Question

What is common ion effect?

Answer 1

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).

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Consider the dissociation of `"Ca"("OH")_2(s)` in water at `25^@ "C"`, with solubility product constant `K_(sp) = 5.5 xx 10^(-6)`.

`"Ca"("OH")_2(s) rightleftharpoons "Ca"^(2+)(aq) + 2"OH"^(-)(aq)`

`"I"" "-" "" "" "" "" "0" "" "" "" "" "0`
`"C"" "-" "" "" "" "+s" "" "" "" "+2s`
`"E"" "-" "" "" "" "" "s" "" "" "" "" "2s`

The mass action expression in pure water is thus:

`K_(sp) = ["Ca"^(2+)]["OH"^(-)]^2`

`= s(2s)^2 = 4s^3`

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

`color(blue)(s) -= ["Ca"("OH")_2(aq)] = ["Ca"^(2+)] = (K_(sp)/4)^(1//3)`

`=` `color(blue)("0.011 M")`

Now suppose the solution already contained `"0.100 M"` `"NaOH"`. Then we would have the common ion `"OH"^(-)`, and the ICE table would be modified:

`"Ca"("OH")_2(s) rightleftharpoons "Ca"^(2+)(aq) + 2"OH"^(-)(aq)`

`"I"" "-" "" "" "" "" "0" "" "" "" "" "0.100`
`"C"" "-" "" "" "" "+s" "" "" "" "+2s`
`"E"" "-" "" "" "" "" "s" "" "" "" "0.100+2s`

The mass action expression in `"0.100 M NaOH"(aq)` is thus:

`K_(sp) = ["Ca"^(2+)]["OH"^(-)]^2`

`= s(0.100 + 2s)^2 ~~ 0.100^2s`

where we have used the small `s` approximation, i.e. that `s` `"<<"` `0.100` because `K_(sp)` ~ `10^(-5)` or less.

The new solubility of `"Ca"("OH")_2(aq)` in `"0.100 M NaOH"` is:

`color(blue)(s) -= ["Ca"("OH")_2(aq)] = ["Ca"^(2+)] = K_(sp)/0.100^2`

`=` `color(blue)("0.00055 M")`

which is only `5%` of what it was in pure water. Hence, the solubility of `"Ca"^(2+)`, the ion NOT in common, was suppressed by the ion that IS in common, `"OH"^(-)`.