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JEE Main levelSingle correct optionDifficulty 2 of 3ia152

Find the solubility of $CaF X_{2}$ ( $K_{s p} = 4.0 \times 1 0^{- 11}$ ) in a buffer of pH $= 3.00$ . The fluoride released partly forms HF ( $K_{a} = 6.6 \times 1 0^{- 4}$ ); assume the pH stays constant.

A $2.2 \times 1 0^{- 4} M$
B $2.9 \times 1 0^{- 4} M$
C $1.2 \times 1 0^{- 4} M$
D $4.0 \times 1 0^{- 4} M$

Answer: D

[F X^{-}] = α (2 s)

with

α = K_{a} / (K_{a} + [H X^{+}]) = 0.3976

.

K_{s p} = s (2 s α)^{2} = 4 s^{3} α^{2}

, so

s = [K_{s p} / (4 α^{2})]^{1/3} = 4.0 \times 1 0^{- 4} M

(pure water:

2.2 \times 1 0^{- 4} M

).

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