# What is common ion effect?

May 17, 2018

It is simply when an ion in-common with one about to be introduced into solution is already there, thus suppressing the solubility of the other ion(s).

Consider the dissociation of "Ca"("OH")_2(s) in water at ${25}^{\circ} \text{C}$, with solubility product constant ${K}_{s p} = 5.5 \times {10}^{- 6}$.

${\text{Ca"("OH")_2(s) rightleftharpoons "Ca"^(2+)(aq) + 2"OH}}^{-} \left(a q\right)$

$\text{I"" "-" "" "" "" "" "0" "" "" "" "" } 0$
$\text{C"" "-" "" "" "" "+s" "" "" "" } + 2 s$
$\text{E"" "-" "" "" "" "" "s" "" "" "" "" } 2 s$

The mass action expression in pure water is thus:

${K}_{s p} = {\left[{\text{Ca"^(2+)]["OH}}^{-}\right]}^{2}$

$= s {\left(2 s\right)}^{2} = 4 {s}^{3}$

and the solubility of "Ca"("OH")_2(aq) in pure water is:

$\textcolor{b l u e}{s} \equiv \left[{\text{Ca"("OH")_2(aq)] = ["Ca}}^{2 +}\right] = {\left({K}_{s p} / 4\right)}^{1 / 3}$

$=$ $\textcolor{b l u e}{\text{0.011 M}}$

Now suppose the solution already contained $\text{0.100 M}$ $\text{NaOH}$. Then we would have the common ion ${\text{OH}}^{-}$, and the ICE table would be modified:

${\text{Ca"("OH")_2(s) rightleftharpoons "Ca"^(2+)(aq) + 2"OH}}^{-} \left(a q\right)$

$\text{I"" "-" "" "" "" "" "0" "" "" "" "" } 0.100$
$\text{C"" "-" "" "" "" "+s" "" "" "" } + 2 s$
$\text{E"" "-" "" "" "" "" "s" "" "" "" } 0.100 + 2 s$

The mass action expression in $\text{0.100 M NaOH} \left(a q\right)$ is thus:

${K}_{s p} = {\left[{\text{Ca"^(2+)]["OH}}^{-}\right]}^{2}$

$= s {\left(0.100 + 2 s\right)}^{2} \approx {0.100}^{2} s$

where we have used the small $s$ approximation, i.e. that $s$ $\text{<<}$ $0.100$ because ${K}_{s p}$ ~ ${10}^{- 5}$ or less.

The new solubility of "Ca"("OH")_2(aq) in $\text{0.100 M NaOH}$ is:

$\textcolor{b l u e}{s} \equiv \left[{\text{Ca"("OH")_2(aq)] = ["Ca}}^{2 +}\right] = {K}_{s p} / {0.100}^{2}$

$=$ $\textcolor{b l u e}{\text{0.00055 M}}$

which is only 5% of what it was in pure water. Hence, the solubility of ${\text{Ca}}^{2 +}$, the ion NOT in common, was suppressed by the ion that IS in common, ${\text{OH}}^{-}$.