Question

The conductivity of saturated solution of $BaS{O}_4$ is $3.06\times {10}^{-6}$ mho $c{m}^{-1}$ and its equivalent conductance is $1.53$ mho $c{m}^2$ $e{q}^{-1}$. The $K_{sp}$ for $BaS{O}_4$ will be:

• A. $4\times {10}^{-12}$M
• B. $4\times {10}^{-6}$M
• C. $4\times {10}^{-12}{M}^2$
• D. $1\times {10}^{-6}{M}^2$

Answer 1

The correct option is B $4\times {10}^{-6}$M

Let us consider the problem:

$BaS{O}_4\rightleftharpoons B{a}^{2+}+S{O}_4^{2-}$

$\Lambda forN{a}_{2}S{O}_4=3.06\times {10}^{-6}{\Omega }^{-1}$

${\Lambda }_{eq}forN{a}_{2}S{O}_4=1.53\times {\Omega }^{-1}e{q}^{-1}$

Solubility calculated -

$s=\frac{\Lambda \times 1000}{{\Lambda }_{eq}}=\frac{3.06\times {10}^{-6}\times 100}{1.53}=2\times {10}^{-3}mol{L}^{-1}$

$K_{SP}={\left[N{a}^{+}\right]}^2\left[S{O}_4^{2-}\right]=s^2$

$K_{SP}={\left(2\times {10}^{-3}mol{L}^{-1}\right)}^2=4\times {10}^{-6}mo{l}^3{L}^{-3}$