Question

The conductivity of a saturated solution of $BaS{O}_4$ is $3.06\times {10}^{-6}ohm{}^{-1}cm{}^{-1}$ and its molar conductance is $1.53ohm{}^{-1}c{m}^{2}mol{}^{-1}$. What will be the the $K_{sp}\text{of}BaS{O}_4$ ?

• 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

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

$\text{Conductivity,}\kappa =3.06\times {10}^{-6}oh{m}^{-1}c{m}^{-1}$

$BaS{O}_4$ is a sparingle soluble salt and it will have very less solubility.
Since, a saturated solution of a sparingly soluble salt is a very dilute solution.

${\Lambda }_m={\Lambda }_m^0$

${\Lambda }_m=\kappa \times \frac{1000\left(c{m}^3{L}^{-1}\right)}{C}$
where, $C$ is concentration in $mol{L}^{-1}$
$\kappa$ is in $oh{m}^{-1}c{m}^{-1}$
$C=\kappa \times \frac{1000}{{\Lambda }_m}$
$C=\frac{3.06\times {10}^{-6}oh{m}^{-1}c{m}^{-1}\times 1000c{m}^3{L}^{-1}}{1.53oh{m}^{-1}c{m}^{2}mo{l}^{-1}}$

$C=2\times {10}^{-3}mol{L}^{-1}$

$\text{Solubility,}S=2\times {10}^{-3}mol{L}^{-1}$
$BaS{O}_4\rightleftharpoons B{a}^{2+}+S{O}_4^{2-}$
$K_{sp}=\left[B{a}^{2+}\right]\left[S{O}_4^{2-}\right]$
$K_{sp}=S\times S=S^2$
$K_{sp}=4\times {10}^{-6}mo{l}^2{L}^{-2}$