If K_{sp} of CaF_2 at 25,^{circ}C is 1.7 time | vedclass

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Question

(A) precipitate forms when the ionic product of the ions in the solution exceeds the solubility product constant $(K_{sp})$.For the dissociation of $C

Vedclass · Accepted Answer

$1 	imes 10^{-2}\,M\,Ca^{2+}$ and $1 	imes 10^{-3}\,M\,F^{-}$

Vedclass · Answer

(A) precipitate forms when the ionic product of the ions in the solution exceeds the solubility product constant $(K_{sp})$.For the dissociation of $CaF_2$: $CaF_2 ightleftharpoons Ca^{2+} + 2F^{-}$.The ionic product is given by: $Q_{sp} = [Ca^{2+}][F^{-}]^2$.For option $(A)$: $[Ca^{2+}] = 1 	imes 10^{-2}\,M$ and $[F^{-}] = 1 	imes 10^{-3}\,M$.$Q_{sp} = (1 	imes 10^{-2}) 	imes (1 	imes 10^{-3})^2 = (1 	imes 10^{-2}) 	imes (1 	imes 10^{-6}) = 1 	imes 10^{-8}$.Since $1 	imes 10^{-8} > 1.7 	imes 10^{-10}$,the ionic product exceeds $K_{sp}$,leading to the formation of a precipitate.Therefore,option $(A)$ is correct.

If $K_{sp}$ of $CaF_2$ at $25\,^{\circ}C$ is $1.7 \times 10^{-10}$,the combination amongst the following which gives a precipitate of $CaF_2$ is

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