Question

The specific conductivity of a saturated solution of $AgCl$ is $3.40\times 10^{-6}$ ohm ${}^{-1}$ $c{m}^{-1}$ at $25^{\circ }C.$ If ${\lambda }_{Ag+}=62.3oh{m}^{-1}c{m}^2$ $mo{l}^{-1}\&{\lambda }_{cl}=67.7oh{m}^{-1}c{m}^{2}mo{l}^{-1}$, the solubility of $AgCl$ at $25^{\circ }C$ is :

• A. $2.6\times 10^{-5}M$
• B. $4.5\times 10^{-3}M$
• C. $3.6\times 10^{-5}M$
• D. $3.6\times 10^{-3}M$

Answer 1

Answer: (A)

${\lambda }_{A{g}^{+}}=62.3Sc{m}^{2}mo{l}^{-1}$,
${\lambda }_{c{l}^{-}}=67.7Sc{m}^{2}mol{e}^{-1}$
$K_{\text{Agcl }}=3.4\times 10^{-6}Sc{m}^{-1}$
${\wedge }_{AgCl}^{\infty }=(62.3+67.5)$
$=\frac{1000\times 3.4\times 10^{-6}}{S}$
$S=\frac{3.4\times 10^{-3}}{(62.3+67.5)}$
$=2.6\times 10^{-5}M$