Question

tion 12 of 27 > In a saturated aqueous solution of CaF , the calcium ion concentration is 2.15 x 10-4 M and the fluoride ion concentration is 4.31 x 10M. Calculate the solubility product, Kıp, for CaF, Kop =

Answer 1

CaF₂ dissolves as follows:

$CaF_2_{(s)} \rightarrow Ca^{2+} _{(aq)} + 2 F^-_{(aq)}$

Hence, given that the solution is saturated, the equilibrium constant K_sp of the above reaction can be written as

Note that CaF₂ concentration does not appear in the equilibrium expression as it is a pure solid with concentration unity.

Now, it is given that the concentrations are

$[Ca^{2+}_{(aq)}] = 2.15 \times 10^{-4} \ M$

$[F^-_{(aq)}] = 4.31 \times 10^{-4} \ M$

Hence, K_sp for CaF₂ can be calculated as

$K_{sp} = [Ca^{2+}_{(aq)}] [F^-_{(aq)}]^2 = 2.15 \times 10^{-4} \ M \times (4.31 \times 10^{-4} \ M)^2 \approx {\color{Red} 3.99 \times 10^{-11}}$

Hence, the K_sp value calculated for CaF₂ is ${\color{Red} 3.99 \times 10^{-11}}$ . ( rounded to three significant figures).