Question

The specific conductance of a saturated solution of $AgCl$ at ${25}^{\circ }C$ is $3.55\times {10}^{-6}{\Omega }^{-1}c{m}^{-1}$ and that of water used for preparing the solution is $1.60\times {10}^{-6}{\Omega }^{-1}c{m}^{-1}$. What is the solubility product of AgCl?
$\left({\Lambda }^{\infty }\right(AgCl)=150.0{\Omega }^{-1}c{m}^2{\text{eqv}}^{-1})$

• A. $2.37\times {10}^{-5}$
• B. $5.60\times {10}^{-10}$
• C. $1.3\times {10}^{-5}$
• D. $1.69\times {10}^{-10}$

Answer 1

The correct option is D $1.69\times {10}^{-10}$

We know,
$K_{sol}=K_{AgCl}+K_{water}$
$K_{AgCl}=(3.55-1.60)\times {10}^{-6}{\Omega }^{-1}c{m}^{-1}$ = $1.95\times {10}^{-6}{\Omega }^{-1}c{m}^{-1}$

Again, ${\Lambda }^{\infty }\left(AgCl\right)=\frac{K\times 1000}{c}$ $\Rightarrow 150.0=\frac{1.95\times {10}^{-6}\times 1000}{c}$
$\Rightarrow c=1.3\times {10}^{-5}M$

Solubility prodcut of $AgCl$ = $K_{sp}=\left[A{g}^{+}\right]\left[C{l}^{-}\right]$
$K_{sp}=c^2$ $=(1.3\times {10}^{-5}{)}^2=1.69\times {10}^{-10}(molc{m}^{-3}{)}^2$