Problem 26 - Chemistry: The Central Science in SI Units, Global Edition

Question

You are asked to prepare a \(\mathrm{pH}=3.00\) buffer starting from \(2.00 \mathrm{~L}\) of \(0.025 \mathrm{M}\) solution of benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)\) and any amount you need of sodium benzoate \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COONa}\right) .(\mathbf{a})\) What is the \(\mathrm{pH}\) of the benzoic acid solution prior to adding sodium benzoate? (b) How many grams of sodium benzoate should be added to prepare the buffer? Neglect the small volume change that occurs when the sodium benzoate is added.

Step-by-Step Solution

Answer

The pH of the 0.025 M benzoic acid solution prior to adding sodium benzoate is approximately 2.80. To prepare a pH 3.00 buffer, 17.8 grams of sodium benzoate should be added to the benzoic acid solution.

Step 1: Calculate the pH of 0.025 M benzoic acid solution

The acid dissociation constant, Ka, is given for benzoic acid: Ka = \(6.3 \times 10^{-5}\). Given the equation: pH = -log[H+] and Ka = \(\frac{[H^+] [A^-]}{[HA]}\), We can write the equation for Ka as: \(6.3 \times 10^{-5} = \frac{[H^+] [A^-]}{[HA]}\), where [HA] is the concentration of benzoic acid, which is 0.025 M, [A^-] is the concentration of the benzoate ion, and [H+] is the hydrogen ion concentration. Since the auto-ionization of water can be neglected initially, we can assume [A^-] = [H+]. Plugging these values into the equation: \(6.3 \times 10^{-5} = \frac{[H^+]^2}{0.025}\), Solve for [H+]: [H+] = \(1.58 \times 10^{-3} \mathrm{M}\). Now calculate the pH: pH = -log(\(1.58 \times 10^{-3}\)) ≈ 2.80.

Step 2: Use Henderson-Hasselbalch equation to find concentration ratio

The Henderson-Hasselbalch equation is given by: pH = pKa + log \(\frac{[A^-]}{[HA]}\), where pKa = -log(Ka). The desired pH of the buffer is 3.00 and we found the pH of benzoic acid solution as 2.80. Therefore: 3.00 = 4.20 + log \(\frac{[A^-]}{0.025}\), Solving for [A^-] (sodium benzoate concentration): [A^-] = \(0.0618 \mathrm{M}\).

Step 3: Calculate moles and grams of sodium benzoate.

To find the required moles of sodium benzoate to prepare the buffer, we can use the equation: Moles of sodium benzoate = [A^-] * volume In this case, the volume is 2.00 L: Moles of sodium benzoate = \(0.0618 \times 2.00 = 0.1236 \mathrm{mol}\). Now, we can find the mass of sodium benzoate to be added by using its molar mass (M = \(144.10 \mathrm{g/mol}\)): Mass of sodium benzoate = moles * molar mass Mass of sodium benzoate = \(0.1236 \times 144.10 = 17.8 \mathrm{g}\). So, 17.8 grams of sodium benzoate need to be added to the benzoic acid solution to prepare a pH 3.00 buffer.

Key Concepts

pH calculation

The concept of pH is crucial in understanding the acidity or basicity of a solution. It is a measure of the concentration of hydrogen ions \([H^+]\) in a solution, where pH is defined by the formula:
\[\text{pH} = -\log[H^+]\]This logarithmic scale means that a decrease in one pH unit reflects a tenfold increase in hydrogen ion concentration. In the case of a 0.025 M benzoic acid solution, the pH can be calculated using the dissociation constant (Ka) of benzoic acid, which is given as \(6.3 \times 10^{-5}\).
The expression for this equilibrium state is:

• Ka = \(\frac{[H^+][A^-]}{[HA]}\)

Given that \( [A^-] = [H^+]\) at equilibrium, you can substitute to find \[H^+]\]. Once determined, calculate the pH by substituting \[H^+]\] into the pH formula. This approach highlights the relationship between the concentration of ions and the pH of the solution. Standard pH calculation methods are vital for predicting the behavior of acids and their response in buffering solutions.

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is an invaluable tool for calculating the pH of a buffer solution. It relates the pH to the concentration of its acid and conjugate base. The equation is expressed as:
\[\text{pH} = \text{pKa} + \log \frac{[\text{A}^-]}{[\text{HA}]}\]Here, pKa is the negative logarithm of the acid dissociation constant (Ka), providing insight into the strength of the acid. To achieve a desired pH of 3.00, the equation helps determine the required ratio of the base (sodium benzoate) to the acid (benzoic acid).
Using the Henderson-Hasselbalch equation, you can adjust the buffer components to maintain a stable pH even when small amounts of acids or bases are added. In the exercise, calculation showed that a sodium benzoate concentration of \(0.0618 \, \text{M}\) was required to achieve this balance. This concept is essential in laboratory settings and for biochemical applications where precise pH control is critical.

Benzoic acid dissociation

Benzoic acid, represented chemically as C\(_6\)H\(_5\)COOH, is an organic acid that partially dissociates in water. The dissociation process involves benzoic acid releasing a hydrogen ion (\(H^+\)) and forming its conjugate base, the benzoate ion (C\(_6\)H\(_5\)COO\(^-\)). This can be expressed by the equilibrium reaction:

• C\(_6\)H\(_5\)COOH \( \rightleftharpoons \) C\(_6\)H\(_5\)COO\(^-\) + H\(^+\)

The extent of this dissociation is governed by the acid dissociation constant (Ka). For benzoic acid, \(Ka = 6.3 \times 10^{-5}\), indicating it is a weak acid with partial ionization in solution. Calculating the degree of dissociation helps in understanding the concentration of each species present, thereby influencing the resultant pH of the solution.
In practical scenarios, understanding the dissociation equilibrium of weak acids like benzoic acid is crucial, especially when designing buffers. These buffers rely on the equilibrium between the weak acid and its conjugate base to resist changes in pH upon addition of external acids or bases.