Question

The conductivity of a saturated solution of $CaF_{2}$ after deducting the contribution from water is $4.0\times 10^{- 5}Scm^{- 1}$ . If molar conductivity of $CaF_{2}$ at infinite dilution is $200Scm^{2}mol^{- 1},$ then the solubility product of $CaF_{2}$ 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)

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