At 310 K , the solubility of $\mathrm{CaF}_{2}$ in water is $2.34 \times 10^{-3} \mathrm{~g} / 100 \mathrm{~mL}$. The solubility product of $\mathrm{CaF}_{2}$ is $\_\_\_\_$ $\times 10^{-8}(\mathrm{~mol} / \mathrm{L})^{3}$. (Given molar mass : $\mathrm{CaF}_{2}=78 \mathrm{~g} \mathrm{~mol}^{-1}$ )

JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Equilibrium

At 310 K , the solubility of

CaF_{2}

in water is

2.34 \times 1 0^{- 3} g /100 mL

. The solubility product of

CaF_{2}

____

\times 1 0^{- 8} (mol / L)^{3}

. (Given molar mass :

CaF_{2} = 78 g mol^{- 1}

)

Answer: 0

First, convert solubility to mol/L:

Solubility (S) =

\frac{2.34 \times 1 0 ^{- 3} g}{100 mL} \times \frac{1000 mL}{1 L} \times \frac{1 mol}{78 g}

S = \frac{2.34 \times 1 0 ^{- 3} \times 10}{78} = \frac{2.34 \times 1 0 ^{- 2}}{78} = 3 \times 1 0^{- 4}

mol/L

For

CaF X_{2}

dissociation:

CaF X_{2} Ca X^{2 +} + 2 F X^{-}

If solubility = S, then

[Ca X^{2 +}] = S

and

[F X^{-}] = 2 S

K_{s p} = [Ca X^{2 +}] [F X^{-}]^{2} = S \times (2 S)^{2} = 4 S^{3}

K_{s p} = 4 \times (3 \times 1 0^{- 4})^{3} = 4 \times 27 \times 1 0^{- 12} = 108 \times 1 0^{- 12}

K_{s p} = 1.08 \times 1 0^{- 10} = 0.0108 \times 1 0^{- 8}

(mol/L)

^{3}

Rounding to the nearest integer: 0 × 10

^{- 8}

(mol/L)

^{3}

Therefore, the answer is 0.

JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Integer answer