7.69 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 ).

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,

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

Ionic product of copper iodate:

$=\left[{\mathrm{Cu}}^{2+}\right]{\left[{{10}_{3}}^{-}\right]}^{2}\phantom{\rule{0ex}{0ex}}=\left(0.001\right){\left(0.001\right)}^{2}\phantom{\rule{0ex}{0ex}}=1×{10}^{-9}$

Since the ionic product $\left(1×{10}^{-9}\right)$ is less than ${K}_{sp}{\left(7.4×10\right)}^{-8}$, precipitation will not occur.