Question

Calculate the solubility product of the sparingly soluble salt $Ca{F}_2$ from the following data:
The molar ionic conductances (at infinite dilution) of $C{a}^{2+}$ and $F^{-}$ ions are $104\times {10}^{-4}$ and $48\times {10}^{-4}S{m}^{2}mo{l}^{-1}$respectively. The specific conductance of the saturated solution of $Ca{F}_2$ at room temperature is $4.25\times {10}^{-3}S{m}^{-1}$ and the specific conductance of water used for preparing the solution is $2\times {10}^{-4}S{m}^{-1}$.

• A. $8.48\times {10}^{-11}mol{}^{3}d{m}^{-9}$
• B. $16.6\times {10}^{-12}mol{}^{3}d{m}^{-9}$
• C. $3.32\times {10}^{-11}mol{}^{3}d{m}^{-9}$
• D. $21.1\times {10}^{-11}mol{}^{3}d{m}^{-9}$

Answer 1

The correct option is C $3.32\times {10}^{-11}mol{}^{3}d{m}^{-9}$

Let $C$ mol $d{m}^{-3}$ be the solubility of $Ca{F}_2$.
Accordingly, the solubility equilibrium may be written as

$Ca{F}_2\left(s\right)\rightleftharpoons \underset{C}{C{a}^{2+}\left(aq\right)}+\underset{2C}{2F^{-}\left(aq\right)}$

Then, at equilibrium, the solubility product of $Ca{F}_2$ is given by

$K_{sp}=\left[C{a}^{2+}\right]\left[F^{-}\right]=\left(C\right){\left(2C\right)}^2=4C^3....eqn.1$

Specific conductance due to $Ca{F}_2$ is
$\kappa \left(Ca{F}_2\right)=\kappa \left(solution\right)-\kappa \left(water\right)=(4.25\times {10}^{-3}-2.0\times {10}^{-4})S{m}^{-1}=4.05\times {10}^{-3}S{m}^{-1}$

By Kohlrausch law,
${\Lambda }_{m\left(Ca{F}_2\right)}^0={{\lambda }^0}_{\left(C{a}^{2+}\right)}+2{\lambda }_{\left(F^{-}\right)}^0=(104+2\times 48)\times {10}^{-4}S{m}^{2}mo{l}^{-1}=200\times {10}^{-4}S{m}^{2}mol{}^{-1}$

For saturated solution of sparingly soluble salt, ${\Lambda }_m\approx {\Lambda }_m^0$
Also
${\Lambda }_m^0=\frac{\kappa }{C}$

$\therefore$
$C=\frac{\kappa }{{\Lambda }_m^0}$
$C=\frac{4.05\times {10}^{-3}S{m}^{-1}}{200\times {10}^{-4}S{m}^{2}mo{l}^{-1}}=2.025\times {10}^{-1}mol{m}^{-3}C=2.025\times {10}^{-4}mold{m}^{-3}$

Hence, from Eq.(i)
$K_{sp}=4C^3=4{(2.025\times {10}^{-4}mold{m}^{-3})}^3=3.32\times {10}^{-11}mol{}^{3}d{m}^{-9}$