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. that#s# #"<<"# #0.100# because#K_(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