Question

The conductivity of $0.001MN{a}_{2}S{O}_4$ solution is $2.6\times$ $10^{-4}Sc{m}^{-1}$ and increases to $7.0\times 10^{-4}Sc{m}^{-1}$ when the solution is saturated with $CaS{O}_4$. The molar conductivities of $N{a}^{+}$and $C{a}^{2+}$ are $50$ and $120Sc{m}^{2}mo{l}^{-1}$ respectively. By neglecting conductivity of used water, the solubility product of $CaS{O}_4$ is calculated as $x\times 10^{-6}{M}^2.$ Find the value of $x$.

Answer 1

Answer: 4

Conductivity of $N{a}_{2}S{O}_4=2.6\times 10^{-4}$
${\Lambda }_m\left(N{a}_{2}S{O}_4\right)=\frac{1000\times 2.6\times 10^{-4}}{0.001}$
$=260Sc{m}^{2}mo{l}^{-1}$
${\lambda }_m\left(S{O}_4^{-2}\right)={\Lambda }_m\left(N{a}_{2}S{O}_4\right)-2{\lambda }_m\left(N{a}^{+}\right)$
$=260-2\times 50$
$=160Sc{m}^{2}mo{l}^{-1}$
Conductivity of $CaS{O}_4$ solution $=7\times 10^{-4}=2.6\times 10^{-4}$
$=4.4\times 10^{-4}Sc{m}^{-1}$
${\Lambda }_m\left(CaS{O}_4\right)={\lambda }_m\left(C{a}^{2+}\right)+{\lambda }_m\left(S{O}_4{}^{2-}\right)$
$=120+160$
$=260Sc{m}^{2}mo{l}^{-1}$
Solubility $C=\frac{1000\times k}{{\Lambda }_m}=\frac{1000\times 4.4\times 10^{-4}}{280}$ $=1.57\times 10^{-3}M$
$K_{sp}$ of $CaS{O}_4=\left[C{a}^{2+}\right]\left[S{O}_4{}^{2-}\right]$
$=\left(1.57\times 10^{-3}\right)\left(1.57\times 10^{-3}+1\times 10^{-3}\right)$
$=4\times 10^{-6}{M}^2$