Find the solubility of CaF2 in a buffer of fixed pH 3.00 (in mol L^-1). K_sp(CaF2) = 3.9 x 10^-11; K_a(HF) = 6.6 x 10^-4. Assume all fluoride is present as F- or HF.

JEE Main levelSingle correct optionDifficulty 2 of 3ia079

Find the solubility of $CaF X_{2}$ in a buffer of fixed pH 3.00 (in $mol L^{- 1}$ ). $K_{s p} (CaF X_{2}) = 3.9 \times 1 0^{- 11}$ ; $K_{a} (HF) = 6.6 \times 1 0^{- 4}$ . Assume all fluoride is present as $F X^{-}$ or $HF$ .

Fraction present as

F X^{-}

α = K_{a} / (K_{a} + [H X^{+}]) = 0.398

. Total fluoride is

2 s

, so

[F X^{-}] = 2 s α

and

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

. Hence

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

(compared with

2.1 \times 1 0^{- 4} M

in pure water).