The conductivity of 0.001 M ${\mathrm{Na}}_2{\mathrm{SO}}_4$ solution is $2.6\times {10}^{-4}{\mathrm{Scm}}^{-1}$ and increases to $7.0\times {10}^{-4}{\mathrm{Scm}}^{-1},$when the solution is saturated with ${\mathrm{CaSO}}_4,$The molar conductivities of ${\mathrm{Na}}^{+}$& ${\mathrm{Ca}}^{+2}$ are 50 &120${\mathrm{Scm}}^{2 }{\mathrm{mol}}^{-1}$ respectively. Neglect conductivity of used water. What is the solubility product of ${\mathrm{CaSO}}_4$

Conductivity of ${\mathrm{Na}}_2{\mathrm{SO}}_4=2.6\times {10}^{-4}$

${\mathrm{\lambda }}_{\mathrm{M}}\left({\mathrm{Na}}_2{\mathrm{SO}}_4\right)=\frac{1000\times 2.6\times {10}^{-4}}{0.001}$

$=260 {\mathrm{Scm}}^2{\mathrm{mol}}^{-1}$

${\mathrm{\lambda }}_{\mathrm{M}\left({\mathrm{SO}}_4^{-2}\right)}={\mathrm{\lambda }}_{\mathrm{M}\left({\mathrm{Na}}_2{\mathrm{SO}}_4\right)}-2{\mathrm{\lambda }}_{\mathrm{M}}\left({\mathrm{Na}}^{+}\right)$

$=260-2\times 50$

$=160 {\mathrm{Scm}}^2{\mathrm{mol}}^{-1}$

Conductivity of ${\mathrm{CaSO}}_4 \mathrm{solu}=7\times {10}^{-4}-2.6\times {10}^{-4}$

$=4.4\times {10}^{-4}{\mathrm{Scm}}^{-1}$

${\mathrm{\lambda }}_{\mathrm{M}}\left({\mathrm{CaSO}}_4\right)={\mathrm{\lambda }}_{\mathrm{M}\left({\mathrm{ca}}^{+2}\right)}+{\mathrm{\lambda }}_{\mathrm{M}\left({\mathrm{SO}}_4^{-2}\right)}$

$=120+160=280{\mathrm{Scm}}^2{\mathrm{mol}}^{-1}$

$\mathrm{Solubility} =\mathrm{C}=\frac{1000\times \mathrm{K}}{{\mathrm{\lambda }}_{\mathrm{M}}}=\frac{1000\times 4.4\times {10}^{-4}}{280}$

$=1.57\times {10}^{-3}\mathrm{M}$

${\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Ca}}^{+2}\right]{\left[{\mathrm{SO}}_4^{-2}\right]}_{\mathrm{total}}$

=(0.00157)(0.00157+0.001)

$=4.0\times {10}^{-6}{\mathrm{M}}^2$