can you please explain to me why a common ion lowers the solubility of an ionic compound
Le Chatelier's Principle at work.
Suppose we prepare a saturated solution of CuS. The equilibrium is
CuS(s) <==> Cu^2+(aq) + S^2-(aq)
It will have a Ksp = (Cu^2+)(S^2-)
Now if we add EITHER Cu^2+ (say from CuCl2 which is soluble) OR S^2- (say from Na2S which is soluble)(and both are common ions), what happens to the equilibrium? Adding Cu^2+ or S^2- increases those ions, the reaction shifts to the left and that means some of the ions ppt to form solid. That makes it less soluble. That is the whole explanation but if you want to calculate it you can show it this way.
............CuS(s) ==> Cu^2+ + S^2-
I............solid......0.......0
C............solid......x.......x
E............solid......x.......x
So at equilibrium what is the solubility of CuS.
Ksp = about 10^-40 or so = x^2 so
solubility = about 10^20 M.
Now see what happens if we add 0.1M Na2S. That makes the S^2- now 0.1M from the Na2S and x from the CuS already there but the amount from CuS at about 10^-20 is so small we can neglect it (that is 0.1 + 10^-20 = 0.1) so the Ksp now looks like this.
Ksp= (Cu^2+)(S^2-)
10^-40 = (x)(0.1)
x = 10^-40/10^-1 = 10^-39 so the (Cu^2+) = 10^-39 which means the solubility of CuS is 10^-39 M and that is much much less than the initial solubility of 10^-20 M. Le Chatelier's Principle says it all but you can calculate it to show that is so also.