Question

Equivalent conductivity of $BaC{l}_2,$ $H_{2}S{O}_4$ and $HCl$ are $x_1,x_2,and{x}_{3}Sc{m}^{2}e{q}^{-1}$
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. $\frac{{10}^6{x}^2}{{(x_1+x_2-2x_3)}^3}$
• C. $\frac{2.5\times {10}^5{x}^2}{{(x_1+x_2-2x_3)}^2}$
• D. $\frac{0.25x^2}{{(x_1+x_2-x_3)}^2}$

Answer 1

The correct option is C

$\frac{2.5\times {10}^5{x}^2}{{(x_1+x_2-2x_3)}^2}$

Kohlrausch law of independent migration of ions:

${\Lambda }_m^0={\lambda }_{m^{+}}^0+{\lambda }_{m^{-}}^0$

${\lambda }_{m^{+}}^0\to \text{Limiting molar conductivity of cation}$
${\lambda }_{m^{-}}^0\to \text{Limiting molar conductivity of anion}$

${\Lambda }_{eq}^0={\lambda }_{e{q}^{+}}^0+{\lambda }_{e{q}^{-}}^0$

${\lambda }_{e{q}^{+}}^0\to \text{Limiting equivalent conductivity of cation}$
${\lambda }_{e{q}^{-}}^0\to \text{Limiting equivalent conductivity of anion}$

${\Lambda }_{eq}^0=\frac{{\lambda }_{m^{+}}^0}{\text{Charge on the cation}}+\frac{{\lambda }_{m^{-}}^0}{\text{Charge on the anion}}$

For $BaS{O}_4$

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

${\Lambda }_{eq}^0=\frac{{\lambda }_m^0\left(B{a}^{2+}\right)}{2}+\frac{{\lambda }_m^0\left(S{O}_4^{2-}\right)}{2}$

${\Lambda }_{eq}^0=\frac{{\lambda }_m^0\left(B{a}^{2+}\right)+{\lambda }_m^0\left(S{O}_4^{2-}\right)}{2}$

${\Lambda }_m^0\left(BaS{O}_4\right)={\lambda }_m^0\left(B{a}^{2+}\right)+{\lambda }_m^0\left(S{O}_4^{2-}\right)$

${\Lambda }_{eq}^0=\frac{{\Lambda }_m^0}{2}$

${\Lambda }_m^0\left(BaS{O}_4\right)=2{\Lambda }_{eq}^0\left(BaS{O}_4\right)$

${\Lambda }_{eq}^0\left(BaS{O}_4\right)={\Lambda }_{eq}^0\left(BaC{l}_2\right)+{\Lambda }_{eq}^0\left(H_{2}S{O}_4\right)-2{\Lambda }_{eq}^0\left(HCl\right)$

${\Lambda }_{eq}^0\left(BaS{O}_4\right)=x_1+x_2-2x_3$

${\Lambda }_m^0=2(x_1+x_2-2x_3)$

For sparingly soluble salt
${\Lambda }_m^0=\frac{\kappa }{S}\times 1000$

$S=\frac{x}{2(x_1+x_2-2x_3)}\times 1000$

$\Rightarrow \frac{500x}{(x_1+x_2-2x_3)}$

$BaS{O}_4\rightleftharpoons B{a}^{2+}+S{O}_4^{2-}$
$K_{sp}=S\times S$
$K{}_{sp}=S^2\Rightarrow \frac{2.5\times {10}^5{x}^2}{{(x_1+x_2-2x_3)}^2}$