Question

At 100°C, water dissolves $\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{-2}$g of AgCl per liter. Compute the Ksp of AgCl at this temperature.

Short answer

The value of ${\mathrm{K}}_{\mathrm{sp}}=1.56\times {10}^{-8}$

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.

The solubility is measured in $\mathbf{mol}/\mathbf{L}$.

Solubility of AgCl

Given $1.8\times {10}^{-2}$g of silver chloride dissolves in 1.00 L of water.

The molar mass of AgCl is 143.32 g/mol.

$\begin{array}{c}\mathrm{Solubility}\left(\mathrm{s}\right)=\frac{1.8\times {10}^{-2}\mathrm{g}}{143.32\mathrm{g}/\mathrm{mol}}\times \frac{1}{1.00\mathrm{L}}\\ =1.25\times {10}^{-4}\mathrm{mol}/\mathrm{L}\end{array}$

Calculation of solubility product

The solubility of lead iodate is as shown below:

$\begin{array}{l}\mathrm{AgCl}\left(\mathrm{s}\right)\rightleftharpoons {\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)+{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)\\ \mathrm{s}\mathrm{s}\\ {\mathrm{K}}_{\mathrm{sp}}=\mathrm{s}\times \mathrm{s}\\ ={\mathrm{s}}^2\\ ={(1.25\times {10}^{-4})}^2\\ =1.56\times {10}^{-8}\end{array}$