Common-ion effect
The common-ion effect is the decrease in a salt’s molar solubility \(s\) when the solution
already contains one (or both) of its ions. For a general salt \(\mathrm{A}_m\mathrm{B}_n\):
\(\mathrm{A}_m\mathrm{B}_n(s)\;\rightleftharpoons\; m\,\mathrm{A}^{\,z_A+}(aq)\;+\;n\,\mathrm{B}^{\,z_B-}(aq)\)
With the usual approximation (activities \(\approx\) concentrations, solid omitted):
\(K_{sp}=\big[\mathrm{A}^{\,z_A+}\big]^m\,\big[\mathrm{B}^{\,z_B-}\big]^n\)
ICE setup and master equation
Let the initial ion concentrations be \(\big[\mathrm{A}^{\,z_A+}\big]_0\) and
\(\big[\mathrm{B}^{\,z_B-}\big]_0\). If the molar solubility is \(s\) (mol·L\(^{-1}\)), then at equilibrium:
- \(\big[\mathrm{A}^{\,z_A+}\big]=\big[\mathrm{A}^{\,z_A+}\big]_0 + m\,s\)
- \(\big[\mathrm{B}^{\,z_B-}\big]=\big[\mathrm{B}^{\,z_B-}\big]_0 + n\,s\)
Substituting into the expression for \(K_{sp}\) gives the equation the calculator solves:
\((\,[\mathrm{A}]_0 + m\,s\,)^{m}(\,[\mathrm{B}]_0 + n\,s\,)^{n} = K_{sp}\)
Key special cases
-
No common ions (\([\mathrm{A}]_0=[\mathrm{B}]_0=0\)):
\(K_{sp}=m^{m}n^{n}s^{\,m+n}\),
\(\displaystyle s=\Big(\dfrac{K_{sp}}{m^{m}n^{n}}\Big)^{1/(m+n)}\).
-
Only one common ion, in large excess (e.g. \([\mathrm{B}]_0\gg n\,s\), \([\mathrm{A}]_0\approx 0\)):
\(\displaystyle s \approx \Big(\dfrac{K_{sp}}{m^{m}[\mathrm{B}]_0^{\,n}}\Big)^{1/m}\).
For a 1:1 salt: \(s \approx \dfrac{K_{sp}}{[\mathrm{B}]_0}\).
-
Already saturated by common ions:
if \(([\mathrm{A}]_0)^{m}([\mathrm{B}]_0)^{n} \ge K_{sp}\), then \(s \approx 0\)
(no further dissolution; precipitation is favored).
Assumes pure water, activities ≈ concentrations, and no side reactions (complexation, pH effects, etc.).
The calculator solves the full equation for \(s \ge 0\) and may also show the simple common-ion
approximation when valid.
