Question

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

• A. $4\times {10}^{-6}$
• B. $1.57\times {10}^{-3}$
• C. $4\times {10}^{-4}$
• D. $2.46\times {10}^{-6}$

Answer 1

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

$\text{Conductivity of}N{a}_{2}S{O}_4\left(K_{N{a}_{2}S{O}_4}\right)=2.6\times {10}^{-4}Sc{m}^{-1}$

$\text{Molar conductivity of}N{a}_{2}S{O}_4({\wedge }_{m\left(N{a}_{2}S{O}_4\right)}=\frac{\text{conductivity (K)}}{\text{concentration (C)}}$

$=\frac{2.6\times {10}^{-4}\times 1000}{0.001}$
$=260Sc{m}^{2}mo{l}^{-1}$
${\wedge }_{m\left(S{O}_4^{2-}\right)}={\wedge }_{m\left(N{a}_{2}S{O}_4\right)}-2{\wedge }_{m\left(N{a}^{+}\right)}$
$\wedge m\left(S{O}_4^{2-}\right)=260-(2\times 50)$
$=160Sc{m}^{2}mo{l}^{-1}$
Conductivity of $CaS{O}_4$ solution $=7\times {10}^{-4}-2.6\times {10}^{-4}=4.4\times {10}^{-4}Sc{m}^{-1}$
${\wedge }_{m\left(CaS{O}_4\right)}={\wedge }_{m\left(C{a}^{2+}\right)}+{\wedge }_{m\left(S{O}_4^{2-}\right)}$
$=120+160$
$=280Sc{m}^{2}mo{l}^{-1}$
$\text{Solubility of}C{a}^{2+}=\frac{1000\times K}{{\wedge }_m}=\frac{1000\times 4.4\times {10}^{-4}}{280}=1.57\times {10}^{-3}M$
$\text{Solubility product}\left(K_{SP}\right)ofCaS{O}_4=\left[C{a}^{2+}\right]{\left[S{O}_4^{2-}\right]}_{\text{total}}$
$=\left(0.00157\right)(0.00157+0.001)$
$=4.0\times {10}^{-6}{M}^2$