Question

The conductivity of a saturated solution of $Ca{F}_2$ after deducting the contribution from water is $4.0\times 10^{-5}Sc{m}^{-1}$ . If molar conductivity of $Ca{F}_2$ at infinite dilution is $200Sc{m}^{2}mo{l}^{-1},$ then the solubility product of $Ca{F}_2$ is :

• A. $3.2\times 10^{-20}{M}^3$
• B. $8.0\times 10^{-12}{M}^3$
• C. $3.2\times 10^{-11}{M}^3$
• D. $1.6\times 10^{-8}{M}^3$

Answer 1

Answer: (C)

${\left(K_{sp}\right)}_{{\left(CaF\right)}_2}=4s^3$
${\Lambda }_m^{\infty }=\frac{\kappa \times 1000}{s}$
$200=\frac{4\times 10^{-5}\times 10^3}{s}$
$s=2\times 10^{-4}$
$4s^3=4\times {\left(2\times {\left(10\right)}^{-4}\right)}^3$
$=4\times 8\times 10^{-12}$
$=32\times 10^{-12}$
${\left(K_{sp}\right)}_{{\left(CaF\right)}_2}=3.2\times {\left(10\right)}^{-11}{M}^3$