QuestionJuly 5, 2025

Calculate the solubility at 25^circ C of CaF_(2) in pure water and in a 0.0110M NaF solution. You'll find K_(sp) data in the ALEKS Data tab. Round both of your answers to 2 significant digits. solubility in pure water: square (g)/(L) solubility in 0.0110 M NaF solution: square (g)/(L)

Calculate the solubility at 25^circ C of CaF_(2) in pure water and in a 0.0110M NaF solution. You'll find K_(sp) data in the ALEKS Data tab. Round both of your answers to 2 significant digits. solubility in pure water: square (g)/(L) solubility in 0.0110 M NaF solution: square (g)/(L)
Calculate the solubility at 25^circ C of CaF_(2) in pure water and in a 0.0110M NaF
solution. You'll find K_(sp)
data in the ALEKS Data tab.
Round both of your answers to 2 significant digits.
solubility in pure water:
square (g)/(L)
solubility in 0.0110 M NaF solution:
square (g)/(L)

Solution
4.6(323 votes)

Answer

Solubility in pure water: 2.0 \times 10^{-4} \frac{g}{L} ### Solubility in 0.0110 M NaF solution: 3.2 \times 10^{-7} \frac{g}{L} Explanation 1. Determine K_{sp} for CaF_2 Assume K_{sp} = 3.9 \times 10^{-11} (typical value for CaF_2). 2. Calculate solubility in pure water For CaF_2, dissociation is CaF_2 \rightleftharpoons Ca^{2+} + 2F^-. Let solubility be s. Then, [Ca^{2+}] = s and [F^-] = 2s. **K_{sp} = [Ca^{2+}][F^-]^2 = s(2s)^2 = 4s^3**. Solve for s: 4s^3 = 3.9 \times 10^{-11}, so s = \sqrt[3]{\frac{3.9 \times 10^{-11}}{4}}. 3. Calculate solubility in 0.0110 M NaF solution In the presence of 0.0110 M NaF, [F^-] = 0.0110 + 2s. **K_{sp} = [Ca^{2+}][F^-]^2 = s(0.0110 + 2s)^2**. Assuming 2s \ll 0.0110, approximate K_{sp} = s(0.0110)^2. Solve for s: s = \frac{3.9 \times 10^{-11}}{(0.0110)^2}.

Explanation

1. Determine $K_{sp}$ for $CaF_2$<br /> Assume $K_{sp} = 3.9 \times 10^{-11}$ (typical value for $CaF_2$).<br /><br />2. Calculate solubility in pure water<br /> For $CaF_2$, dissociation is $CaF_2 \rightleftharpoons Ca^{2+} + 2F^-$. Let solubility be $s$. Then, $[Ca^{2+}] = s$ and $[F^-] = 2s$. **$K_{sp} = [Ca^{2+}][F^-]^2 = s(2s)^2 = 4s^3$**. Solve for $s$: $4s^3 = 3.9 \times 10^{-11}$, so $s = \sqrt[3]{\frac{3.9 \times 10^{-11}}{4}}$.<br /><br />3. Calculate solubility in 0.0110 M NaF solution<br /> In the presence of 0.0110 M NaF, $[F^-] = 0.0110 + 2s$. **$K_{sp} = [Ca^{2+}][F^-]^2 = s(0.0110 + 2s)^2$**. Assuming $2s \ll 0.0110$, approximate $K_{sp} = s(0.0110)^2$. Solve for $s$: $s = \frac{3.9 \times 10^{-11}}{(0.0110)^2}$.
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