Question

The conductivity of a saturated solution of $A{g}_{3}P{O}_4$ is $9\times {10}^{-6}S{m}^{-1}$ and its equivalent conductivity is $1.50\times {10}^{-4}S{m}^2$ equivalent${}^{-1}$. The $K_{sp}$ of $A{g}_{3}P{O}_4$ is:

• A. $4.32\times {10}^{-18}$
• B. $1.8\times {10}^{-9}$
• C. $8.64\times {10}^{-13}$
• D. None of these

Answer 1

The correct option is A $4.32\times {10}^{-18}$

Given,

$k=9\times {10}^{-6}s/m$

${\lambda }_{eq}=1.5\times {10}^{-4}S{m}^2$

equivalent conductivity,

${\lambda }_{eq}=\frac{k\times 1000}{N}$

Normality, $N=\frac{k\times 1000}{{\lambda }_{eq}}=\frac{9\times {10}^{-6}\times {10}^{-2}\times 1000}{1.4\times {10}^{-4}\times {10}^4}=6\times {10}^{-5}$

$A{g}_{3}P{O}_4\to 3A{g}^{+}+P{O}_4^{-}$

so, 3 electron transfer, $n=3$

solubility, $s=\frac{N}{n}=\frac{6\times {10}^{-6}}{3}=2\times {10}^{-5}mol/L$

$K_{sp}$ of $A{g}_{3}P{O}_4$ is

$K_{sp}=(3s{)}^3.s=27s^4$

$K_{sp}=27\times (2\times {10}^{-5}{)}^4=4.32\times {10}^{-18}$

Hence, the correct option is $\text{A}$