Question

The conductivity of a saturated solution of $Ba{SO}_4$ is $306\times {10}^{-6}{ohm}^{-1}{cm}^{-1}$ and its equivalent conductance is $1.53{ohm}^{-1}{cm}^2{equiv}^{-1}$.

The $K_{sp}$ for ${BaSO}_4$ will be :

• A. $4\times {10}^{-12}$
• B. $2.5\times {10}^{-9}$
• C. $2.5\times {10}^{-13}$
• D. $4\times {10}^{-6}$

Answer 1

Answer: (D)

$4\times {10}^{-6}$

Given : $\wedge eq=1.53{\Omega }^{-1}e{q}^{-1}$

$\wedge =3.06\times {10}^{-6}{\Omega }^{-1}$

Solubilicty, $s=\frac{\wedge \times 1000}{\wedge eq}=\frac{3.06\times {10}^{-6}\times {10}^3}{1.53}=2\times {10}^{-3}M$

$K_{\wedge p}=\left[B{a}^{2+}\right]\left[S{O}_4^{2-}\right]=S^2$

$K_{\wedge p}=(2\times {10}^{-3}{)}^2=4\times {10}^{-6}{M}^2$

Therefore, the correct option is D.