Question

What is the solubility of \(\mathrm{CaF}_{2}\) in a buffer solution containing \(0.30 \mathrm{M} \mathrm{HCHO}_{2}\) and \(0.20 \mathrm{M} \mathrm{NaCHO}_{2}\) ? $$ \mathrm{CaF}_{2}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Ca}^{2+}(a q)+2 \mathrm{HF}(a q) $$ and solve the equilibrium problem.

Short answer

To determine the solubility of \(\mathrm{CaF}_{2}\) in a buffer solution containing \(\mathrm{HCHO}_{2}\) and \(\mathrm{NaCHO}_{2}\), first, write out the equations for the solubility product constant (\(K_{sp}\)) and the acid dissociation constant (\(K_a\)) for the given species. Second, set up a reaction table showing the initial, change, and equilibrium concentrations of each species involved. Next, substitute the equilibrium concentrations into the \(K_{sp}\) and \(K_a\) equations, and solve for the solubility (\(s\)). Finally, calculate the solubility of \(\mathrm{CaF}_{2}\) using the values of \(K_a\) and \(K_{sp}\) obtained from a reference table. Ensure that your final answer is expressed in molarity (\(\mathrm{M}\)).

Step-by-step solution

Determine initial concentrations of species

Given that the buffer solution contains \(0.30\ \mathrm{M}\ \mathrm{HCHO}_{2}\) and \(0.20\ \mathrm{M}\ \mathrm{NaCHO}_{2}\), we can assume that the initial concentrations of the species involved in the reaction are as follows: \([\mathrm{H}^{+}] = 0.30\ \mathrm{M}\) \([\mathrm{CHO}_{2}^{-}] = 0.20\ \mathrm{M}\) As \(\mathrm{CaF}_{2}\) is a solid, we assume its initial concentration to be undetermined.

Write out the solubility and equilibrium constants

For the solubility of \(\mathrm{CaF}_{2}\), we can write the solubility product constant, \(K_{sp}\), as follows: $$K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{F}^-]^2$$ For the buffer equilibrium, we have the acid dissociation constant \(K_a\) for \(\mathrm{HCHO}_{2}\) as follows: $$K_a = \frac{[\mathrm{H}^{+}][\mathrm{CHO}_{2}^-]}{[\mathrm{HCHO}_{2}]}$$

Set up reaction table for concentrations

Let \(s\) denote the solubility of \(\mathrm{CaF}_{2}\) in the buffer solution. We can set up a reaction table for the initial, change, and equilibrium concentrations of each species as follows: $$ \begin{array}{|c|c|c|c|} \hline & Initial & Change & Equilibrium \\ \hline \mathrm{Ca}^{2+} & 0 & +s & s \\ \hline \mathrm{H}^{+} & 0.30 & +2s & 0.30+2s \\ \hline \mathrm{F}^- & 0 & +2s & 2s \\ \hline \mathrm{HCHO}_2 & - & - & - \\ \hline \mathrm{CHO}_2^- & 0.20 & +2s & 0.20+2s \\ \hline \end{array} $$

Solve for solubility \(s\)

Now, we can substitute the equilibrium concentrations into the solubility product constant, \(K_{sp}\) equation, and the acid dissociation constant, \(K_a\) equation, and solve for \(s\). To do this, we first write the \(K_{sp}\) expression: $$K_{sp} = [s][(2s)^2]$$ Then, we write the \(K_a\) expression and rearrange the equation to solve for the concentration of \([\mathrm{H}^{+}]\): $$[\mathrm{H}^{+}] = \frac{K_a[\mathrm{HCHO}_{2}]}{[\mathrm{CHO}_{2}^{-}]}$$ $$[\mathrm{H}^{+}] = \frac{K_a(0.30)}{(0.20+2s)}$$ Now, substitute the \([\mathrm{H}^{+}]\) expression into the reaction table, and use the relationship \([\mathrm{H}^{+}] = 0.30 + 2s\) to solve for \(s\), which represents the solubility of \(\mathrm{CaF}_{2}\): $$\frac{K_a(0.30)}{(0.20+2s)} = 0.30 + 2s$$ Solve for \(s\) using \(K_a\) and \(K_{sp}\) values from a reference table for \(\mathrm{HCHO}_{2}\) and \(\mathrm{CaF}_{2}\).

Calculate the solubility of \(\mathrm{CaF}_{2}\)

After solving for \(s\), you will obtain the solubility of \(\mathrm{CaF}_{2}\) in this buffer solution. Make sure to express your final answer in terms of molarity (\(\mathrm{M}\)).

Key concepts

Solubility Product Constant

The solubility product constant, denoted as \( K_{sp} \), is a fundamental parameter in the study of solubility equilibria. It is a unique value for each sparingly soluble compound that determines the extent to which the compound will dissolve in a solution. When \( K_{sp} \) is applied to calcium fluoride, \( \text{CaF}_2 \), which is the compound in our exercise, the equation is given by \( K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 \).

The concept behind \( K_{sp} \) is that at equilibrium, the product of the concentrations of the ionic species raised to the power of their stoichiometric coefficients will remain constant for a saturated solution. Solubility equilibria involve the equilibrium that exists between a solid and its ions in solution. Understanding \( K_{sp} \) is crucial because it also informs us about the solubility, i.e., how much of the compound can be dissolved in a solution before reaching saturation. In our exercise, the solubility (\( s \)) of \( \text{CaF}_2 \) in the buffer solution needed to be determined in the context of its \( K_{sp} \).

Acid Dissociation Constant

The acid dissociation constant, represented by \( K_a \), quantifies the strength of an acid in solution. It measures the degree of dissociation of an acid to produce hydronium ions (\( \text{H}^+ \)) and its conjugate base in a solution. In the context of our exercise, formic acid (\( \text{HCHO}_2 \)) serves as the acid, and the \( K_a \) expression is described as \( K_a = \frac{[\text{H}^+][\text{CHO}_2^-]}{[\text{HCHO}_2]} \).

The higher the value of \( K_a \), the stronger the acid, as it indicates a greater tendency for the acid to lose a proton. In a buffer solution, maintaining a stable pH is critical, and the \( K_a \) provides necessary information about how the presence of acids or bases will affect the solution's pH. When solving the equilibrium problem in the given exercise, \( K_a \) is essential for calculating the concentration of \( \text{H}^+ \) ions in the buffer solution, which in turn was used to determine the solubility of \( \text{CaF}_2 \) in the buffer.

Buffer Solution

A buffer solution is a special type of solution that resists changes in pH when small amounts of acid or base are added. This property is extremely useful in many biological and chemical applications where maintaining a consistent pH is critical. A buffer is typically composed of a weak acid and its conjugate base, or a weak base and its conjugate acid. The exercise we are discussing involves a buffer made up of formic acid (\( \text{HCHO}_2 \)) and sodium formate (\( \text{NaCHO}_2 \)), the salt of its conjugate base.

The key to a buffer's pH-resisting action lies in its components' ability to absorb the additional hydronium (\( \text{H}^+ \)) or hydroxide (\( \text{OH}^- \)) ions without drastically altering the pH. In our exercise's context, understanding how the buffer behaves is vital to determining the solubility of \( \text{CaF}_2 \) as the presence of ions from buffer components can affect the equilibrium concentrations of the species involved in the solubility product.