Question

At $298K$, the conductivity of a saturated solution of $AgCl$ in water is $2.6\times 10^{-6}oh{m}^{-1}c{m}^{-1}$. Given ${\lambda }_m^{\infty }\left(A{g}^{+}\right)=63oh{m}^{-1}c{m}^{2}mo{l}^{-1}$ and ${\lambda }_m^{\infty }\left(C{l}^{-}\right)=67oh{m}^{-1}c{m}^{2}mo{l}^{-1}$ 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)

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