Acids, Bases & pH

Which species is acting as a Brønsted–Lowry base in the forward reaction HCO₃⁻ + H₂O ⇌ H₂CO₃ + OH⁻?

• AH₂O
• BHCO₃⁻✓
• CH₂CO₃
• DOH⁻

Answer: A Brønsted–Lowry base is a proton (H⁺) acceptor. Here HCO₃⁻ accepts a proton from water to become H₂CO₃, so HCO₃⁻ is the base and water is the acid in this forward reaction.

Calculate the pH of 0.0500 mol dm⁻³ hydrochloric acid (a strong monoprotic acid). Use pH = −log₁₀[H⁺].

• ApH = 1.30✓
• BpH = 2.00
• CpH = 1.70
• DpH = 0.05

Answer: HCl is a strong acid that fully dissociates, so [H⁺] = 0.0500 mol dm⁻³. pH = −log₁₀(0.0500) = 1.30.

Calculate the pH of 0.0200 mol dm⁻³ sodium hydroxide at 298 K (Kw = 1.0 × 10⁻¹⁴ mol² dm⁻⁶).

• ApH = 1.70
• BpH = 2.00
• CpH = 12.30✓
• DpH = 12.00

Answer: NaOH is a strong base, so [OH⁻] = 0.0200 mol dm⁻³. [H⁺] = Kw/[OH⁻] = 1.0 × 10⁻¹⁴ / 0.0200 = 5.0 × 10⁻¹³ mol dm⁻³. pH = −log₁₀(5.0 × 10⁻¹³) = 12.30. (Equivalently pOH = 1.70, pH = 14 − 1.70.)

A weak monoprotic acid HA has Ka = 1.8 × 10⁻⁵ mol dm⁻³. Calculate the pH of a 0.100 mol dm⁻³ solution, assuming [H⁺] = √(Ka × c).

• ApH = 4.74
• BpH = 5.74
• CpH = 1.00
• DpH = 2.87✓

Answer: [H⁺] = √(Ka × c) = √(1.8 × 10⁻⁵ × 0.100) = √(1.8 × 10⁻⁶) = 1.34 × 10⁻³ mol dm⁻³. pH = −log₁₀(1.34 × 10⁻³) = 2.87. The 4.74 distractor is the pKa, not the pH.

The acid dissociation constant of ethanoic acid is Ka = 1.8 × 10⁻⁵ mol dm⁻³. What is its pKa?

• A5.00
• B1.80
• C4.74✓
• D9.26

Answer: pKa = −log₁₀(Ka) = −log₁₀(1.8 × 10⁻⁵) = 4.74. A smaller Ka (weaker acid) corresponds to a larger pKa.

Which mixture would act as an effective acidic buffer solution?

• AHydrochloric acid and sodium chloride
• BSodium hydroxide and sodium chloride
• CEthanoic acid and sodium ethanoate✓
• DHydrochloric acid and sodium hydroxide in equal moles

Answer: An acidic buffer needs a weak acid and the salt of that weak acid (its conjugate base). Ethanoic acid (weak acid) with sodium ethanoate provides a reservoir of both CH₃COOH and CH₃COO⁻ to resist pH change. A strong acid plus its salt (A) cannot buffer.

A buffer is made from a weak acid HA (pKa = 4.76) and its sodium salt. The concentration of the salt A⁻ is 0.20 mol dm⁻³ and that of the acid HA is 0.10 mol dm⁻³. Using pH = pKa + log₁₀([A⁻]/[HA]), calculate the pH.

• A5.06✓
• B4.46
• C4.76
• D9.52

Answer: pH = pKa + log₁₀([A⁻]/[HA]) = 4.76 + log₁₀(0.20/0.10) = 4.76 + log₁₀(2) = 4.76 + 0.30 = 5.06. Because there is more conjugate base than acid, the pH is above the pKa.