The solubility of CaF_2 in water is 1.7 times | vedclass

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(B) Step $1$: Calculate molar solubility $(S)$.Solubility in $g/L = 1.7 	imes 10^{-3} \ g/100 \ mL 	imes 10 = 0.017 \ g/L$.Molar solubility $S = \fr

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$4.14 	imes 10^{-11}$

Vedclass · Answer

(B) Step $1$: Calculate molar solubility $(S)$.Solubility in $g/L = 1.7 	imes 10^{-3} \ g/100 \ mL 	imes 10 = 0.017 \ g/L$.Molar solubility $S = \frac{0.017 \ g/L}{78 \ g/mol} \approx 2.18 	imes 10^{-4} \ mol/L$.Step $2$: Write the dissociation equation for $CaF_2$.$CaF_2(s) ightleftharpoons Ca^{2+}(aq) + 2F^-(aq)$.Step $3$: Express $K_{sp}$ in terms of $S$.$K_{sp} = [Ca^{2+}][F^-]^2 = (S)(2S)^2 = 4S^3$.Step $4$: Calculate $K_{sp}$.$K_{sp} = 4 	imes (2.18 	imes 10^{-4})^3 \approx 4 	imes 1.036 	imes 10^{-11} \approx 4.14 	imes 10^{-11}$.

The solubility of $CaF_2$ in water is $1.7 \times 10^{-3} \ g/100 \ mL$ at $298 \ K$. Calculate the solubility product $(K_{sp})$ of $CaF_2$. (Molar mass of $CaF_2 = 78 \ g/mol$)

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