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

The expression for the acid dissociation constant (Kₐ) for propanoic acid is given by:

$K_a = \frac{[H^+][C_2H_5COO^-]}{[C_2H_5COOH]}$

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

1. Calculate the moles of propanoic acid:

$\text{Moles} = 0.500 \times 0.025 = 0.0125 \text{ mol}$

2. Find the new concentration in 100.0 cm³:

$[C_2H_5COOH] = \frac{0.0125}{0.1} = 0.125 \, \text{mol dm}^{-3}$

3. Set up the equilibrium expression where $x$ is the concentration of $H^+$ produced:

For the weak acid dissociation:

$C_2H_5COOH \rightleftharpoons H^+ + C_2H_5COO^-$

At equilibrium:

• $[C_2H_5COOH] = 0.125 - x$
• $[H^+] = x$
• $[C_2H_5COO^-] = x$

Apply the equilibrium constant expression:

$K_a = \frac{x \cdot x}{0.125 - x} = 1.35 \times 10^{-5}$

Assuming $x$ is small compared to 0.125, we simplify to:

$1.35 \times 10^{-5} = \frac{x^2}{0.125}$

Solving for $x$ gives:

$x^2 = 1.35 \times 10^{-5} \cdot 0.125$

$x^2 = 1.6875 \times 10^{-6}$

Taking the square root:

$x \approx 0.0013$

4. Calculate the pH:

$pH = -\log(0.0013) \approx 2.89$

The buffer solution is made with a final concentration of propanoic acid of 0.250 mol dm⁻³ in a total volume of 500 cm³.

1. Calculate moles of propanoic acid:

$\text{Moles of propanoic acid} = 0.250 \, \text{mol dm}^{-3} \times 0.500 \, \text{dm}^3 = 0.125 \, \text{mol}$

2. Using the Henderson-Hasselbalch equation:

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

Where $pK_a = -\log(1.35 \times 10^{-5}) \approx 4.87$.

3. Rearranging gives:

$4.50 = 4.87 + \log\left( \frac{x}{0.125} \right)$

4. Solving for $(x)$:

$\log\left( \frac{x}{0.125} \right) = 4.50 - 4.87 \Rightarrow \log\left( \frac{x}{0.125} \right) = -0.37$

5. Converting from logarithm:

$\frac{x}{0.125} = 10^{-0.37} \Rightarrow x = 0.125 \times 0.426 = 0.05325 \, \text{mol}$

6. Calculate mass:

Molar mass of sodium propanoate (C₂H₅COONa) is approximately 110.08 g/mol:

$\text{Mass} = 0.05325 \, \text{mol} \times 110.08 \, \text{g/mol} \approx 5.85 \, \text{g}$