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"^(-)#.