Question

At $298 \,K$, the conductivity of a saturated solution of $AgCl$ in water is $2.6 \times 10^{-6} \,ohm ^{-1} \,cm ^{-1}$. Given $\lambda_{m}^{\infty}\left( Ag ^{+}\right)=63 \,ohm ^{-1} \,cm ^{2} \,mol ^{-1}$ and $\lambda_{m}^{\infty}\left( Cl ^{-}\right)=67 \,ohm ^{-1}\, cm ^{2} \,mol ^{-1}$ Therefore solubility product of $AgCl$ is

• A. $2 \times 10^{-5}$
• B. $4 \times 10^{-10}$
• C. $4 \times 10^{-16}$
• D. $2 \times 10^{-8}$

Answer 1

Answer: (B)

$s=\frac{10^{3} \times \kappa}{\lambda_{m}^{\infty}( AgCl )}=\frac{10^{3} \times 2.6 \times 10^{-6}}{(63+67)}$
$=2 \times 10^{-5}( M )$
$\therefore K_{ sp }=\left(2 \times 10^{-5}\right)^{2}=4 \times 10^{-10}$