

Answer: $1.8 \times 10^{-9}\ \text{M}$.

Worked example: A 1:2 salt with a common anion

Find the molar solubility of $\mathrm{PbI_2}$ ($K_{sp} = 9.8 \times 10^{-9}$) in $0.10\ \text{M}$ KI.

$[\mathrm{Pb^{2+}}] = s$, $[\mathrm{I^-}] = 0.10 + 2s \approx 0.10$.
$K_{sp} = s(0.10)^2$, so $s = \dfrac{9.8 \times 10^{-9}}{0.010} = 9.8 \times 10^{-7}\ \text{M}$.

Answer: $9.8 \times 10^{-7}\ \text{M}$.

Worked example: A common cation

Find the molar solubility of $\mathrm{CaF_2}$ ($K_{sp} = 3.5 \times 10^{-11}$) in $0.010\ \text{M}$ $\mathrm{Ca(NO_3)_2}$.

$[\mathrm{Ca^{2+}}] = 0.010 + s \approx 0.010$; $[\mathrm{F^-}] = 2s$.
$K_{sp} = (0.010)(2s)^2 = 0.040s^2$, so $s = \sqrt{\dfrac{3.5 \times 10^{-11}}{0.040}} = 3.0 \times 10^{-5}\ \text{M}$.
Check: $3.0 \times 10^{-5}$ is 0.3% of 0.010, so dropping it was fine. (In pure water $s = 2.1 \times 10^{-4}\ \text{M}$.)

Answer: $3.0 \times 10^{-5}\ \text{M}$.

Chemists use the common-ion effect on purpose. To remove a metal ion from solution as completely as possible, they add a modest excess of the precipitating ion: the extra anion pushes the leftover metal ion concentration far below what it would be in a saturated solution of the pure salt.

Common mistake

Using the molar solubility in pure water even when a common ion is already present. The ion already in solution counts in $K_{sp}$. Check: the solubility with a common ion must come out smaller than in pure water.

The Common-Ion Effect

Worked examples

Practice

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Sources and further reading