Question

The conductivity of $0.001\text{M}$ $N{a}_{2}S{O}_4$ solution is $2.6\times {10}^{–4}{\text{S cm}}^{–1}$ and increases to $7\times {10}^{–4}{\text{S cm}}^{–1}$ , when the solution is saturated with $CaS{O}_4$. The molar conductivity of $N{a}^{+}$ and $C{a}^{2+}$ are $50$ and $120{\text{S cm}}^2{\text{mol}}^{–1}$, respectively. Neglect the conductivity of water, find the solubility product of $CaS{O}_4$.

• A. $1\times {10}^{-6}$
• B. $2\times {10}^{-4}$
• C. $3\times {10}^{-5}$
• D. $4\times {10}^{-6}$

Answer 1

The correct option is D $4\times {10}^{-6}$

Conductivity of $N{a}_{2}S{O}_4=2.6\times {10}^{-4}$
${\Lambda }_m\left(N{a}_{2}S{O}_4\right)=\frac{1000\times 2.6\times {10}^{-4}}{0.001}=260{\text{S cm}}^2{\text{mol}}^{-1}$
${\Lambda }_m\left(S{O}_4^{2-}\right)={\Lambda }_m\left(N{a}_{2}S{O}_4\right)-2{\Lambda }_m\left(N{a}^{+}\right)$
$=260-2\times 50=160{\text{S cm}}^2{\text{mol}}^{-1}$

Conductivity of $CaS{O}_4$ solution
$=7\times {10}^{-4}-2.6\times {10}^{-4}=4.4\times {10}^{-4}{\text{S cm}}^{-1}$

${\Lambda }_m\left(CaS{O}_4\right)={\Lambda }_m\left(C{a}^{2+}\right)+{\Lambda }_m\left(S{O}_4^{2-}\right)$
$=120+160=280{\text{S cm}}^2{\text{mol}}^{-1}$

$\text{Solubility, S}=\frac{1000\times K}{{\Lambda }_m}=\frac{1000\times 4.4\times {10}^{-4}}{280}$
$=1.57\times {10}^{-3}\text{cm}$

$K_{sp}=\left[C{a}^{2+}\right][S{O}_4^{2-}{]}_{\text{total}}=(0.00157\left)\right(0.00157+0.001)$
$\approx 4\times {10}^{-6}{\text{M}}^2$