Question

At 25°C, 400 mL of water can dissolve 0.00896 g of lead iodate $\mathbf{Pb}{\left({\mathbf{IO}}_3\right)}_2$. Calculate ${\mathbf{K}}_{\mathrm{sp}}$for lead iodate.

Short answer

The value of ${\mathrm{K}}_{\mathrm{sp}}=6.50\times {10}^{-14}$

Step-by-step solution

Solubility

The solubility of a solution can be defined as the maximum amount of solute that can be dissolved in 100g of solvent.

For some solids, the solubility increases with the increase in temperature, and for some solids, the solubility decreases with the increase in temperature.

Solubility of lead iodate

Given 0.00896 g of lead acetate dissolves in 400 mL of water.

The molar mass of lead iodate is 557.0 g/mol.

$\begin{array}{c}\mathrm{Solubility}\left(\mathrm{s}\right)=\frac{0.00896\mathrm{g}}{557.0\mathrm{g}/\mathrm{mol}}\times \frac{1}{0.400\mathrm{L}}\\ =4.02\times {10}^{-5}\mathrm{mol}/\mathrm{L}\end{array}$

Calculation of solubility product

The solubility of lead iodate is as shown below:

$\begin{array}{l}\mathrm{Pb}{\left({\mathrm{IO}}_3\right)}_2\left(\mathrm{s}\right)\rightleftharpoons {\mathrm{Pb}}^{2+}\left(\mathrm{aq}\right)+2{{\mathrm{IO}}_3}^{-}\left(\mathrm{aq}\right)\\ \mathrm{s}2\mathrm{s}\\ {\mathrm{K}}_{\mathrm{sp}}=\mathrm{s}{\left(2\mathrm{s}\right)}^2\\ =4{\mathrm{s}}^3\\ =4{(4.02\times {10}^{-5}\mathrm{mol}/\mathrm{L})}^3\\ =6.50\times {10}^{-14}\end{array}$