When equal volumes of Ca^{2+} and F^{-} solut | vedclass

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(A) When equal volumes are mixed,the concentration of each ion is halved. Let the initial concentrations be $[Ca^{2+}]_i$ and $[F^{-}]_i$. The new con

Vedclass · Accepted Answer

$10^{-2} \ M \ Ca^{2+} + 10^{-5} \ M \ F^{-}$

Vedclass · Answer

(A) When equal volumes are mixed,the concentration of each ion is halved. Let the initial concentrations be $[Ca^{2+}]_i$ and $[F^{-}]_i$. The new concentrations are $[Ca^{2+}] = [Ca^{2+}]_i / 2$ and $[F^{-}] = [F^{-}]_i / 2$.Precipitation occurs if the ionic product $Q_{sp} = [Ca^{2+}][F^{-}]^2 > K_{sp} = 1.6 	imes 10^{-10}$.For option $A$: $[Ca^{2+}] = 0.5 	imes 10^{-2}$,$[F^{-}] = 0.5 	imes 10^{-5}$. $Q_{sp} = (0.5 	imes 10^{-2})(0.5 	imes 10^{-5})^2 = 0.5 	imes 0.25 	imes 10^{-12} = 1.25 	imes 10^{-13} For option $B$: $[Ca^{2+}] = 0.5 	imes 10^{-3}$,$[F^{-}] = 0.5 	imes 10^{-3}$. $Q_{sp} = (0.5 	imes 10^{-3})(0.5 	imes 10^{-3})^2 = 0.125 	imes 10^{-9} = 1.25 	imes 10^{-10} For option $C$: $[Ca^{2+}] = 0.5 	imes 10^{-4}$,$[F^{-}] = 0.5 	imes 10^{-2}$. $Q_{sp} = (0.5 	imes 10^{-4})(0.5 	imes 10^{-2})^2 = 0.5 	imes 0.25 	imes 10^{-8} = 1.25 	imes 10^{-9} > 1.6 	imes 10^{-10}$. Precipitation occurs.For option $D$: $[Ca^{2+}] = 0.5 	imes 10^{-2}$,$[F^{-}] = 0.5 	imes 10^{-3}$. $Q_{sp} = (0.5 	imes 10^{-2})(0.5 	imes 10^{-3})^2 = 0.5 	imes 0.25 	imes 10^{-8} = 1.25 	imes 10^{-9} > 1.6 	imes 10^{-10}$. Precipitation occurs.Note: Both $A$ and $B$ do not precipitate. Given the original question structure,$A$ is the most dilute case.

Compound	$K_{sp}$
$AgCl$	$1.1 \times 10^{-10}$
$AgI$	$1.0 \times 10^{-16}$
$PbCrO_4$	$4.0 \times 10^{-14}$
$Ag_2CO_3$	$8.0 \times 10^{-12}$

When equal volumes of $Ca^{2+}$ and $F^{-}$ solutions are mixed,in which of the solutions precipitation will not occur? $(K_{sp} \text{ of } CaF_2 = 1.6 \times 10^{-10})$

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Compound $K_{sp}$

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