Question

When 0.001 M Na2SO4 solution is saturated with CaSO4, its conductivity increases from 2.6 x 10^-4 S cm^-1 to 7.0 x 10^-4 S cm^-1. If the molar conductivities of Na⁺ and Ca2⁺ respectively are 50 S cm² mol^-1 and 120 S cm² mol^-1, then the solubility product of CaSO4 is:[Assume that conductivity of water used is negligible.]

• (A) 7.0 x 10^-6 M²
• (B) 4.0 x 10^-6 M²
• (C) 3.5 x 10^-6 M²
• (D) 2.0 x 10^-6 M²

Answer 1

The Correct Option is B

Explanation:
Conductivity of ${\mathrm{Na}}_2{\mathrm{SO}}_4=2.6\times {10}^{-4}\mathrm{S}{\mathrm{cm}}^{-1}$${\mathrm{\Lambda }}_m\left(N_{2}S{O}_4\right)=\frac{1000\times 26\times {10}^{-4}}{0.001}=260{\mathrm{cm}}^2{\mathrm{mof}}^{-1}$${\mathrm{\Lambda }}_m\left(S{O}_4^{2-}\right)={\mathrm{\Lambda }}_m\left({\mathrm{Na}}_2{\mathrm{SO}}_4\right)-2{\mathrm{\Lambda }}_m\left({\mathrm{Na}}^{+}\right)$$=260-2\times 50=160\mathrm{S}{\mathrm{cm}}^2{\mathrm{mol}}^{-1}$Conductivity of CaSO ${}_4$ solution$=7\times {10}^{-4}-2.6\times {10}^{-4}$$=4.4\times {10}^{-4}{\mathrm{Scm}}^{-1}$${\mathrm{\Lambda }}_m\left({\mathrm{CaSO}}_4\right)={\mathrm{\Lambda }}_m\left({\mathrm{Ca}}^{2+}\right)+{\mathrm{\Lambda }}_m\left(S{O}_4^{2-}\right)$${\mathrm{\Lambda }}_m=120+160=280\mathrm{S}{\mathrm{cm}}^2{\mathrm{mol}}^{-1}$$M=\frac{1000\times K}{{\mathrm{\Lambda }}_m}=\frac{1000\times 4.4\times {10}^{-4}}{280}$$M=1.57\times {10}^{-3}\mathrm{M}$$K_{SP}=\left[C{a}^{2+}\right]{\left[S{O}_4^{2-}\right]}_{\text{total }}$$=(0.00157)(0.00157+0.001)=4.0\times {10}^{-6}{\mathrm{M}}^2$