Question

Molar conductances of $Ba{Cl}_2,H_2{SO}_4$ and $HCl$ are $x_1,x_2$ and $x_3$ $S{cm}^2$ ${mol}^{-1}$ at infinite dilution. If specific conductance of saturated $Ba{ZO}_4$ solution is of $y$ $S{cm}^{-1}$, then $k_{sp}$ of $Ba{SO}_4$ is:

• A. $\frac{{10}^{3}y}{2(x_1+x_2-2x_3)}$
• B. $\frac{{10}^6{y}^2}{{(x_1+x_2-2x_3)}^2}$
• C. $\frac{{10}^6{y}^2}{{4(x_1+x_2-2x_3)}^2}$
• D. $\frac{x_1+x_2-2x_3}{{10}^6{y}^2}$

Answer 1

The correct option is C $\frac{{10}^6{y}^2}{{4(x_1+x_2-2x_3)}^2}$

$\lambda =$ equivalence conductance

$\lambda BaC{l}_2=x_1$

$\lambda H_{2}S{o}_4=x_2$

$\lambda HCl=x_3$

Specific conductance (k) = y

of $BaS{o}_4$

So $\lambda BaS{o}_4=\lambda BaC{l}_2+\lambda H_{2}S{o}_4-2\left(\lambda HCl\right)$

$=x_1+x_2-2x_3$

$\lambda BaS{o}_4=\frac{{10}^3\times y}{solubility}$

$x_1+x_2-2x_3=\frac{1000y}{Solubility}$

$\because$ solubility of $BaS{o}_4=1000yN$ $\{N=M/2\}$

So solubility $=\frac{1000yM}{2(x_1+x_2-2x_3)}$ $(N=2)$

$BaS{o}_4\to B{a}^{2+}+S{o}_4^{2-}$

$Ksp\left(BaS{o}_4\right)=\left(\right[B{a}^{2+}\left]\right[S{o}_4^{2-}]M{)}^2$

$=\frac{{10}^6{y}^2}{4(x_1+x_2-2x_3{)}^2}$