Question

The conductivity of a saturated solution of $BaSO_{4}$ is $3.06 \times 10^{-6}\, ohm^{-1}\, cm^{-1}$ and its equivalent conductance is $1.53\, ohm^{-1}\,cm^{-1}$ ${\text{equivalent}}^{-1}$ The $K_{sp}$ of the $BaSO_{4}$ 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)

$\lambda_{m}=\frac{1000\,K}{S}$
$=\frac{1000\times3.06\times10^{-6}}{S}=1.53$
$S=2\times10^{-3} \frac{\text{mol}}{\text{litre}}$
$K_{sp\left(BaSO_4\right)}=S_{2}=\left(2\times10^{-3}\right)^{2}=4\times10^{-6}$