Question

Calculate the mass of AgCl that can dissolve in 100 mL of 0.150 M NaCl solution.

Short answer

The amount of AgCl that dissolves in 100 mL NaCl solution is $1.5\times {10}^{-8}\mathrm{g}.$

Step-by-step solution

Common ion effect

The solubility of a sparingly soluble salt decreases further by adding a common ion to it. This is due to the increase in the concentration of ions on the right side of the equilibrium will allow the equilibrium to shift towards the unionised salt.

Calculation

The solubility product value of AgCl is $1.6\times {10}^{-10}$.

The solubility equation is as shown below:

$\begin{array}{l}\mathrm{AgCl}\left(\mathrm{s}\right)\rightleftharpoons {\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)+{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)\\ -\mathrm{s}\mathrm{s}+0.150\\ {\mathrm{K}}_{\mathrm{sp}}=\mathrm{s}\times (\mathrm{s}+0.150)\\ 1.6\times {10}^{-10}=\mathrm{s}\left(0.150\right)\\ \mathrm{s}=1.07\times {10}^{-9}\end{array}$

That means $1.07\times {10}^{-9}$mol of AgCl dissolves in 100 mL of 0.150 M NaCl solution.

The mass of AgCl is obtained by multiplying with its molar mass.

$\begin{array}{c}{\mathrm{m}}_{\mathrm{AgCl}}=1.07\times {10}^{-9}\mathrm{mol}\times 143.3\mathrm{g}/\mathrm{mol}\\ =1.5\times {10}^{-8}\mathrm{g}\end{array}$