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AS & A-Level Chemistry 13 — Quantitative equilibria and kinetics

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Independent Deckloop A Level Chemistry study material aligned with Cambridge International 9701 (2025–2027). Deck 13 of 18: Quantitative equilibria and kinetics. Original explanations, worked applications and practice. Not affiliated with or endorsed by Cambridge International Education.

Chemistry EN A-Level
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pH Calculations and Buffer Solutions

Calculating the pH of strong acids requires assuming complete dissociation, so [H+][\mathrm{H^+}] equals the concentration of a monoprotic acid. Strong alkalis also dissociate fully; [H+][\mathrm{H^+}] is found by dividing KwK_\mathrm{w} by [OH][\mathrm{OH^-}]. For weak acids, partial dissociation means [H+][\mathrm{H^+}] must be calculated using KaK_\mathrm{a}, often with the approximation that equilibrium [HA][\mathrm{HA}] equals its initial concentration, yielding [H+]=Ka[HA][\mathrm{H^+}] = \sqrt{K_\mathrm{a} [\mathrm{HA}]}.

A buffer solution is a mixture that resists changes to pH upon the addition of small amounts of strong acid or strong alkali. Acidic buffers are commonly made from a weak acid and its conjugate base (e.g., sodium ethanoate). When acid is added, the conjugate base reacts with the incoming H+\mathrm{H^+}; when alkali is added, the weak acid neutralises the incoming OH\mathrm{OH^-}. An important blood buffer near pH 7.4 is the H2CO3/HCO3\mathrm{H_2CO_3} / \mathrm{HCO_3^-} buffer system, operating alongside physiological regulation. Buffer pH calculations use the KaK_\mathrm{a} expression directly, applying the standard assumption that initial concentrations of acid and salt reliably approximate their equilibrium values. The strong-acid and alkali shortcuts require their ion concentrations to dominate water auto-ionisation. For an isolated weak acid, check that the calculated dissociation is small compared with its initial concentration and that water contributes negligible hydrogen ions. After a buffer reacts with added strong acid or alkali, calculate the remaining acid and conjugatebase- \text{base} amounts by stoichiometry before applying the buffer expression.

Key points

  • Strong acids and bases fully dissociate; weak acids partially dissociate.
  • For a weak acid HA\mathrm{HA}, [H+]Ka[HA]initial[\mathrm{H^+}] \approx \sqrt{K_\mathrm{a} [\mathrm{HA}]_\mathrm{initial}}.
  • Buffers resist pH change when small amounts of H+\mathrm{H^+} or OH\mathrm{OH^-} are added.
  • A buffer requires both a weak acid and a substantial concentration of its conjugate base.
  • Human blood pH is buffered by the hydrogencarbonate ion, HCO3\mathrm{HCO_3^-}.

Worked example

Question

A buffer solution is prepared by mixing 40.0cm340.0 \, \mathrm{cm^3} of 0.200moldm30.200 \, \mathrm{mol \, dm^{-3}} propanoic acid (Ka=1.35×105moldm3K_\mathrm{a} = 1.35 \times 10^{-5} \, \mathrm{mol \, dm^{-3}}) with 60.0cm360.0 \, \mathrm{cm^3} of 0.100moldm30.100 \, \mathrm{mol \, dm^{-3}} sodium propanoate. Calculate the pH of the resulting solution at 298K298 \mathrm{K}.

Solution

1. Moles of propanoic acid (HA) = (40.0/1000)×0.200=0.00800mol(40.0 / 1000) \times 0.200 = 0.00800 \, \mathrm{mol}.

2. Moles of propanoate (A\mathrm{A^-}) = (60.0/1000)×0.100=0.00600mol(60.0 / 1000) \times 0.100 = 0.00600 \, \mathrm{mol}.

3. The total volume is 100.0cm3100.0 \, \mathrm{cm^3}, so [HA]=0.0800moldm3[\mathrm{HA}] = 0.0800 \, \mathrm{mol \, dm^{-3}} and [A]=0.0600moldm3[\mathrm{A^-}] = 0.0600 \, \mathrm{mol \, dm^{-3}}.

4. Using the given KaK_\mathrm{a}, [H+]=Ka×[HA][A]=1.35×105×0.08000.0600=1.80×105moldm3[\mathrm{H^+}] = K_\mathrm{a} \times \frac{[\mathrm{HA}]}{[\mathrm{A^-}]} = 1.35 \times 10^{-5} \times \frac{0.0800}{0.0600} = 1.80 \times 10^{-5} \, \mathrm{mol \, dm^{-3}}.

5. pH=log10(1.80×105)=4.74\mathrm{pH} = -\log_{10}(1.80 \times 10^{-5}) = 4.74.

pH=4.74\mathrm{pH} = 4.74

Common pitfalls

  • Calculating the pH of a buffer by simply averaging the pH of the component solutions. The correct approach uses the stoichiometric ratio of moles of weak acid to moles of conjugate base in the KaK_\mathrm{a} expression.
  • Forgetting to multiply the concentration of a Group 2 strong alkali (like Ba(OH)2\mathrm{Ba(OH)_2}) by 2 when finding [OH][\mathrm{OH^-}].

Prerequisites

  • Study Equilibria, acids and reaction rates first.
  • Study Thermodynamics and electrochemistry first.