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

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Question

(B) When equal volumes are mixed,the concentration of each ion is halved.For option $(A)$: $[Ca^{2+}] = 0.5 	imes 10^{-4} \ M$,$[F^-] = 0.5 	imes 10

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) When equal volumes are mixed,the concentration of each ion is halved.For option $(A)$: $[Ca^{2+}] = 0.5 	imes 10^{-4} \ M$,$[F^-] = 0.5 	imes 10^{-4} \ M$.Ionic Product $(IP)$ = $[Ca^{2+}][F^-]^2 = (0.5 	imes 10^{-4}) 	imes (0.5 	imes 10^{-4})^2 = 1.25 	imes 10^{-13}$.Since $IP For option $(B)$: $[Ca^{2+}] = 0.5 	imes 10^{-2} \ M$,$[F^-] = 0.5 	imes 10^{-3} \ M$.Ionic Product $(IP)$ = $[Ca^{2+}][F^-]^2 = (0.5 	imes 10^{-2}) 	imes (0.5 	imes 10^{-3})^2 = 0.5 	imes 10^{-2} 	imes 0.25 	imes 10^{-6} = 1.25 	imes 10^{-9}$.Since $IP > K_{sp}$ $(1.25 	imes 10^{-9} > 1.7 	imes 10^{-10})$,precipitation occurs.

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

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