What is common ion effect?

1 Answer
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^@ "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"^(-).