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 added: ext{Moles of HCl} = 0.100 ext{ mol dm}^{-3} imes (10.35 ext{ cm}^3 imes rac{1 ext{ dm}^3}{1000 ext{ cm}^3}) = 0.001035 ext{ mol}

2. Calculate moles of Ba(OH)₂ in solution: ext{Moles of Ba(OH)}_2 = 0.150 ext{ mol dm}^{-3} imes (25 ext{ cm}^3 imes rac{1 ext{ dm}^3}{1000 ext{ cm}^3}) = 0.00375 ext{ mol}

3. Since Ba(OH)₂ fully dissociates:

   • Each mole produces 2 moles of OH⁻: $ext{Moles of OH}^- = 2 imes 0.00375 = 0.0075 ext{ mol}$

4. Total volume of the solution: $ext{Total volume} = 10.35 ext{ cm}^3 + 25 ext{ cm}^3 = 35.35 ext{ cm}^3 = 0.03535 ext{ dm}^3$

5. Calculate concentration of OH⁻: [ ext{OH}^-] = rac{0.0075 ext{ mol}}{0.03535 ext{ dm}^3} = 0.212 ext{ mol dm}^{-3}

6. Calculate pOH: $ext{pOH} = - ext{log}[ ext{OH}^-] = - ext{log}(0.212) = 0.672$

7. Calculate pH: $ext{pH} = 14 - ext{pOH} = 14 - 0.672 = 13.328$

   Therefore, the pH to 2 decimal places is 13.33.

Water is neutral at 30 °C because its pH at this temperature equals 7, which is characteristic of pure water, indicating that the concentration of H⁺ ions is equal to the concentration of OH⁻ ions, maintaining a balanced condition.

SO₂ can react with water to form sulfurous acid (H₂SO₃), which is an acidic solution and would yield a pH around 2.

The expression for the acid dissociation constant ($K_a$) for ethanoic acid is given by: K_a = rac{[H^+][CH_3COO^-]}{[CH_3COOH]}

1. Calculate moles of ethanoic acid (to find its concentration): $0.700 ext{ mol dm}^{-3} imes 0.500 ext{ dm}^{3} = 0.350 ext{ mol}$

2. Calculate moles of sodium ethanoate: $0.025 ext{ mol}$

3. After adding HCl:

   • Moles of HCl added: rac{2.00 ext{ mol dm}^{-3} imes 0.0050 ext{ dm}^3}{1} = 0.010 ext{ mol}
   • Subtract this from sodium ethanoate:
   • Moles of CH₃COO⁻ after reaction: $0.025 - 0.010 = 0.015 ext{ mol}$

4. New total volume: $500 ext{ cm}^3 + 5.00 ext{ cm}^3 = 505.00 ext{ cm}^3 = 0.505 ext{ dm}^3$

5. Calculate concentrations:

   • For CH₃COOH: [ ext{CH}_3 ext{COOH}] = rac{0.350}{0.505} = 0.692 ext{ mol dm}^{-3}
   • For CH₃COO⁻: [ ext{CH}_3 ext{COO}^-] = rac{0.015}{0.505} = 0.030 ext{ mol dm}^{-3}

6. Use the Henderson-Hasselbalch equation: ext{pH} = ext{pK}_a + ext{log}rac{[ ext{CH}_3 ext{COO}^-]}{[ ext{CH}_3 ext{COOH}]}

   • Where $pK_a = - ext{log}(1.76 × 10^{-5}) = 4.75.$

   Therefore:

   $$$$

ightarrow ext{pH} ext{ calculated leads to approximately } 4.28.$$