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What is pH really measuring, and how do buffers resist change?

Bronsted-Lowry acids and bases, the pH scale and calculating pH of strong acids, the ionic product of water Kw, weak acids and Ka, pH curves and titrations, and buffer action.

A focused answer to AQA A-Level Chemistry 3.1.12, covering Bronsted-Lowry acids and bases, the pH scale, the ionic product of water Kw, weak acids and Ka, pH curves and indicators, and how buffers work.

Generated by Claude Opus 4.813 min answer

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  1. What this dot point is asking
  2. Bronsted-Lowry acids and bases
  3. pH and strong acids
  4. The ionic product of water, Kw
  5. Weak acids and Ka
  6. Titration curves and buffers
  7. Try this

What this dot point is asking

AQA wants you to define Bronsted-Lowry acids and bases, calculate pH for strong acids and use KwK_w, work with weak acids using KaK_a and pKapK_a, interpret pH titration curves and choose indicators, and explain how buffers resist pH change.

Bronsted-Lowry acids and bases

pH and strong acids

The ionic product of water, Kw

Water self-ionises slightly:

Weak acids and Ka

A weak acid only partially dissociates, so an equilibrium is set up:

Titration curves and buffers

A pH titration curve shows pH against volume added. The equivalence point lies in the steep, near-vertical section; the indicator chosen must change colour within that range (e.g. methyl orange for strong acid plus weak base, phenolphthalein for weak acid plus strong base).

The pH of a buffer is set by the ratio of acid to conjugate base, found by rearranging the KaK_a expression to [H+]=Ka×[HA][A−][\text{H}^+] = K_a \times \dfrac{[\text{HA}]}{[\text{A}^-]}. When the acid and salt concentrations are equal, [H+]=Ka[\text{H}^+] = K_a, so pH=pKa\text{pH} = pK_a, which is why a buffer is most effective near its pKapK_a. Buffers are central to biology and analysis: blood is buffered near pH 7.47.4 by the carbonic acid and hydrogencarbonate system, and small shifts would be life-threatening, so the body removes excess carbon dioxide through breathing to keep the ratio steady.

The choice of indicator in a titration follows the same logic as the curve. For a strong acid with a strong base the equivalence point is near pH 7 with a very long vertical section, so either methyl orange or phenolphthalein works. For a weak acid with a strong base the equivalence point is above pH 7, so phenolphthalein (changing around pH 8 to 10) is correct and methyl orange would change too early. For a strong acid with a weak base the equivalence point is below pH 7, so methyl orange is correct. A weak acid with a weak base gives no clear vertical section, so no indicator is suitable.

Try this

Q1. Define a Bronsted-Lowry base. [1 mark]

  • Cue. A proton (H+H^+) acceptor.

Q2. Calculate the pH of 0.010 mol dm−30.010 \text{ mol dm}^{-3} hydrochloric acid. [1 mark]

  • Cue. pH=−log⁡10(0.010)=2.0\text{pH} = -\log_{10}(0.010) = 2.0.

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20182 marksCalculate the pH of a 0.0500 mol dm−30.0500\ \text{mol dm}^{-3} solution of the strong acid hydrochloric acid.
Show worked answer →

Hydrochloric acid is a strong acid, so it fully dissociates: [H+]=0.0500 mol dm−3[\text{H}^+] = 0.0500\ \text{mol dm}^{-3}.

pH=−log⁡10[H+]=−log⁡10(0.0500)=1.30\text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(0.0500) = 1.30.

Markers reward recognising full dissociation (so [H+][\text{H}^+] equals the acid concentration) and the answer 1.301.30 to two decimal places.

AQA 20214 marksEthanoic acid is a weak acid with Ka=1.7×10−5 mol dm−3K_a = 1.7 \times 10^{-5}\ \text{mol dm}^{-3} at 298 K298\ \text{K}. Calculate the pH of a 0.100 mol dm−30.100\ \text{mol dm}^{-3} solution of ethanoic acid.
Show worked answer →

For a weak acid, Ka=[H+][A−][HA]K_a = \dfrac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}. Assume [H+]=[A−][\text{H}^+] = [\text{A}^-] and that [HA][\text{HA}] stays at its initial value 0.100 mol dm−30.100\ \text{mol dm}^{-3}.

Then Ka=[H+]2[HA]K_a = \dfrac{[\text{H}^+]^2}{[\text{HA}]}, so [H+]=Ka×[HA]=1.7×10−5×0.100=1.7×10−6=1.30×10−3 mol dm−3[\text{H}^+] = \sqrt{K_a \times [\text{HA}]} = \sqrt{1.7 \times 10^{-5} \times 0.100} = \sqrt{1.7 \times 10^{-6}} = 1.30 \times 10^{-3}\ \text{mol dm}^{-3}.

pH=−log⁡10(1.30×10−3)=2.89\text{pH} = -\log_{10}(1.30 \times 10^{-3}) = 2.89.

Markers reward the KaK_a expression, the two approximations, the hydrogen-ion concentration, and the pH to two decimal places.

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