What is the solubility of calcium fluoride (C | vedclass

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Question

(A) The dissociation of calcium fluoride is given by: $CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2F^{-}(aq)$.Let the solubility be $S \ mol/L$. Then $

Vedclass · Accepted Answer

$2.0 	imes 10^{-4} \ mol/L$

Vedclass · Answer

(A) The dissociation of calcium fluoride is given by: $CaF_2(s) ightleftharpoons Ca^{2+}(aq) + 2F^{-}(aq)$.Let the solubility be $S \ mol/L$. Then $[Ca^{2+}] = S$ and $[F^{-}] = 2S$.The solubility product expression is: $K_{sp} = [Ca^{2+}][F^{-}]^2$.Substituting the values: $K_{sp} = (S)(2S)^2 = 4S^3$.Given $K_{sp} = 3.2 	imes 10^{-11}$,we have $4S^3 = 3.2 	imes 10^{-11}$.$S^3 = \frac{3.2 	imes 10^{-11}}{4} = 0.8 	imes 10^{-11} = 8 	imes 10^{-12}$.Taking the cube root: $S = \sqrt[3]{8 	imes 10^{-12}} = 2 	imes 10^{-4} \ mol/L$.

What is the solubility of calcium fluoride $(CaF_2)$ in a saturated solution,if its solubility product $(K_{sp})$ is $3.2 \times 10^{-11}$?

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