Question

The specific conductance and the equivalent conductance of a saturated solution of $BaS{O}_4$ is $8\times {10}^{-5}oh{m}^{-1}c{m}^{-1}$ and $8000oh{m}^{-1}c{m}^{2}mo{l}^{-1}$ respectively. The $K_{sp}$ of $BaS{O}_4$ is:

• A. $4\times {10}^{-8}{M}^2$
• B. $1\times {10}^{-8}{M}^2$
• C. $2.5\times {10}^{-11}{M}^2$
• D. $1\times {10}^{-4}{M}^2$

Answer 1

Answer: (C)

$2.5\times {10}^{-11}{M}^2$

The molar conductance is ${\Lambda }_m=2\times {\Lambda }_{eq}=2\times 8000=16000{ohm}^{-1}{cm}^2{mol}^{-1}.$
The relationship between molar conductance and specific conductance is ${\Lambda }_m=\frac{1000\times \kappa }{C}.$

Substitute values in the above expression, we get:
$16000{ohm}^{-1}{cm}^2{mol}^{-1}=\frac{1000\times 8\times {10}^{-5}{ohm}^{-1}{cm}^{-1}}{C}$
$⟹C=5\times {10}^{-6}M.$

Hence, the solubility product is $K_{sp}={(5\times {10}^{-6}M)}^2=2.5\times {10}^{-11}{M}^2.$

Hence, option C is correct.