Question

Specific conductance of a saturated solution of $AgBr$ is $8.486\times {10}^{-7}{ohm}^{-1}$ ${cm}^{-1}$ at ${25}^{o}C$. Specific conductance of pure water at ${25}^{o}C$ is $0.75\times {10}^{-6}{ohm}^{-1}$ ${cm}^{-2}$. ${\Lambda }_m^{\infty }$ for $KBr,Ag{NO}_3$ and $K{NO}_3$ are $137.4,133,131$ ($S$ ${cm}^2$ ${mol}^{-1}$) respectively. Calculate the solubility of $AgBr$ in gm/litre.

Answer 1

$KBr+Ag{NO}_3\to AgBr+K{NO}_3$

$\therefore A_m^{\infty }AgBr=A_m^{\infty }\left(KBr\right)+\left(Ag{NO}_3\right)-{KNO}_3$

$=137.4+133-131$

$A_m^{\infty }=139.4{Scm}^2{mol}^{-1}$

$K$ of $AgBR=$ saturated solution $-$ pure water

$=8.486\times {16}^{-7}-0.75\times {10}^{-6}$

$=9.86\times {10}^{-8}oh{m}^{-1}{cm}^{-1}$

we know, $A_m=\frac{K}{C}\times 1000$

if C is the solubility

$C=\frac{K}{A_m}\times 1000$

$=\frac{9.86\times {10}^{-8}\times 1000}{139.4}$

$C=7.07\times {10}^{-7}mol/L$