Question

A 1.0 L solution is prepared by dissolving 2.0 g of benzoic acid and 4.0 g of sodium benzoate in water. The pH of the resulting solution is _______. (rounded off to one decimal place)
Given: Molar mass of benzoic acid is 122 g mol⁻¹ 
Molar mass of sodium benzoate is 144 g mol⁻¹ 
p𝐾_a of benzoic acid is 4.2

Answer 1

Correct Answer: 4.4 - 4.5

This is a buffer solution formed by benzoic acid (a weak acid) and sodium benzoate (its conjugate base). To calculate the pH of this buffer, we use the Henderson-Hasselbalch equation:

pH = \( pK_a + \log \left( \frac{[A^-]}{[HA]} \right) \)

Where:

• pK_a = 4.2 (given)
• [A^-] is the concentration of the conjugate base (benzoate ion)
• [HA] is the concentration of the weak acid (benzoic acid)

First, calculate the moles of benzoic acid and sodium benzoate:

Moles of benzoic acid = \( \frac{2.0 \, \text{g}}{122 \, \text{g/mol}} = 0.01639 \, \text{mol} \)

Moles of sodium benzoate = \( \frac{4.0 \, \text{g}}{144 \, \text{g/mol}} = 0.02778 \, \text{mol} \)

The total volume of the solution is 1.0 L, so the concentrations are:

• [HA] = \( \frac{0.01639 \, \text{mol}}{1.0 \, \text{L}} = 0.01639 \, \text{M} \)
• [A^-] = \( \frac{0.02778 \, \text{mol}}{1.0 \, \text{L}} = 0.02778 \, \text{M} \)

Now, use the Henderson-Hasselbalch equation:

pH =\( 4.2 + \log \left( \frac{0.02778}{0.01639} \right)\)

pH \(= 4.2 + \log(1.695) = 4.2 + 0.230\)

pH = 4.4

Thus, the pH of the solution is 4.4.