Question

In the determination of the solubility for copper (II) iodate, what common ion of the same concentration (for example, 0.010 M Cu2+ or 0.010 M IO3-) would have a greater effect on reducing the solubility of Cu(IO3)2? Why?

Answer 1

Adding IO₃^- as the common ion would result in a greater solubility decrease of the sparingly soluble salt of copper, Cu(IO₃)₂ than on adding Cu²⁺of same concentration as the dissolution of copper(II) iodate produces IO₃^- twice as much as that of Cu²⁺ ion.  A simple calculation of solubility would show the same.

ICE table for addition of 0.010 M Cu²⁺.     ICE table for addition of 0.010 M IO₃^-.

I: x x 2x I: x x 2x

C: +0 +0.010 +0 C: +0 +0 +0.010

E: x x + 0.010 2x E: x x 2x+0.010

K_sp = [Cu²⁺][ IO₃^-]² = 6.94* 10^-8

Now, for the initial condition when there is no added common ion, the solubility is x.

Now, if we add 0.010 M IO₃^- ion as common ion,

Since the solubility of a sparingly soluble salt, x is very little,i.e. very much less than 0.010 we can use the approximation

2x+0.010 = 0.010 M

Therefore,

Therefore solubility in presence of IO₃^- as common ion =

Again, if we add 0.010 M Cu²⁺ as common ion,

Again, the solubility of a sparingly soluble salt, x is very little,i.e. very much less than 0.010 we can use the approximation

x+0.010 = 0.010 M

Now,

Therefore solubility in presence of Cu²⁺as common ion =

As seen above, the solubility of copper(II) iodate is much less when iodate ion is added compared to that when copper(II) ion is added.