Hydrochloric acid is a strong acid and ethanoic acid is a weak acid - AQA - A-Level Chemistry - Question 5 - 2018 - Paper 1

1. Calculate moles of HCl:

   Moles of HCl = concentration × volume = 0.100 mol dm⁻³ × 10.35 cm³ × (1 dm³ / 1000 cm³) = 0.001035 mol.

   2. Calculate moles of Ba(OH)₂:

   Moles of Ba(OH)₂ = 0.150 mol dm⁻³ × 25.0 cm³ × (1 dm³ / 1000 cm³) = 0.00375 mol.

   3. Moles of OH⁻ from Ba(OH)₂:

   Moles of OH⁻ = 2 × moles of Ba(OH)₂ = 2 × 0.00375 = 0.0075 mol.

   4. Calculate the excess moles of OH⁻ after neutralization:

   Moles of OH⁻ = 0.0075 - 0.001035 = 0.006465 mol.

   5. Calculate concentration of OH⁻ in final solution:

   Total volume = 10.35 cm³ + 25.0 cm³ = 35.35 cm³ = 0.03535 dm³

   Concentration of OH⁻ = 0.006465 mol / 0.03535 dm³ = 0.18286 mol dm⁻³.

   6. Calculate pOH:

   pOH = -log[OH⁻] = -log(0.18286) = 0.737.

   7. Calculate pH:

   pH = 14 - pOH = 14 - 0.737 = 13.263.

   8. Final answer:

   pH = 13.26 (to 2 decimal places).

Water is neutral at this temperature because it has equal concentrations of hydrogen ions [H⁺] and hydroxide ions [OH⁻], resulting in a pH of 7. The pH can vary with temperature due to changes in the ion product (K_w), but at 30 °C, pH neutrality corresponds to a pH of approximately 6.92.

The correct oxide is Na₂O.

The expression for the acid dissociation constant (K_a) for ethanoic acid is given by:

$K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]}$

1. Calculate moles of sodium ethanoate and ethanoic acid:

   Moles of sodium ethanoate = 0.025 mol. Moles of ethanoic acid = 0.700 mol dm⁻³ × 0.500 dm³ = 0.350 mol.

   2. After adding HCl, calculate moles:

   Moles of HCl added = 2.00 mol dm⁻³ × (5.00 cm³ / 1000 cm³) = 0.010 mol.

   3. Determine the new concentrations:

   Moles of ethanoic acid after reaction = 0.350 mol + 0.010 mol = 0.360 mol. Moles of sodium ethanoate after reaction = 0.025 mol - 0.010 mol = 0.015 mol.

   4. Calculate the total volume:

   Total volume = 500 cm³ + 5.00 cm³ = 505 cm³ = 0.505 dm³.

   5. New concentrations:

   [CH₃COOH] = 0.360 mol / 0.505 dm³ = 0.7129 mol dm⁻³. [CH₃COO⁻] = 0.015 mol / 0.505 dm³ = 0.0297 mol dm⁻³.

   6. Use the Henderson-Hasselbalch equation:

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

   7. Calculate pK_a:

   pK_a = -log(1.76 × 10⁻⁵) = 4.752.

   8. Substitute values:

   $pH = 4.752 + \log \frac{0.0297}{0.7129}$

   9. Final calculation:

   pH = 4.752 - 1.146 = 3.606.

   pH of the solution formed = 3.61 (to 2 decimal places).