Question

Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed. Will it lead to the precipitation of copper iodate? (For cupric iodate Ksp = 7.4 × 10–8).

Answer 1

When equal volumes of sodium iodate and cupric chlorate solutions are mixed, then the molar concentrations of both solutions are reduced to half i.e., 0.001 M.
Then, NaI03 $\rightarrow$ Na+ + 10-3
0.001M 0.001M
Cu(CIO3)2$\rightarrow$ Cu2+ + 2CIO3-
0.001M 0.001M
Now, the solubility equilibrium for copper iodate can be written as: Cu(I03)2$\rightarrow$Cu2+(aq) + 2I03-(aq)
Ionic product of copper iodate: = [Cu2+] [I03-]2 = (0.001)(0.001)2 = 1 × 10-9
Since the ionic product (1 × 10-9) is less than Ksp (7.4 × 10-8), precipitation will not occur.