A precipitate of CaF_2 (K_{sp} = 1.7 times 10 | vedclass

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(B) For precipitation to occur,the ionic product $(Q_{sp})$ must be greater than the solubility product $(K_{sp})$.When equal volumes are mixed,the co

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

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(B) For precipitation to occur,the ionic product $(Q_{sp})$ must be greater than the solubility product $(K_{sp})$.When equal volumes are mixed,the concentration of each ion is halved.$K_{sp} = [Ca^{2+}][F^{-}]^2 = 1.7 	imes 10^{-10}$.For option $A$: $Q_{sp} = (\frac{10^{-4}}{2}) 	imes (\frac{10^{-4}}{2})^2 = 1.25 	imes 10^{-13} For option $B$: $Q_{sp} = (\frac{10^{-2}}{2}) 	imes (\frac{10^{-3}}{2})^2 = (0.5 	imes 10^{-2}) 	imes (0.25 	imes 10^{-6}) = 1.25 	imes 10^{-9} > K_{sp}$.For option $C$: $Q_{sp} = (\frac{10^{-5}}{2}) 	imes (\frac{10^{-3}}{2})^2 = 1.25 	imes 10^{-12} For option $D$: $Q_{sp} = (\frac{10^{-3}}{2}) 	imes (\frac{10^{-5}}{2})^2 = 1.25 	imes 10^{-14} Since $Q_{sp} > K_{sp}$ only in option $B$,a precipitate will form.

$A$ precipitate of $CaF_2$ $(K_{sp} = 1.7 \times 10^{-10})$ will be obtained when equal volumes of the following are mixed:

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