Question

The solubility of $AgCl$ in water at $298K$ is $1.06\times {10}^{-5}$ mole per litre. Calculate its solubility product at this temperature.

• A. $1.12\times {10}^{10}mo{l}^2{L}^{-2}$
• B. $1.12\times {10}^{-10}mo{l}^2{L}^{-2}$
• C. $1.12\times {10}^{-10}mo{l}^{-2}{L}^2$
• D. $1.12\times {10}^{10}mo{l}^{-2}{L}^2$

Answer 1

The correct option is B $1.12\times {10}^{-10}mo{l}^2{L}^{-2}$

The equilibrium in the saturated solution will be as :
$\begin{array}{ccccc}AgCl\left(s\right)& \rightleftharpoons & A{g}^{+}& +& C{l}^{-}\\ 1& & 0& & 0\\ 1-s& & s& & s\end{array}$
$K_{sp}=\left[A{g}^{+}\right(aq\left)\right]\left[C{l}^{-}\right(aq\left)\right]$
$=s\times s=s^2$

Given: the solubility of $AgCl$ is $1.06\times {10}^{-5}$ mole per litre.
$\Rightarrow s=1.06\times {10}^{-5}mol{L}^{-1}$

putting the values,
$K_{sp}=(1.06\times {10}^{-5}{)}^2$
$=1.12\times {10}^{-10}mo{l}^2{L}^{-2}$