Question

A solution is prepared by dissolving 0.56 g benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}, K_{\mathrm{a}}=6.4 \times 10^{-5}\right)\) in enough water to make \(1.0 \mathrm{L}\) of solution. Calculate \(\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\right],\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}\right],\left[\mathrm{H}^{+}\right],\left[\mathrm{OH}^{-}\right]\) and the pH of this solution.

Short answer

In the solution, the concentrations of the species are approximately as follows: \(\left[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}\mathrm{H}\right] \approx 0.00458\, \mathrm{M}\), \(\left[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}^{-}\right] \approx 7.4 \times 10^{-6}\, \mathrm{M}\), \(\left[\mathrm{H}^{+}\right] \approx 7.4 \times 10^{-6}\, \mathrm{M}\), and \(\left[\mathrm{OH}^{-}\right] \approx 1.35 \times 10^{-9}\, \mathrm{M}\). The pH of the solution is approximately 5.13.

Step-by-step solution

Calculate initial concentration of benzoic acid

: We'll first calculate the initial concentration of benzoic acid \([\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}\mathrm{H}]\). To do this, we need to convert the grams of benzoic acid to moles and then divide by the volume of the solution. The molar mass of \(\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}\mathrm{H}\) is \(12 \times 7 + 6 \times 6 + 16 \times 2 = 122\) g/mol. Therefore, the initial concentration of \(\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}\mathrm{H}\) is: \[\frac{0.56 \, \mathrm{g}}{122\, \mathrm{g/mol}} \cdot \frac{1}{1.0\, \mathrm{L}} = 0.00459\, \mathrm{M}\]

Write the equilibrium expression

: Next, we'll write the ionization equilibrium for benzoic acid as follows: \[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}\mathrm{H} \longleftrightarrow \mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}^{-} + \mathrm{H}^{+}\] The equilibrium constant expression for this reaction is: \[K_a = \frac{[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}^{-}][\mathrm{H}^{+}]}{[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}\mathrm{H}]}\]

Assumptions and equation for \([\mathrm{H}^+]\)

: We'll assume that the initial concentration of benzoic acid is approximately equal to the equilibrium concentration. With this assumption, we can now write the equation for the concentrations of the species in terms of \(x\), where \(x\) is the concentration of \(\mathrm{H}^{+}\): \([\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}\mathrm{H}] = 0.00459 - x\) \([\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}^{-}] = x\) \([\mathrm{H}^{+}] = x\) Substituting these expressions into the equilibrium constant equation and solving for \(x\): \(K_a = \frac{x^2}{0.00459 - x}\) \(6.4 \times 10^{-5} = \frac{x^2}{0.00459 - x}\)

Calculate concentrations of benzoic acid, benzoate ion, and hydroxide ion

: Now, we solve for \(x\) to get the concentration of \(\mathrm{H}^{+}\): \(x^2 + 6.4 \times 10^{-5}x - 0.00459 \times 6.4 \times 10^{-5} = 0\) We can disregard the negative root and solve for the positive root to get \(x \approx 7.4 \times 10^{-6}\): \([\mathrm{H}^{+}] \approx 7.4 \times 10^{-6}\, \mathrm{M}\) Now, we can find the other concentrations: \([\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}\mathrm{H}] \approx 0.00459 - 7.4\times10^{-6} \approx 0.00458\, \mathrm{M}\) \([\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}^{-}] \approx 7.4 \times 10^{-6}\, \mathrm{M}\) Lastly, we'll calculate the concentration of hydroxide ions, \([\mathrm{OH}^{-}]\), using the ion product of water: \(K_w = [\mathrm{H}^{+}][\mathrm{OH}^{-}]\) \(1.0 \times 10^{-14} = (7.4 \times 10^{-6})[\mathrm{OH}^{-}]\) \([\mathrm{OH}^{-}] \approx 1.35 \times 10^{-9}\, \mathrm{M}\)

Calculate the pH of the solution

: Finally, we'll calculate the pH of the solution using the equation pH = \(- \log [\mathrm{H}^{+}]\): pH \(\approx -\log(7.4 \times 10^{-6}) \approx 5.13\) The concentrations of the species in the solution are approximately as follows: \(\left[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}\mathrm{H}\right] \approx 0.00458\, \mathrm{M}\), \(\left[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}^{-}\right] \approx 7.4 \times 10^{-6}\, \mathrm{M}\), \(\left[\mathrm{H}^{+}\right] \approx 7.4 \times 10^{-6}\, \mathrm{M}\), and \(\left[\mathrm{OH}^{-}\right] \approx 1.35 \times 10^{-9}\, \mathrm{M}\). The pH of the solution is approximately 5.13.

Key concepts

Benzoic Acid

Benzoic acid, with a chemical formula of \(\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}\mathrm{H}\), is a simple aromatic carboxylic acid. It's commonly used as a food preservative due to its antimicrobial properties. In an aqueous solution, benzoic acid can undergo ionization, acting as a weak acid, which partially dissociates to release hydrogen ions \((\mathrm{H}^{+})\). This process establishes an acid-base equilibrium that is crucial for understanding its behavior in solution.
When benzoic acid dissolves in water, not all molecules ionize—only a small fraction will. The chemical equation representing this ionization is:
\(\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}\mathrm{H} \rightleftharpoons \mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}^{-} + \mathrm{H}^{+}\)
This reaction highlights the establishment of equilibrium where both the ionized and unionized forms coexist. Understanding this concept is important for calculating concentrations of different species and predicting the properties of the solution under different conditions.

Ionization Constant

The ionization constant, often represented as \(K_a\), is a crucial parameter in acid-base chemistry. It quantifies the strength of an acid in solution. For benzoic acid, \(K_a\) is given as \(6.4 \times 10^{-5}\). This small value indicates that benzoic acid is a weak acid, which only partially ionizes in aqueous solution.
To express this equilibrium in terms of \(K_a\), we utilize the equilibrium constant expression:
\[K_a = \frac{[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}^{-}][\mathrm{H}^{+}]}{[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}\mathrm{H}]}\]
This setup allows us to solve for unknown concentrations, assuming initial known values. The smaller the \(K_a\), the lesser the degree of ionization and the greater the concentration of undissociated acid in the solution. Grasping the concept of \(K_a\) aids in making accurate calculations regarding the acidic properties of various solutions and predicting their equilibria behavior.

pH Calculation

Calculating the pH of a solution involves determining the concentration of \(\mathrm{H}^{+}\) ions and then applying the pH formula. The pH is a mathematical measure indicating the acidity or basicity of a solution. The lower the pH value, the higher the acidity; conversely, higher pH values indicate basicity. For benzoic acid, we calculated \([\mathrm{H}^{+}] \approx 7.4 \times 10^{-6} \, \mathrm{M}\).
The pH is calculated using:
\[\text{pH} = -\log[\mathrm{H}^{+}]\]
In this exercise, substituting the value results in:
pH \(\approx -\log(7.4 \times 10^{-6}) \approx 5.13\)
This pH reflects a slightly acidic solution, common for weak acids like benzoic acid. Understanding pH is critical for professions in chemistry, biology, and environmental science, as it directly influences reaction mechanisms and living organisms' biological processes.

Equilibrium Concentration

Equilibrium concentrations involve calculating how the concentrations of reactants and products are distributed once chemical equilibrium in a solution is established. In the case of benzoic acid, the reaction does not go to completion, leading to residual concentrations of both \([\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}\mathrm{H}]\) and its ionized form \([\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}^{-}]\) when equilibrium is reached.
The assumption that the initial concentration is close to the equilibrium concentration of undissociated acid simplifies our calculations, allowing us to express all equilibrium concentrations in terms of the change, \(x\), which represents \([\mathrm{H}^{+}]\).
Here, we calculated:
\([\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}\mathrm{H}] \approx 0.00458 \, \mathrm{M}\)
\([\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}^{-}] \approx 7.4 \times 10^{-6} \, \mathrm{M}\)
Equilibrium calculations are pivotal for predicting how concentration changes affect reaction dynamics, essential for laboratory and industrial applications.