Question

The conductivity of a saturated solution of ${BaSO}_4$ is $3.06\times {10}^{-6}{ohm}^{-1}{cm}^{-1}$ and its molar conductance is $1.53{ohm}^{-1}{cm}^2{mol}^{-1}$. The $K_{sp}$ of ${BaSO}_4$ 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

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

$\because$ ${\Lambda }_m$ =$\frac{K\times 1000}{M}$

$\Rightarrow$ $M$ = $\left(\frac{3.06\times {10}^{-6}\times 1000}{1.53}\right)$

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

$\therefore K_{sp}$$\left(BaS{O}_4\right)$ = ${\left(M\right)}^2$

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

Hence, option D is correct.