Question

The conductivity of a saturated solution of $BaS{O}_4$ is $3.06\ast {10}^{-8}$ $oh{m}^{-1}c{m}^{-1}$ and its equivalent conductance is $1.53$ $oh{m}^{-1}c{m}^{2}e{q}^{-1}$.

The $K_{sp}$ of $BaS{O}_4$ is:

• A. $4\ast {10}^{-12}$
• B. $2.5\ast {10}^{-9}$
• C. $2.5\ast {10}^{-13}$
• D. $1.0\ast {10}^{-10}$

Answer 1

The correct option is D $1.0\ast {10}^{-10}$

The relationship between equivalent conductance and conductivity is ${\Lambda }_{eq}=\frac{1000\kappa }{C}$
Substitute values in above expression.
$1.53{ohm}^{-1}{cm}^2{eq}^{-1}=\frac{1000\times 3.06\times {10}^{-8}{ohm}^{-1}{cm}^{-1}}{C}C=2\times {10}^{-5}eqd{m}^{-3}or1\times {10}^{-5}mold{m}^{-3}$
Hence, $\left[{Ba}^{2+}\right]=\left[{SO}_4^{2-}\right]$

$=1\times {10}^{-5}mold{m}^{-3}$.
The expression for the solubility product is $K_{sp}=\left[{Ba}^{2+}\right]\left[{SO}_4^{2-}\right]$. Substitute values in the above expression
$K_{sp}=\left[{Ba}^{2+}\right]\left[{SO}_4^{2-}\right]$

$=(1\times {10}^{-5})\times (1\times {10}^{-5})$
$=1\times {10}^{-10}$