Question

The conductivity of $0.001\, M\, Na _{2} SO _{4}$ solution is $2.6 \times$ $10^{-4} S\,cm ^{-1}$ and increases to $7.0 \times 10^{-4} S\,cm ^{-1}$ when the solution is saturated with $CaSO _{4}$. The molar conductivities of $Na ^{+}$and $Ca ^{2+}$ are $50$ and $120\, S\,cm ^{2} mol ^{-1}$ respectively. By neglecting conductivity of used water, the solubility product of $CaSO _{4}$ is calculated as $x \times 10^{-6} M ^{2} .$ Find the value of $x$.

Answer 1

Conductivity of $Na _{2} SO _{4}=2.6 \times 10^{-4}$
$\Lambda_{ m }\left( Na _{2} SO _{4}\right)=\frac{1000 \times 2.6 \times 10^{-4}}{0.001}$
$=260\, S\, cm ^{2} mol ^{-1}$
$\lambda_{ m }\left( SO _{4}^{-2}\right)=\Lambda_{ m }\left( Na _{2} SO _{4}\right)-2\lambda_{ m }\left( Na ^{+}\right)$
$=260-2 \times 50$
$=160\, S\, cm ^{2} mol ^{-1}$
Conductivity of $CaSO _{4}$ solution $=7 \times 10^{-4}=2.6 \times 10^{-4}$
$=4.4 \times 10^{-4} S\, cm ^{-1}$
$\Lambda_{ m }\left( CaSO _{4}\right)=\lambda_{ m }\left( Ca ^{2+}\right)+\lambda_{ m }\left( SO _{4}{ }^{2-}\right)$
$=120+160$
$=260\, S\, cm ^{2} mol ^{-1}$
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 $CaSO _{4}=\left[ Ca ^{2+}\right]\left[ SO _{4}{ }^{2-}\right]$
$=\left(1.57 \times 10^{-3}\right)\left(1.57 \times 10^{-3}+1 \times 10^{-3}\right)$
$=4 \times 10^{-6} M ^{2}$