Question

A silver wire dipped in 0.1 M HCl solution saturated with AgCl develops a potential -0.25V. If $E{{}^0}_{Ag/A{g}^{+}}=-0.799V,K_{SP}$ of AgCl in pure water will be :

• A. $2.95\times {10}^{-11}$
• B. $5.30\times {10}^{-11}$
• C. $3.95\times {10}^{-11}$
• D. $1.95\times {10}^{-11}$

Answer 1

The correct option is B $5.30\times {10}^{-11}$

The expression for the electrode potential is $E_{el}=E_{el}^0-\frac{0.0592}{n}log\left[A{g}^{+}\right]$
Substitute values in the above expression.
$-0.25V=-0.799V-\frac{0.0592}{1}log\left[A{g}^{+}\right]log\left[A{g}^{+}\right]=-9.2736\left[A{g}^{+}\right]=5.3\times {10}^{-10}$
Now calculate the solubility product.
$K_{sp}=\left[A{g}^{+}\right]\left[C{l}^{-}\right]=5.3\times {10}^{-10}\times 0.1=5.3\times {10}^{-11}.$