Question

At $298K$, the conductivity of a saturated solution of $AgCl$ in water is $2.6\times 10^{-6}Sc{m}^{-1}$. Its solubility product at $298K$ is:
[Given: ${\lambda }^{\infty }\left(A{g}^{+}\right)=63.0Sc{m}^{2}mo{l}^{-1}{\lambda }^{\infty }\left(C{l}^{-}\right)=67.0Sc{m}^{2}mo{l}^{-1}$]

• 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)

${\lambda }_{AgC1}^{\circ }={\lambda }_{A{g}^{+}}^{\circ }+{\lambda }_{C{1}^{-}}^{\circ }$
Solubility, $S=\frac{1000k}{{\lambda }_{AgCl}^{\circ }}$
$=\frac{1000\times 2.6\times 10^{-6}}{{\lambda }_{A{g}^{+}}^{\circ }+{\lambda }_{C{l}^{-}}^{\circ }}$
$=\frac{2.6\times 10^{-3}}{63+67}$
$=2\times 10^{-5}mol{L}^{-1}$
$K_{sp}=S^2$
$={\left(2\times 10^{-5}\right)}^2{M}^2$
$=4.0\times 10^{-10}{M}^2$