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How do we measure and control the position of an equilibrium, including for weak acids and buffers?

The equilibrium constant K and its expression, Le Chatelier's principle, the dissociation of weak acids in terms of Ka and pKa, the calculation of pH for weak acids, the action of buffer solutions, and the selection of indicators for titrations.

An SQA Advanced Higher Chemistry answer on chemical equilibrium, covering the equilibrium constant K and its expression, Le Chatelier's principle, weak acids in terms of Ka and pKa, calculating the pH of a weak acid, the action and pH of buffer solutions, and selecting an indicator for a titration.

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  1. What this key area is asking
  2. The equilibrium constant
  3. Le Chatelier's principle
  4. Weak acids: Ka and pKa
  5. Buffer solutions
  6. Choosing an indicator
  7. Examples in context
  8. Try this

What this key area is asking

The SQA wants you to write and use the equilibrium constant KK, apply Le Chatelier's principle, treat weak acids using KaK_a and pKapK_a, calculate the pH of a weak acid, explain how buffer solutions resist pH change, and select a suitable indicator for a titration. The weak-acid pH calculation and the buffer explanation are reliable exam earners.

The equilibrium constant

A large value of KK (much greater than 1) means the equilibrium lies to the right, favouring products; a small value means it lies to the left. KK depends only on temperature: changing concentration or pressure shifts the position but does not change KK, whereas changing temperature changes KK itself.

Le Chatelier's principle

Weak acids: Ka and pKa

A strong acid is fully dissociated in water, whereas a weak acid only partly dissociates, setting up an equilibrium:

HAH++A\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-

The acid dissociation constant is Ka=[H+][A][HA]K_a = \dfrac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}, and pKa=log10KapK_a = -\log_{10} K_a. A smaller KaK_a (larger pKapK_a) means a weaker acid. For a weak acid, the approximation [H+]=Kac[\text{H}^+] = \sqrt{K_a c} holds because dissociation is slight, so the undissociated concentration is taken as the original concentration cc.

Buffer solutions

The pH of a buffer is set by the ratio of acid to conjugate base and by the acid's pKapK_a:

pH=pKa+log10[conjugate base][acid]\text{pH} = pK_a + \log_{10}\frac{[\text{conjugate base}]}{[\text{acid}]}

When the concentrations of acid and conjugate base are equal, the pH equals the pKapK_a of the weak acid.

Choosing an indicator

An indicator is itself a weak acid (written HIn\text{HIn}) whose dissociated and undissociated forms have different colours. It changes colour over a range roughly pKa±1pK_a \pm 1. For a titration, a suitable indicator is one whose colour-change range falls on the steep part of the pH curve at the equivalence point. For a strong acid with a weak base, methyl orange is suitable; for a weak acid with a strong base, phenolphthalein is suitable.

Examples in context

Equilibrium control runs through industry and biology. The Haber process for ammonia and the Contact process for sulfuric acid both balance temperature, pressure and catalyst to maximise yield and rate, a direct application of Le Chatelier and equilibrium. Buffers keep the pH of blood near 7.47.4 using a carbonic-acid-hydrogencarbonate system, and laboratory buffers calibrate pH meters and run enzyme assays. Weak-acid behaviour explains why ethanoic acid (vinegar) and carbonic acid (in fizzy drinks) are far less corrosive than strong acids of the same concentration, and indicator choice is exactly the skill needed when designing the titrations in the Researching Chemistry area.

Try this

Q1. Write the equilibrium constant expression for N2(g)+3H2(g)2NH3(g)\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g). [1 mark]

  • Cue. K=[NH3]2[N2][H2]3K = \dfrac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}.

Q2. State the effect of adding a catalyst on the position of an equilibrium. [1 mark]

  • Cue. No effect on the position; it only reaches equilibrium faster.

Q3. A weak acid has Ka=1.0×104K_a = 1.0 \times 10^{-4}. Calculate the pH of a 0.10 mol l10.10 \text{ mol l}^{-1} solution. [2 marks]

  • Cue. [H+]=1.0×104×0.10=3.16×103[\text{H}^+] = \sqrt{1.0 \times 10^{-4} \times 0.10} = 3.16 \times 10^{-3}, so pH=2.50\text{pH} = 2.50.

Exam-style practice questions

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

SQA AH 20193 marksEthanoic acid is a weak acid with Ka=1.7×105 mol l1K_a = 1.7 \times 10^{-5} \text{ mol l}^{-1}. Calculate the pH of a 0.10 mol l10.10 \text{ mol l}^{-1} solution of ethanoic acid.
Show worked answer →

Markers reward the approximation, the hydrogen ion concentration and the pH.

For a weak acid, [H+]=Ka×c[\text{H}^+] = \sqrt{K_a \times c}, using the approximation that the acid is only slightly dissociated so cc is unchanged.

[H+]=1.7×105×0.10=1.7×106=1.30×103 mol l1[\text{H}^+] = \sqrt{1.7 \times 10^{-5} \times 0.10} = \sqrt{1.7 \times 10^{-6}} = 1.30 \times 10^{-3} \text{ mol l}^{-1}

pH=log10[H+]=log10(1.30×103)=2.89\text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(1.30 \times 10^{-3}) = 2.89

A common loss is using the full strong-acid formula and forgetting that a weak acid is only partly dissociated.

SQA AH specimen2 marksA buffer solution contains ethanoic acid and sodium ethanoate. Explain how this buffer resists a change in pH when a small amount of acid is added.
Show worked answer →

The answer must name the reservoir species and the neutralising reaction.

A buffer contains a reservoir of a weak acid (ethanoic acid) and its conjugate base (the ethanoate ion from sodium ethanoate). When a small amount of acid (H+\text{H}^+) is added, the conjugate base reacts with it:

CH3COO+H+CH3COOH\text{CH}_3\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{COOH}

The added hydrogen ions are removed, so the pH barely changes. The large reservoirs of both species mean the ratio of acid to conjugate base, which sets the pH, is only slightly altered.

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