Question

The molar solubility of silver sulphate is $1.5\times {10}^{-2}$ mol $L^{-1}$. The solubility product of the salt will be:

• A. $2.25\times {10}^{-4}$
• B. $1.35\times {10}^{-5}$
• C. $1.7\times {10}^{-6}$
• D. $3.0\times {10}^{-3}$

Answer 1

Answer: (B)

$1.35\times {10}^{-5}$

Solution:- (B) $1.35\times {10}^{-5}$

$\underset{\left(\text{Silver sulphate}\right)}{{Ag}_{2}S{O}_4}\rightleftharpoons \underset{2s}{2{Ag}^{+}}+\underset{s}{{S{O}_4}^{-2}}$

For the above reaction-

$K_{sp}={\left(2s\right)}^2\times s$

$\Rightarrow K_{sp}=4s^3$

$\Rightarrow K_{sp}=4\times {(1.5\times {10}^{-2})}^3(\because s=1.5\times {10}^{-2}mol/L)$

$\Rightarrow K_{sp}=4\times 3.375\times {10}^{-6}=1.35\times {10}^{-5}$

Hence the solubility product of the salt is $1.35\times {10}^{-5}$.