If the $[F^{-}] = 2.0 \times 10^{-5} \ M$ in water,how many grams of $CaCl_{2}$ must be added to initiate the precipitation of $CaF_{2}$? Given: $K_{sp}$ for $CaF_{2} = 1.7 \times 10^{-10}$ and the molar mass of $CaCl_{2} = 111 \ g \ mol^{-1}$. Assume the volume of the solution is $1 \ L$.
(N/A) The precipitation of $CaF_{2}$ occurs when the ionic product exceeds the solubility product constant $(K_{sp})$.
$K_{sp} = [Ca^{2+}][F^{-}]^{2}$
Given $[F^{-}] = 2.0 \times 10^{-5} \ M$ and $K_{sp} = 1.7 \times 10^{-10}$.
Substituting the values: $1.7 \times 10^{-10} = [Ca^{2+}] \times (2.0 \times 10^{-5})^{2}$
$[Ca^{2+}] = \frac{1.7 \times 10^{-10}}{4.0 \times 10^{-10}} = 0.425 \ M$.
Since $1 \ mol$ of $CaCl_{2}$ provides $1 \ mol$ of $Ca^{2+}$,the concentration of $CaCl_{2}$ required is $0.425 \ M$.
For $1 \ L$ of solution,the mass of $CaCl_{2}$ required = $0.425 \ mol \times 111 \ g \ mol^{-1} = 47.175 \ g$.
Therefore,more than $47.175 \ g$ of $CaCl_{2}$ must be added to initiate precipitation.
$K_{sp} = [Ca^{2+}][F^{-}]^{2}$
Given $[F^{-}] = 2.0 \times 10^{-5} \ M$ and $K_{sp} = 1.7 \times 10^{-10}$.
Substituting the values: $1.7 \times 10^{-10} = [Ca^{2+}] \times (2.0 \times 10^{-5})^{2}$
$[Ca^{2+}] = \frac{1.7 \times 10^{-10}}{4.0 \times 10^{-10}} = 0.425 \ M$.
Since $1 \ mol$ of $CaCl_{2}$ provides $1 \ mol$ of $Ca^{2+}$,the concentration of $CaCl_{2}$ required is $0.425 \ M$.
For $1 \ L$ of solution,the mass of $CaCl_{2}$ required = $0.425 \ mol \times 111 \ g \ mol^{-1} = 47.175 \ g$.
Therefore,more than $47.175 \ g$ of $CaCl_{2}$ must be added to initiate precipitation.
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