Question

At $298K$, the conductivity of a saturated solution of $AgCl$ in water is $2.6\times {10}^{-6}S{cm}^{-1}$. Its solubility product at $298K$ is:

[Given: ${\lambda }^{\infty }\left({Ag}^{+}\right)=63.0S{cm}^2{mol}^{-1}$
${\lambda }^{\infty }\left({Cl}^{-}\right)=67.0S{cm}^2{mol}^{-1}$]

• 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

The correct option is B $4.0\times {10}^{-10}{M}^2$

${\lambda }_{AgCl}^o={\lambda }_{{Ag}^{+}}^o+{\lambda }_{{Cl}^{-}}^o$

Solubility, $S=\frac{1000k}{{\lambda }_{AgCl}^o}$

$=\frac{1000\times 2.6\times {10}^{-6}}{{\lambda }_{{Ag}^{+}}^o+{\lambda }_{{Cl}^{-}}^o}$

$=\frac{2.6\times {10}^{-3}}{63+67}$

$=2\times {10}^{-5}mol{L}^{-1}$

$K_{sp}=S^2$

$=(2\times {10}^{-5}{)}^2{M}^2$

$=4.0\times {10}^{-10}{M}^2$

Hence, option B is correct.