If the solubility product of calcium fluoride | 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 of $CaF_2$ be $s \ mol/L

Vedclass · Accepted Answer

$2.0 	imes 10^{-4} \ M$

Vedclass · Answer

(A) The dissociation of calcium fluoride is given by: $CaF_2(s) ightleftharpoons Ca^{2+}(aq) + 2F^-(aq)$.Let the solubility of $CaF_2$ 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 $K_{sp} = 3.2 	imes 10^{-11}$.So,$4s^3 = 3.2 	imes 10^{-11}$.$s^3 = 0.8 	imes 10^{-11} = 8 	imes 10^{-12}$.$s = \sqrt[3]{8 	imes 10^{-12}} = 2 	imes 10^{-4} \ M$.

If the solubility product of calcium fluoride $(CaF_2)$ is $3.2 \times 10^{-11}$,then its solubility will be ..........

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