SATHEE: Equilibrium - Result Question 65

Equilibrium - Result Question 65

####68. The molar solubility of $C a F_{2} (K_{s p} = 5.3 \times 10^{- 11})$ in $0.1 M$ solution of $N a F$ will be

[NEET Odisha 2019]

(a) $5.3 \times 10^{- 10} m o l L^{- 1}$

(b) $5.3 \times 10^{- 11} m o l L^{- 1}$

(d) $5.3 \times 10^{- 9} m o l L^{- 1}$

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Solution:

(d) $C a F_{2} \to C a^{2 +} + 2 F^{-}$ $t = 0 0 0$

At eqm. $s 2 s$

$N a F ⟶ N a^{+} + F^{-}$

$\begin{matrix} 0.1 & 0.1 & 0.1 \end{matrix}$

Due to common ion effect of $N a F$ , solubility of $C a F_{2}$ is further supressed. Therefore, the concentration of $F^{-}$ will be mainly due to $N a F$ .

$K_{s p} = [C a^{2 +}] [F^{-}]_{2}$

$K_{s p}^{s p} = (s) (0.1 + 2 s)^{2} 0.1 + 2 s \approx 0.1$

$s = \frac{K_{s p}}{(0.1)^{2}} = \frac{5.3 \times 10^{- 11}}{(0.1)^{2}} = 5.3 \times 10^{- 9} m o l L^{- 1}$

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Equilibrium - Result Question 65

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