\begin{tabular}{l} Calculate the solubility at $$ 25^{\circ} \mathrm{C} $$ of $$ \mathrm{CaF}_{2} $$ in pure water and in a 0.0100 M NaF solution. You'll find $$ K_{s p} $$ data in the ALEKS Data tab. \ Round both of your answers to 2 significant digits. \ solubility in pure water: \ solubility in 0.0100 M NaF solution: \ \hline$$ \frac{\mathrm{g}}{\mathrm{L}} $$\end{tabular}

Question

UpStudy AI · Accepted Answer

We start with the dissolution equilibrium for calcium fluoride:

$$
\text{CaF}_{2}(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\,\text{F}^{-}(aq)
$$

Let the molar solubility in pure water be $$ s $$ (in mol/L). Then at equilibrium the concentrations are:

- $$[ \text{Ca}^{2+}] = s$$
- $$[ \text{F}^{-}] = 2s$$

The solubility product is given by

$$
K_{sp} = s (2s)^2 = 4s^3
$$

### Solubility in Pure Water

Solve for $$ s $$:

$$
s = \sqrt[3]{\frac{K_{sp}}{4}}
$$

Using the provided ALEKS value (commonly used is $$K_{sp} = 3.9\times10^{-11}$$):

$$
s = \sqrt[3]{\frac{3.9 \times 10^{-11}}{4}} = \sqrt[3]{9.75 \times 10^{-12}}
$$

A cube-root evaluation gives approximately

$$
s \approx 2.14 \times 10^{-4}\,\text{M}
$$

To convert to grams per liter, we use the molar mass of $$\text{CaF}_{2}$$. With

- $$ M_{\text{Ca}} \approx 40.08\,\text{g/mol} $$
- $$ M_{\text{F}} \approx 19.00\,\text{g/mol} $$

the molar mass is

$$
M_{\text{CaF}_{2}} = 40.08 + 2(19.00) \approx 78.08\,\text{g/mol}
$$

Thus, the solubility in $$\text{g/L}$$ is

$$
\text{Solubility} = s \times M_{\text{CaF}_{2}} \approx \left(2.14 \times 10^{-4}\right) \times 78.08 \approx 0.0167\,\frac{\text{g}}{\text{L}}
$$

Rounded to 2 significant digits:

$$
\text{Solubility in pure water} \approx 0.017\,\frac{\text{g}}{\text{L}}
$$

---

### Solubility in a $$ 0.0100\,\text{M} $$ NaF Solution

In a solution of sodium fluoride, the common ion $$\text{F}^{-}$$ is already present. Let the new solubility be $$ s' $$ (in mol/L). The initial $$\text{F}^{-}$$ concentration from NaF is $$0.0100\,\text{M}$$. When $$\text{CaF}_{2}$$ dissolves, it adds $$ 2s' $$ to the $$\text{F}^{-}$$ concentration. However, because $$ s' $$ will be very small compared to $$0.0100\,\text{M}$$, we approximate:

$$
[\text{F}^{-}] \approx 0.0100\,\text{M}
$$

At equilibrium, the concentrations are:

- $$[ \text{Ca}^{2+}] = s'$$
- $$[ \text{F}^{-}] \approx 0.0100\,\text{M}$$

The solubility product expression becomes

$$
K_{sp} = s' \,(0.0100)^2
$$

Solving for $$ s' $$:

$$
s' = \frac{K_{sp}}{(0.0100)^2} = \frac{3.9 \times 10^{-11}}{1.0 \times 10^{-4}} = 3.9 \times 10^{-7}\,\text{M}
$$

Convert this molar solubility to $$\text{g/L}$$:

$$
\text{Solubility} = s' \times M_{\text{CaF}_{2}} \approx \left(3.9 \times 10^{-7}\right) \times 78.08 \approx 3.0 \times 10^{-5}\,\frac{\text{g}}{\text{L}}
$$

---

### Final Answers

- Solubility in pure water: $$0.017\,\frac{\text{g}}{\text{L}}$$
- Solubility in $$ 0.0100\,\text{M} $$ NaF solution: $$3.0 \times 10^{-5}\,\frac{\text{g}}{\text{L}}$$
- Solubility in pure water: 0.017 g/L - Solubility in 0.0100 M NaF solution: 3.0 × 10⁻⁵ g/L

Calculate the solubility at of in pure water and in a 0.0100 M NaF solution. You'll find data in the ALEKS Data tab.

Round both of your answers to 2 significant digits.

solubility in pure water:

solubility in 0.0100 M NaF solution:

Solución de inteligencia artificial de Upstudy

Responder

Solución

Solubility in Pure Water

Solubility in a NaF Solution

Final Answers

Mind Expander

preguntas relacionadas

Latest Chemistry Questions

Calculate the solubility at of in pure water and in a 0.0100 M NaF solution. You'll find data in the ALEKS Data tab. Round both of your answers to 2 significant digits. solubility in pure water: solubility in 0.0100 M NaF solution: