Question

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

Answer 1

When equal volumes of sodium iodate and cupric chlorate solutions are mixed together, then the molar concentrations of both solutions are reduced to half i.e., 0.001 M.

Then,

\[\ce{\underset{0.001 M}{{Na10}_3} -> Na+ + \underset{0.001 M}{10^-_3}}\]

\[\ce{\underset{0.001 M}{Cu(ClO3)2} -> Cu^{2+} + \underset{0.001 M}{2ClO^-_3}}\]

Now, the solubility equilibrium for copper iodate can be written as:

\[\ce{Cu(10_3)_2 -> Cu^{2+}_{(aq)} + 210^-_{3(aq)}}\]

Ionic product of copper iodate:

`= ["Cu"^(2+)][10_3^-]^2`

`= (0.001)(0.001)^2`

`= 1xx 10^(-9)`

Since the ionic product (1 × 10^–9) is less than K_sp (7.4 × 10^–8), precipitation will not occur.