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 Ksp = $7.4\times {10}^{-8}$).

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,
$\begin{array}{ccccc}NaI{O}_3& \to & N{a}^{+}& +& I{O}_3^{-}\\ 0.001M& & & & 0.001M\\ Cu(Cl{O}_3{)}_2& \to & C{u}^{2+}& +& 2Cl{O}_3^{-}\\ 0.001M& & & & 0.001M\end{array}$
Now, the solubility equilibrium for copper iodate can be written as:
$Cu({10}_3{)}_2\to C{u}_{\left(aq\right)}^{2-}+{210}_{3\left(aq\right)}^{-}$
Ionic product of copper iodate:
$=\left[C{u}^{2+}\right][{10}_3^{-}{]}^2=(0.001\left)\right(0.001{)}^2=1\times {10}^{-9}$
Since the ionic product $(1\times {10}^{-9})$ is less than $K_{sp}(7.4\times {10}^{-8})$, precipitation will not occur.