Question

The $K_{sp}$ for AgCl is $2.8\times {10}^{-10}$ at a given temperature. The solubility of AgCl in 0.01 molar HCl solution at this temperature will be :

• A. $2.8\times {10}^{-12}mol{L}^{-1}$
• B. $2.8\times {10}^{-8}mol{L}^{-1}$
• C. $5.6\times {10}^{-8}mol{L}^{-1}$
• D. $2.8\times {10}^{-4}mol{L}^{-1}$

Answer 1

The correct option is B $2.8\times {10}^{-8}mol{L}^{-1}$

The chloride ion concentration in 0.01M HCl will be 0.01 M.
The chloride ion concentration due to dissociation of AgCl is neglected due to very low value of solubility product of AgCl.
The expression for the solubility product is as shown below.
$K_{sp}=\left[A{g}^{+}\right]\left[C{l}^{-}\right]$
Substitute values in the above expression.
$2.8\times {10}^{-10}=\left[A{g}^{+}\right]\times 0.01$
Hence, $\left[A{g}^{+}\right]=\frac{2.8\times {10}^{-10}}{0.01}=2.8\times {10}^{-8}$mol/L.