Question

A conductivity of $0.001MN{a}_{2}S{O}_4$ solution is $2.6\times {10}^{-4}Sc{m}^{-1}$ and increase to $7\times {10}^{-4}Sc{m}^{-1}$, when the solution is saturated with $CaS{O}_4$. The molar conductivities of $Na$ and $C{a}^{+2}$ are $50$ and $120Sc{m}^{2}mol{e}^{-1}$, respectively. Neglect conductivity of used water. What is the solublity product of $CaS{O}_4$?

• A. $4\times {10}^{-6}$
• B. $2.57\times {10}^{-3}$
• C. $3\times {10}^{-4}$
• D. $6.46\times {10}^{-6}$

Answer 1

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

Conductivity of $N{a}_{2}S{O}_4=2.6\times {10}^{-4}$
${\Lambda }_m\left(N{a}_{2}S{O}_4\right)=\frac{1000\times 2.6\times {10}^{-4}}{0.001}$

$=260Sc{m}^{2}mo{l}^{-1}$
${\lambda }_m\left(S{O}_4^{2-}\right)={\Lambda }_m\left(N{a}_{2}S{O}_4\right)-2{\lambda }_m\left(N{a}^{+}\right)$
$=260-2\times 50$
$=160Sc{m}^{2}mo{l}^{-1}$
Conductivity of $CaS{O}_4$ solution
$=7\times {10}^{-4}-2.6\times {10}^{-4}$

$=4.4\times {10}^{-4}Sc{m}^{-1}$

${\Lambda }_m\left(CaS{O}_4\right)={\lambda }_m\left(C{a}^{2+}\right)+{\lambda }_m\left(S{O}_4^{2-}\right)$
$=120+160$
$=280Sc{m}^{2}mo{l}^{-1}$
Solubility $C=\frac{1000\times \kappa }{{\Lambda }_m}=\frac{1000\times 4.4\times {10}^{-4}}{280}$

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

$K_{sp}=\left[C{a}^{2+}\right][S{O}_4^{2-}{]}_{total}$
$=\left(0.00157\right)(0.00157+0.001)$

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

Hence option (A) is correct.