Question

Equivalent conductivity of $BaC{l}_2,H_{2}S{O}_4$ and $HCl$ are $x_1,x_2$ and $x_{3}Sc{m}^{-1}e{q}^{-1}$ at infinite dilution. If conductivity of saturated $BaS{O}_4$ solution is $xSc{m}^{-1}$, then $K_{sp}$ of $BaS{O}_4$ is:

• A. $\frac{500x}{(x_1+x_2-2x_3)}$
• B. $\frac{{10}^6{x}^2}{(x_1+x_2-2x_3{)}^2}$
• C. $\frac{2.5\times {10}^5{x}^2}{(x_1+x_2-x_3{)}^2}$
• D. $\frac{0.25x^2}{(x_1+x_2-x_3{)}^2}$

Answer 1

The correct option is C $\frac{2.5\times {10}^5{x}^2}{(x_1+x_2-x_3{)}^2}$

${\Lambda }_m^{\infty }\left(BaS{O}_4\right)=2{\Lambda }_{eq.}^{\infty }\left(BaS{O}_4\right)$
${\Lambda }_m^{\infty }\left(BaS{O}_4\right)={\Lambda }_{eq.}^{\infty }\left(B{a}^{2+}\right)+{\Lambda }_{eq}^{\infty }\left(S{O}_4^{2-}\right)$
$={\Lambda }_{eq.}^{\infty }\left(BaC{l}_2\right)+{\Lambda }_{eq.}^{\infty }\left(H_{2}S{O}_4\right)-{\Lambda }_{eq.}^{\infty }\left(HCl\right)$
${\Lambda }_{eq.}^{\infty }\left(BaS{O}_4\right)=x_1+x_2-x_3$
${\Lambda }_{eq.}^{\infty }=2(x_1+x_2-x_3)$
for sparingly soluble salt
${\Lambda }_m^{\infty }=\frac{\kappa }{M}\times 1000$
or $M=\frac{x}{2(x_1+x_2-x_3)}\times 1000$
$\Rightarrow \frac{500x}{(x_1+x_2-x_3)}$
$K_{sp}=M^2$ $\Rightarrow \frac{2.5\times {10}^5{x}^2}{(x_1+x_2-x_3{)}^2}$ [Option C]