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

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Question

(D) precipitate forms when the ionic product $(Q_{sp})$ exceeds the solubility product $(K_{sp})$.The dissociation of $CaF_2$ is: $CaF_{2(s)} \rightle

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) precipitate forms when the ionic product $(Q_{sp})$ exceeds the solubility product $(K_{sp})$.The dissociation of $CaF_2$ is: $CaF_{2(s)} ightleftharpoons Ca^{2+}_{(aq)} + 2F^{-}_{(aq)}$.The ionic product is calculated as: $Q_{sp} = [Ca^{2+}][F^-]^2$.For option $D$: $Q_{sp} = (10^{-2}) 	imes (10^{-3})^2 = 10^{-2} 	imes 10^{-6} = 10^{-8}$.Since $10^{-8} > 1.7 	imes 10^{-10}$,the ionic product is greater than $K_{sp}$,resulting in the formation of a precipitate.

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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