Question

At $298 \,K$, the conductivity of a saturated solution of $AgCl$ in water is $2.6 \times 10^{-6} \,Scm ^{-1}$. Its solubility product at $298 \, K$ is:
[Given: $\lambda^{\infty}\left( Ag ^{+}\right)=63.0 \,S \, cm ^{2} \,mol ^{-1}\,\,\, \lambda^{\infty}\left( Cl ^{-}\right)=67.0 \,S \, cm ^{2} \, mol ^{-1}$]

• A. $2.0 \times 10^{-5} \, M^2 $
• B. $4.0 \times 10^{-10} \, M^2 $
• C. $4.0 \times 10^{-16} \, M^2 $
• D. $2 \times 10^{-8} \, M^2 $

Answer 1

Answer: (B)

$\lambda_{ AgC 1}^{\circ}=\lambda_{ Ag ^{+}}^{\circ}+\lambda_{ C 1^{-}}^{\circ}$
Solubility, $S =\frac{1000 k }{\lambda_{ AgCl }^{\circ}}$
$=\frac{1000 \times 2.6 \times 10^{-6}}{\lambda_{ Ag ^{+}}^{\circ}+\lambda_{ Cl ^{-}}^{\circ}}$
$=\frac{2.6 \times 10^{-3}}{63+67}$
$=2 \times 10^{-5} \,mol \,L ^{-1}$
$K _{ sp }= S ^{2}$
$=\left(2 \times 10^{-5}\right)^{2} \,M ^{2}$
$=4.0 \times 10^{-10}\, M ^{2}$