A 2.901 L saturated solution is formed by 0. | vedclass

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(A) $1$. Calculate the moles of $CaF_2$: $n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{0.08 \ g}{78 \ g \ mol^{-1}} \approx 1.0256 	imes 10^{-3}

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$1.715 	imes 10^{-10}$

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(A) $1$. Calculate the moles of $CaF_2$: $n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{0.08 \ g}{78 \ g \ mol^{-1}} \approx 1.0256 	imes 10^{-3} \ mol$. $2$. Calculate the molar solubility $(s)$ in $mol \ L^{-1}$: $s = \frac{n}{V} = \frac{1.0256 	imes 10^{-3} \ mol}{2.901 \ L} \approx 3.535 	imes 10^{-4} \ mol \ L^{-1}$. $3$. The dissociation equilibrium is $CaF_2(s) ightleftharpoons Ca^{2+}(aq) + 2F^-(aq)$. $4$. The solubility product expression is $K_{sp} = [Ca^{2+}][F^-]^2 = (s)(2s)^2 = 4s^3$. $5$. Substitute the value of $s$: $K_{sp} = 4 	imes (3.535 	imes 10^{-4})^3 \approx 4 	imes 4.416 	imes 10^{-11} \approx 1.766 	imes 10^{-10}$.

$A$ $2.901 \ L$ saturated solution is formed by $0.08 \ g$ of $CaF_2$ at $298 \ K$. Calculate the $K_{sp}$ of $CaF_2$. (Molar mass of $CaF_2$ is $78 \ g \ mol^{-1}$)

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