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H2 Chemistry · JC 1 and 2

Acid and Base Equilibria

Acid and base equilibria apply the ideas of chemical equilibrium to proton transfer in water. In H2 Chemistry you move beyond strong and weak as labels to calculating pH from Ka, Kb and Kw, explaining how buffers resist pH change and interpreting titration curves to choose indicators. The topic is calculation-heavy, but every calculation rests on a handful of relationships, and the same skills reappear in organic chemistry when comparing the acidity of acids, phenols and alcohols.

What the syllabus expects

  • Use the Brønsted-Lowry theory to identify acids, bases and conjugate acid-base pairs
  • Explain the difference between strong and weak acids and bases in terms of Ka, Kb and pKa
  • Calculate the pH of solutions of strong acids, strong bases, weak acids and weak bases
  • Use Kw = [H⁺][OH⁻] and Ka × Kb = Kw for a conjugate pair
  • Explain how a buffer solution controls pH and calculate the pH of a buffer
  • Interpret pH titration curves and select a suitable indicator
  • Explain why aqueous solutions of some salts are acidic or alkaline

Brønsted-Lowry acids and conjugate pairs

A Brønsted-Lowry acid is a proton (H⁺) donor and a base is a proton acceptor. When ethanoic acid dissolves in water, CH₃COOH + H₂O ⇌ CH₃COO⁻ + H₃O⁺, the acid donates a proton to water. CH₃COOH and CH₃COO⁻ form a conjugate acid-base pair, as do H₃O⁺ and H₂O. The pairs differ by one proton. Strong acids such as HCl ionise completely in water, while weak acids such as ethanoic acid ionise only partially, so the equilibrium lies well to the left. The stronger the acid, the weaker its conjugate base. Water itself can act as an acid or a base, and it ionises slightly: H₂O ⇌ H⁺ + OH⁻, with Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶ at 25 °C.

pH of strong acids and bases

pH = −lg[H⁺], so each unit of pH is a tenfold change in [H⁺]. For a strong monoprotic acid, [H⁺] equals the acid concentration, so 0.010 mol dm⁻³ HCl has pH = −lg(0.010) = 2.00. For a strong base, find [OH⁻] first, then use pOH = −lg[OH⁻] and pH + pOH = 14.00 at 25 °C. So 0.010 mol dm⁻³ NaOH has pOH = 2.00 and pH = 12.00. For a base that releases two hydroxide ions per formula unit, such as barium hydroxide, [OH⁻] is twice the concentration of the base. The ionisation of water is endothermic, so Kw increases with temperature. Pure water at a higher temperature has a pH below 7 but is still neutral, because [H⁺] still equals [OH⁻].

Weak acids and weak bases

For a weak acid HA ⇌ H⁺ + A⁻, the acid dissociation constant is Ka = [H⁺][A⁻] ÷ [HA], and pKa = −lg Ka. A larger Ka, or smaller pKa, means a stronger acid. If no other source of H⁺ or A⁻ is present, [H⁺] = [A⁻], and because the acid is only slightly ionised, [HA] at equilibrium is approximately its initial concentration c. So [H⁺]² = Ka × c, giving [H⁺] = √(Ka × c). For 0.10 mol dm⁻³ ethanoic acid with Ka = 1.8 × 10⁻⁵ mol dm⁻³, [H⁺] = √(1.8 × 10⁻⁶) = 1.34 × 10⁻³ mol dm⁻³ and pH = 2.87. Weak bases work the same way with Kb: [OH⁻] = √(Kb × c). For a conjugate pair, Ka × Kb = Kw, so you can find Kb for the ethanoate ion from the Ka of ethanoic acid.

Buffer solutions

A buffer solution resists changes in pH when small amounts of acid or base are added. An acidic buffer contains a weak acid and its conjugate base in significant amounts, such as ethanoic acid with sodium ethanoate. A basic buffer contains a weak base and its conjugate acid, such as ammonia with ammonium chloride. In the ethanoate buffer, added H⁺ is removed by the large reservoir of conjugate base, CH₃COO⁻ + H⁺ gives CH₃COOH, and added OH⁻ is removed by the weak acid, CH₃COOH + OH⁻ gives CH₃COO⁻ + H₂O. The ratio [CH₃COO⁻] ÷ [CH₃COOH] changes only slightly, so the pH changes only slightly. Rearranging Ka gives [H⁺] = Ka × [HA] ÷ [A⁻], or pH = pKa + lg([A⁻] ÷ [HA]). When the two concentrations are equal, pH = pKa, and the buffer is equally good at resisting added acid and added base.

Titration curves, indicators and salt hydrolysis

A pH titration curve shows how pH changes as a base is added to an acid, or the reverse. For a strong acid with a strong base, the pH at equivalence is 7 and there is a long steep section, roughly pH 3 to 11, so most indicators work. For a weak acid with a strong base, the equivalence point is above 7 because the conjugate base, such as CH₃COO⁻, reacts with water to give OH⁻. Phenolphthalein, which changes over about pH 8.3 to 10, is suitable. For a strong acid with a weak base, the equivalence point is below 7 because NH₄⁺ is acidic, so methyl orange, about pH 3.1 to 4.4, is suitable. Choose an indicator whose working range lies within the steep section. At half-equivalence in a weak acid titration, [HA] = [A⁻], so pH = pKa, which gives a way to find pKa from a curve. The same reasoning explains salt solutions: NaCl is neutral, CH₃COONa is alkaline and NH₄Cl is acidic.

Worked example

25.0 cm³ of 0.100 mol dm⁻³ ethanoic acid (pKa = 4.76) is titrated with 0.100 mol dm⁻³ sodium hydroxide at 25 °C. Calculate the pH (a) at the start, (b) after 12.5 cm³ of NaOH, (c) at the equivalence point, and suggest a suitable indicator.

  1. Ka = 10^(−4.76) = 1.74 × 10⁻⁵ mol dm⁻³.
  2. (a) [H⁺] = √(Ka × c) = √(1.74 × 10⁻⁵ × 0.100) = √(1.74 × 10⁻⁶) = 1.32 × 10⁻³ mol dm⁻³, so pH = 2.88.
  3. (b) Moles of acid = 0.00250 and moles of NaOH = 0.00125, so half the acid has become CH₃COO⁻. [HA] = [A⁻], so pH = pKa = 4.76.
  4. (c) At equivalence, 0.00250 mol of CH₃COO⁻ is in 50.0 cm³, so [CH₃COO⁻] = 0.00250 ÷ 0.0500 = 0.0500 mol dm⁻³.
  5. Kb of CH₃COO⁻ = Kw ÷ Ka = 1.0 × 10⁻¹⁴ ÷ 1.74 × 10⁻⁵ = 5.75 × 10⁻¹⁰ mol dm⁻³.
  6. [OH⁻] = √(5.75 × 10⁻¹⁰ × 0.0500) = √(2.88 × 10⁻¹¹) = 5.36 × 10⁻⁶ mol dm⁻³, so pOH = 5.27 and pH = 14.00 − 5.27 = 8.73.
  7. The equivalence point is at pH 8.73, so choose phenolphthalein, whose range of about 8.3 to 10 lies in the steep section.

Answer: (a) pH 2.88. (b) pH 4.76. (c) pH 8.73. Phenolphthalein is a suitable indicator.

Common mistakes

  • Using [H⁺] = √(Ka × c) for a strong acid or for a buffer. It only applies to a weak acid on its own in water.
  • Forgetting to convert from [OH⁻] to pH for a base. Find pOH first, then pH = 14.00 − pOH at 25 °C.
  • Forgetting the dilution at the equivalence point. The conjugate base is now in the combined volume of both solutions.
  • Assuming every equivalence point is at pH 7. Only strong acid with strong base gives pH 7.
  • Explaining buffer action without equations. Show how the conjugate base removes added H⁺ and how the weak acid removes added OH⁻.
  • Saying pure water at 50 °C is acidic because its pH is below 7. It is still neutral because [H⁺] = [OH⁻].

Exam tips

  • Before any pH calculation, classify the solution first: strong acid, strong base, weak acid, weak base, buffer or salt. Each has its own method, and naming it prevents using the wrong formula.
  • For a mixture of a weak acid and a strong base, work in moles first to see what is left after reaction, then decide whether you have a buffer, a salt solution or excess base.
  • When explaining an indicator choice, quote both the pH at equivalence and the indicator's working range.

Practice questions

1. Easy: Calculate the pH of 0.020 mol dm⁻³ barium hydroxide, Ba(OH)₂, at 25 °C, assuming it is fully ionised.

Show answer

pH 12.60. Each Ba(OH)₂ gives two OH⁻, so [OH⁻] = 0.040 mol dm⁻³. pOH = −lg(0.040) = 1.40, so pH = 14.00 − 1.40 = 12.60.

2. Medium: Calculate the pH of 0.20 mol dm⁻³ aqueous ammonia at 25 °C. Kb = 1.8 × 10⁻⁵ mol dm⁻³.

Show answer

pH 11.28. [OH⁻] = √(Kb × c) = √(1.8 × 10⁻⁵ × 0.20) = √(3.6 × 10⁻⁶) = 1.90 × 10⁻³ mol dm⁻³. pOH = 2.72, so pH = 14.00 − 2.72 = 11.28.

3. Harder: 25.0 cm³ of 0.10 mol dm⁻³ NaOH is added to 50.0 cm³ of 0.20 mol dm⁻³ ethanoic acid (pKa = 4.76). Calculate the pH of the resulting buffer.

Show answer

pH 4.28. Moles of acid = 0.0100 and moles of OH⁻ = 0.00250. After reaction, CH₃COO⁻ = 0.00250 mol and CH₃COOH = 0.0100 − 0.00250 = 0.00750 mol. Both are in the same volume, so pH = pKa + lg(0.00250 ÷ 0.00750) = 4.76 + lg(1/3) = 4.76 − 0.48 = 4.28.

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