A precipitate of CaF_{2} (K_{sp} = 1.7 times | vedclass

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(B) precipitate forms when the ionic product $(Q_{sp})$ exceeds the solubility product $(K_{sp})$.When equal volumes are mixed,the concentration of ea

Vedclass · Accepted Answer

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

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(B) precipitate forms when the ionic product $(Q_{sp})$ exceeds the solubility product $(K_{sp})$.When equal volumes are mixed,the concentration of each ion is halved.For option $B$:$[Ca^{2+}]_{new} = \frac{10^{-2}}{2} = 5 	imes 10^{-3} \ M$$[F^{-}]_{new} = \frac{10^{-3}}{2} = 5 	imes 10^{-4} \ M$$Q_{sp} = [Ca^{2+}] [F^{-}]^2 = (5 	imes 10^{-3}) 	imes (5 	imes 10^{-4})^2 = 5 	imes 10^{-3} 	imes 25 	imes 10^{-8} = 1.25 	imes 10^{-9}$.Since $1.25 	imes 10^{-9} > 1.7 	imes 10^{-10}$,a precipitate will be obtained.

$A$ 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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