What is common ion effect?
1 Answer
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) 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
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
"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
K_(sp) = ["Ca"^(2+)]["OH"^(-)]^2
= s(0.100 + 2s)^2 ~~ 0.100^2s where we have used the small
s approximation, i.e. thats "<<" 0.100 becauseK_(sp) ~10^(-5) or less.
The new solubility of
color(blue)(s) -= ["Ca"("OH")_2(aq)] = ["Ca"^(2+)] = K_(sp)/0.100^2
= color(blue)("0.00055 M")
which is only