Question

If it takes \(0.048 \mathrm{~g} \mathrm{BaF}_{2}\) to saturate \(15.0 \mathrm{~mL}\) of water, what is the \(K_{\mathrm{sp}}\) of \(\mathrm{BaF}_{2}\) ?

Short answer

\(K_{\mathrm{sp}} = 2.45 \times 10^{-5}\)

Step-by-step solution

- Write the Dissociation Equation

The dissociation of \(\mathrm{BaF_{2}}\) in water is given by the equation: \[ \mathrm{BaF_{2}(s)} \rightarrow \mathrm{Ba^{2+}(aq)} + 2 \mathrm{F^{-}(aq)} \]

- Calculate the Molar Mass of \( \mathrm{BaF_{2}} \)

Find the molar mass of \(\mathrm{BaF_{2}} \) using the atomic masses: Barium (\(Ba\)) = 137.33 \(g/mol\), Fluorine (\(F\)) = 19.00 \(g/mol\). Thus, the molar mass of \(\mathrm{BaF_{2}} \) is: \[ 137.33 \ g/mol + 2 \times 19.00 \ g/mol = 175.33 \ g/mol \]

- Convert Grams to Moles

Convert the given mass of \(\mathrm{BaF_{2}} \) into moles: \[ 0.048 \mathrm{~g \ BaF_{2}} \div 175.33 \mathrm{\ g/mol} = 2.74 \times 10^{-4} \mathrm{\ mol} \mathrm{\ BaF_{2}} \]

- Find the Concentration of \( \mathrm{BaF_{2}} \)

Use the volume of water to find the concentration in the solution: \[ \text{Concentration} = \frac{2.74 \times 10^{-4} \mathrm{\ mol}}{15.0 \mathrm{\ mL}} = \frac{2.74 \times 10^{-4} \mathrm{\ mol}}{0.015 \mathrm{\ L}} = 1.83 \times 10^{-2} \mathrm{\ M} \]

- Determine Ion Concentrations

For every one \( \mathrm{BaF_{2}} \) that dissociates, it produces one \( Ba^{2+} \) ion and two \( F^{-} \) ions. Thus: \[ [\mathrm{Ba^{2+}}] = 1.83 \times 10^{-2} \mathrm{\ M} \] \[ [\mathrm{F^{-}}] = 2 \times 1.83 \times 10^{-2} \mathrm{\ M} = 3.66 \times 10^{-2} \mathrm{\ M} \]

- Calculate \( K_{\text{sp}} \)

The solubility product constant \(K_{\mathrm{sp}} \) can be found using the ion concentrations: \[ K_{\mathrm{sp}} = [\mathrm{Ba^{2+}}] \times [\mathrm{F^{-}}]^2 \] Plugging in the values: \[ K_{\mathrm{sp}} = (1.83 \times 10^{-2}) \times (3.66 \times 10^{-2})^2 \] \[ K_{\mathrm{sp}} = (1.83 \times 10^{-2}) \times (1.34 \times 10^{-3}) \] \[ K_{\mathrm{sp}} = 2.45 \times 10^{-5} \]

Key concepts

Dissociation Equation

The first step in solving solubility product constant problems involves writing the dissociation equation.
This equation helps describe how a compound splits into its constituent ions when dissolved in water.
For \( \text{BaF}_2\), the dissociation can be written as: \[ \text{BaF}_2 (s) \rightarrow \text{Ba}^{2+} (aq) + 2 \text{F}^{−} (aq) \]
This tells us that one formula unit of \( \text{BaF}_2\) breaks down into one barium ion (\( \text{Ba}^{2+} \)) and two fluoride ions (\( \text{F}^{−} \)). Understanding this relationship is crucial for calculating the ion concentrations in a solution.

Molar Mass Calculation

Next, we need to find the molar mass of the compound to convert grams to moles.
Molar mass is the mass of one mole of a substance. For \( \text{BaF}_2\), the formula to calculate the molar mass is:
\[ M_{\text{BaF}_2} = M_{\text{Ba}} + 2 \times M_{\text{F}} \]
where \[ M_{\text{Ba}} = 137.33 \ g/mol \] and \[ M_{\text{F}} = 19.00 \ g/mol \].
So, the molar mass of \( \text{BaF}_2 \) is: \[ 137.33 \ g/mol + 2 \times 19.00 \ g/mol = 175.33 \ g/mol \]
Understanding molar mass is essential for any calculation that involves converting between mass and moles.

Convert Grams to Moles

Now, we convert the given mass of \( \text{BaF}_2 \) into moles. This is achieved using the formula:
\[ \text{moles} = \frac{\text{mass}}{\text{molar mass}} \]
Given \ 0.048\ g \ of \( \text{BaF}_2\), and the molar mass \ 175.33\ g/mol \, we calculate: \[ \text{moles of } \text{BaF}_2 = \frac{0.048 \text{g}}{175.33 \ g/mol} = 2.74 \times 10^{-4} \ mol \]
This step is crucial because it transforms the mass of a substance into a usable quantity (moles) for further calculations.

Ion Concentration

Determining ion concentration involves finding out how much of each ion is present in the solution.
For this, we use the volume of the solution and the moles of \( \text{BaF}_2 \).
The concentration formula is:
\[ \text{Concentration} = \frac{\text{moles}}{\text{volume}} \]
Given that we have \ 2.74 \times 10^{-4} \text{moles} \ of \( \text{BaF}_2\) in \ 15.0 \text{mL} \, we convert the volume to liters: \ 15.0\ mL = 0.015\ L \.
So, the concentration is: \[ \frac{2.74 \times 10^{-4} \text{moles}}{0.015 \text{L}} = 1.83 \times 10^{-2} \ M \].
This concentration applies to the dissociation equation, where \( 1\ M \) of \( \text{BaF}_2 \) yields \( 1\ M \) of \( \text{Ba}^{2+} \) and \( 2\ M \) of \( \text{F}^{-} \).
Thus:
\[ [\text{Ba}^{2+}] = 1.83 \times 10^{-2} \ M \]
\[ [\text{F}^{-}] = 2 \times 1.83 \times 10^{-2} \ M = 3.66 \times 10^{-2} \ M \]
Knowing ion concentrations helps us in calculating the solubility product constant.

Ksp Calculation

Finally, we calculate the solubility product constant, \( K_{\text{sp}} \). This constant provides insights into the solubility of ionic compounds.
The formula for \( K_{\text{sp}} \) using ion concentrations is:
\[ K_{\text{sp}} = [\text{Ba}^{2+}] \times [\text{F}^{−}]^2 \]
Plugging in the values:
\[ K_{\text{sp}} = (1.83 \times 10^{-2}) \times (3.66 \times 10^{-2})^2 \]
Perform the multiplication:
\[ K_{\text{sp}} = (1.83 \times 10^{-2}) \times (1.34 \times 10^{-3}) \]
Therefore: \[ K_{\text{sp}} = 2.45 \times 10^{-5} \]
This value indicates the solubility of \( \text{BaF}_2 \) in water, which can be used in various chemical equilibrium calculations.