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 HCl = concentration × volume = 0.100 mol dm⁻³ × (10.35 cm³ ÷ 1000) = 0.001035 mol.

2. Calculate moles of Ba(OH)₂:

   Moles Ba(OH)₂ = concentration × volume = 0.150 mol dm⁻³ × (25.0 cm³ ÷ 1000) = 0.00375 mol.

3. Calculate moles of OH⁻ produced:

   1 mole of Ba(OH)₂ produces 2 moles of OH⁻, so moles OH⁻ = 2 × 0.00375 mol = 0.0075 mol.

4. Write the total moles and net reaction. The HCl and OH⁻ neutralize each other:

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

5. Calculate [OH⁻]:

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

   [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 + pOH = 14, so pH = 14 - 0.737 = 13.263, rounded to 13.26.

Water is neutral at this temperature because the concentration of hydrogen ions [H⁺] equals the concentration of hydroxide ions [OH⁻], resulting in a neutral pH of 7.

The oxide that could react with water to form a solution with pH = 2 is SO₂.

The expression for the acid dissociation constant (Kₐ) for ethanoic acid is:

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

1. Calculate initial moles of CH₃COOH and CH₃COO⁻:

   Moles CH₃COOH = 0.700 mol dm⁻³ × (500 cm³ ÷ 1000) = 0.35 mol.

   Moles CH₃COO⁻ = 0.025 mol.

2. Use the Henderson-Hasselbalch equation:

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

   Where: pKₐ = -log(1.76 × 10⁻⁵) ≈ 4.753.

3. Calculate the concentrations after addition of HCl:

   Moles of H⁺ from HCl = 2.00 mol dm⁻³ × (5.00 cm³ ÷ 1000) = 0.01 mol.

   Total volume after adding HCl = 500 cm³ + 5 cm³ = 505 cm³ = 0.505 dm³.

   New concentrations:

   [CH₃COOH] = (0.35 mol - 0.01 mol) / 0.505 dm³ = 0.6604 mol dm⁻³

   [CH₃COO⁻] = (0.025 mol) / 0.505 dm³ = 0.0495 mol dm⁻³.

4. Calculate pH:

   $pH = 4.753 + log \frac{0.0495}{0.6604} \approx 4.28$.