Question

The conductivity of saturated solution of $BaS{O}_4$ is $3.06\times {10}^{-6}mhoc{m}^{-1}$ and its equivalent conductance is $1.53mhoc{m}^{2}e{q}^{-1}$. 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. $4\times {10}^6{M}^2$

Answer 1

Answer: (D)

$4\times {10}^6{M}^2$

Equivalent conductance, ${\lambda }_{eq}=K\frac{1000}{C}$

$\therefore 1.530mho$ $c{m}^{2}e{q}^{-1}=\frac{3.06\times {10}^{-6}mhoc{m}^{-1}\times 1000}{C}$ where, $R$= conductivity, $C$= concentration in normality.

$\Rightarrow C=2\times {10}^{-3}M$

$\therefore \left[B{a}^{2+}\right]=\left[S{O}_4^{2-}\right]=2\times {10}^{-3}M$

$\therefore K_{sp}=\left[B{a}^{2+}\right]\left[S{O}_4^{2-}\right]=(2\times {10}^{-3}M{)}^2=4\times {10}^{-6}{M}^2$ .