Propanoic acid (C₂H₅COOH) is a weak acid - AQA - A-Level Chemistry - Question 4 - 2020 - Paper 1

The acid dissociation constant (Kₐ) for propanoic acid can be expressed as:

$K_a = \frac{[H^+][C₂H₅COO^-]}{[C₂H₅COOH]}$

To find the pH of the diluted solution, we first calculate the concentration of the propanoic acid after dilution:

$C_1V_1 = C_2V_2$

Where:

• $C_1 = 0.500$ mol dm⁻³,
• $V_1 = 25.0$ cm³,
• $V_2 = 100.0$ cm³.

Thus, $C_2 = \frac{C_1V_1}{V_2} = \frac{0.500 imes 25.0}{100.0} = 0.125 \text{ mol dm}^{-3}$

Next, we use the expression for Kₐ: $K_a = \frac{[H^+][C₂H₅COO^-]}{[C₂H₅COOH]}$

Assuming that $ext{[H}^+ ext{]} = ext{[C₂H₅COO}^-]$ and $ext{[C₂H₅COOH]} = 0.125 - x$, where x is very small:

$K_a = \frac{x^2}{0.125 - x}$

Given Kₐ = 1.35 × 10⁻⁵, $1.35 imes 10^{-5} = \frac{x^2}{0.125}$

Solving for x yields: $x^2 = 1.35 imes 10^{-5} imes 0.125 = 1.6875 imes 10^{-6}$ $x = ext{sqrt}(1.6875 imes 10^{-6}) = 0.001297$

Now, to find the pH: $pH = -\log_{10}(0.001297) = 2.89$

Thus, the pH is 2.89.

For the buffer solution: Using the Henderson-Hasselbalch equation: $pH = pK_a + \log_{10}\left(\frac{[A^-]}{[HA]}\right)$ Where:

• pKₐ = -\log_{10}(1.35 × 10⁻⁵) ≈ 4.87
• [HA] = 0.250 mol dm⁻³
• [A⁻] = unknown concentration of sodium propanoate.

Given pH = 4.50: $4.50 = 4.87 + \log_{10}\left(\frac{[A^-]}{0.250}\right)$

Rearranging yields: $\log_{10}\left(\frac{[A^-]}{0.250}\right) = 4.50 - 4.87 = -0.37$ $\frac{[A^-]}{0.250} = 10^{-0.37}$ $[A^-] = 0.250 \times 10^{-0.37} = 0.250 \times 0.428 = 0.107$

The final volume of the buffer solution is 500 cm³, so: $n(A^-) = [A^-] \times V = 0.107 \times 0.500 = 0.0535 \text{ mol}$

Now calculate the mass of sodium propanoate: $m = n \times M = 0.0535 \times 106.08 = 5.67 ext{ g}$

Thus, x = 5.67 g.