Salam Academy
Sign inCreate account
Learn9701Paper 4 · A Level Structured QuestionsEquilibriapH, Ka and buffers
Equilibria · Subtopic 1 of 2

pH, Ka and Buffers

pH=−log⁡[H+]\mathrm{pH}=-\log[\mathrm{H^+}]. Strong acids: [H+]=[\mathrm{H^+}]= acid concentration (× number of H⁺). Water: Kw=[H+][OH−]=1.00×10−14K_w=[\mathrm{H^+}][\mathrm{OH^-}]=1.00\times10^{-14} at 298 K; for strong bases, [H+]=Kw/[OH−][\mathrm{H^+}]=K_w/[\mathrm{OH^-}]. Weak acids: Ka=[H+][A−][HA]K_a=\frac{[\mathrm{H^+}][\mathrm{A^-}]}{[\mathrm{HA}]}; assume [H+]=[A−][\mathrm{H^+}]=[\mathrm{A^-}], so [H+]=Ka[HA][\mathrm{H^+}]=\sqrt{K_a[\mathrm{HA}]}; pKa=−log⁡Ka\mathrm{p}K_a=-\log K_a. A buffer (weak acid + its salt) resists pH change: [H+]=Ka[HA][A−][\mathrm{H^+}]=K_a\frac{[\mathrm{HA}]}{[\mathrm{A^-}]}, or pH=pKa+log⁡[A−][HA]\mathrm{pH}=\mathrm{p}K_a+\log\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}. Titration curves show a steep section at equivalence; choose an indicator whose colour change lies within it. At half-neutralisation, pH = pKa.

Cambridge 9701 Paper 4Edexcel YCH11 WCH14AQA 7405 Paper 1≈ 55 minRevision 1 · updated 2026-10-30
Section 1 of 14

Learning objectives

  • Calculate pH of strong acids and bases using KwK_w.
  • Calculate pH of weak acids using KaK_a.
  • Explain buffer action and calculate buffer pH.
  • Interpret titration curves and choose indicators.
Pause and think

Blood stays at pH 7.4 even though your cells produce acid all day. How?

Section 2 of 14

Core idea

Start hereYour blood is a buffer

If blood pH drifted from 7.4 to 7.0, you'd be in a coma. Buffers keep it steady while your cells pump out acid all day.

The big idea

pH is −log[H⁺]. Strong acids dissociate fully; weak acids are governed by Ka. Buffers resist pH change because they hold large reserves of a weak acid and its conjugate base.

1. Weak acid pH

From Ka to pH
  1. Ka=[H+][A−][HA]K_a=\frac{[\mathrm{H^+}][\mathrm{A^-}]}{[\mathrm{HA}]}
    Definition.
  2. [H+]=[A−][\mathrm{H^+}]=[\mathrm{A^-}]
    Each HA gives one of each (ignoring water).
  3. [HA]≈c[\mathrm{HA}]\approx c
    Little dissociation.
  4. [H+]=Kac[\mathrm{H^+}]=\sqrt{K_a c}
    Rearrange.
Worked exampleBuffer pH

0.200 mol dm⁻³ CH₃COONa and 0.100 mol dm⁻³ CH₃COOH, pKaK_a = 4.76. pH?

Goal
pH.
Strategy
Henderson–Hasselbalch form.
  1. pH=4.76+log⁡0.2000.100\mathrm{pH}=4.76+\log\frac{0.200}{0.100}
    pH = pKa + log([A⁻]/[HA]).
  2. pH=4.76+0.30=5.06\mathrm{pH}=4.76+0.30=5.06
    log 2 = 0.30.
Answer5.06.
✓ CheckMore base than acid → pH above pKa ✓.
Choosing an indicator

The colour change must fall inside the vertical section.

Summary

  • pH = −log[H⁺]; Kw = 10⁻¹⁴.
  • Weak acid: √(Ka c).
  • Buffer: pKa + log([A⁻]/[HA]).
  • Indicator in the steep section.
Key words
Ka
Acid dissociation constant.
Buffer
Solution that resists pH change on adding small amounts of acid or base.
Equivalence point
Where acid and base have reacted in exact stoichiometric amounts.
HistorySørensen and beer. Søren Sørensen invented the pH scale in 1909 at the Carlsberg Laboratory while studying how acidity affects enzymes in brewing.
Fun factHydrangea colours. Hydrangeas turn blue in acidic soil and pink in alkaline soil: a garden-sized pH indicator.
Keep reading for free

Create a free account to read the rest of this lesson: key formulae, worked examples, explore the graph, common mistakes, exam tip, guided practice and more. Your answers are checked and your progress is saved.

Create free accountSign in