Equivalent conductance of ${\mathrm{BaCl}}_2$, ${\mathrm{H}}_2{\mathrm{SO}}_4$ and $\mathrm{HCl}$ are ${\mathrm{x}}_1$, ${\mathrm{x}}_2$  and ${\mathrm{x}}_3$ ${\mathrm{Scm}}^2{\mathrm{equiv}}^{-1}$  at infinite dilution. Specific conductance of saturated ${\mathrm{BaSO}}_4$ solution is of ‘y’ ${\mathrm{Scm}}^{-1}$.Then ${\mathrm{k}}_{\mathrm{sp}}$ of ${\mathrm{BaSO}}_4$ is

${{\mathrm{\Lambda }}^{\circ }}_{{\mathrm{BaSO}}_4}={{\mathrm{\Lambda }}^{\circ }}_{{\mathrm{BaCl}}_2}+{{\mathrm{\Lambda }}^{\circ }}_{{\mathrm{H}}_2{\mathrm{SO}}_4}-2{{\mathrm{\Lambda }}^{\circ }}_{\mathrm{HCl}}$

$=\left({\mathrm{x}}_1\right)+\left({\mathrm{x}}_2\right)-2\left({\mathrm{x}}_3\right)\to \left(1\right)$

${{\mathrm{\mu }}^{\circ }}_{{\mathrm{BaSO}}_4}=2\times {{\mathrm{\Lambda }}^{\circ }}_{{\mathrm{BaSO}}_4}$

$=2\times ({\mathrm{x}}_1+{\mathrm{x}}_2-2{\mathrm{x}}_3)\to \left(2\right)$

We know that ${\mathrm{\mu }}^{\circ }=\frac{\mathrm{K}\times 1000}{\mathrm{S}}$

For ${\mathrm{BaSO}}_4\Rightarrow {\mathrm{K}}_{\mathrm{SP}}={\mathrm{S}}^2$

$\therefore \mathrm{S}=\sqrt{{\mathrm{K}}_{\mathrm{SP}}}$

$\therefore {\mathrm{\mu }}^{\circ }=\frac{\mathrm{K}\times 1000}{\sqrt{{\mathrm{K}}_{\mathrm{SP}}}}$

$\sqrt{{\mathrm{K}}_{\mathrm{SP}}}=\frac{\mathrm{K}\times 1000}{{\mathrm{\mu }}^0}$

${\mathrm{K}}_{\mathrm{SP}}={\left[\frac{\left(\mathrm{y}\right)\left({10}^3\right)}{2({\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3)}\right]}^2$

${\mathrm{K}}_{\mathrm{SP}}=\frac{{10}^6{\mathrm{y}}^2}{4{({\mathrm{x}}_1+{\mathrm{x}}_2-2{{\mathrm{x}}_3}_{})}^2}$