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How do we measure acidity and control pH?

pH, Ka and pKa, strong and weak acids and bases, the ionic product of water, buffer solutions and pH titration curves.

A focused answer to WJEC A-Level Chemistry Unit 3, covering pH and the ionic product of water, Ka and pKa for weak acids, strong and weak acid and base calculations, buffer solutions and pH titration curves.

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Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

WJEC wants you to calculate pH for strong and weak acids and bases, use KaK_a and pKapK_a, apply the ionic product of water KwK_w, explain and calculate buffers, and interpret pH titration curves.

The answer

pH and the ionic product of water

Strong versus weak

A strong acid is fully dissociated, so [H+][\text{H}^+] equals the acid concentration. A weak acid only partly dissociates, so use Ka=[H+][A][HA]K_a = \dfrac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} and pKa=log10KapK_a = -\log_{10}K_a.

Buffers and titration curves

Working with weak acids

A weak acid such as ethanoic acid only partly dissociates, so its [H+][\text{H}^+] is much smaller than its concentration. The standard approximation is [H+]=Ka×c[\text{H}^+] = \sqrt{K_a \times c}, valid because dissociation is small enough that [HA][\text{HA}] is roughly the original concentration and [H+]=[A][\text{H}^+] = [\text{A}^-]. This is why a 0.10.1 mol dm3^{-3} solution of a weak acid has a pH around 33, while the same concentration of a strong acid has a pH of 11. The smaller the KaK_a (the larger the pKapK_a), the weaker the acid and the higher the pH at a given concentration.

Reading a titration curve

A pH titration curve has four features to identify: the starting pH (set by the acid strength), the gentle buffer region (for a weak acid plus base), the steep vertical jump at the equivalence point, and the plateau toward the pH of excess titrant. The midpoint of the buffer region gives pH=pKa\text{pH} = pK_a, a quick way to read off the acid's strength. The indicator must change colour entirely within the steep region: methyl orange for strong-acid against weak-base titrations, phenolphthalein for weak-acid against strong-base.

Examples in context

Blood as a buffer. The hydrogencarbonate buffer keeps blood pH near 7.47.4; a shift of even 0.10.1 unit is dangerous, illustrating the biological importance of buffering. Indicator choice in titrations. A strong acid with a weak base has its equivalence point below pH 77, so methyl orange (not phenolphthalein) is the correct indicator, read directly from the titration curve.

Try this

Q1. Calculate the pH of 0.01000.0100 mol dm3^{-3} hydrochloric acid. [1 mark]

  • Cue. pH=log10(0.0100)=2.00\text{pH} = -\log_{10}(0.0100) = 2.00.

Q2. State the expression for the acid dissociation constant KaK_a of a weak acid HA. [1 mark]

  • Cue. Ka=[H+][A][HA]K_a = \dfrac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}.

Q3. Name the two components needed to make an acidic buffer. [1 mark]

  • Cue. A weak acid and its conjugate base (its salt).

Q4. State the pH at the midpoint of the buffer region of a weak-acid titration. [1 mark]

  • Cue. pH=pKa\text{pH} = pK_a.

Q5. Calculate the pH of a 0.1000.100 mol dm3^{-3} weak acid with Ka=1.0×105K_a = 1.0 \times 10^{-5} mol dm3^{-3}. [2 marks]

  • Cue. [H+]=1.0×105×0.100=1.0×103[\text{H}^+] = \sqrt{1.0 \times 10^{-5} \times 0.100} = 1.0 \times 10^{-3}; pH=3.00\text{pH} = 3.00.

Exam-style practice questions

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

WJEC 20204 marksCalculate the pH of a 0.150 mol dm30.150\ \text{mol dm}^{-3} solution of ethanoic acid. The acid dissociation constant KaK_a is 1.74×105 mol dm31.74 \times 10^{-5}\ \text{mol dm}^{-3}.
Show worked answer →

For a weak acid, Ka=[H+]2[HA]K_a = \dfrac{[\text{H}^+]^2}{[\text{HA}]}, assuming [H+]=[A][\text{H}^+] = [\text{A}^-] and that dissociation is small.

[H+]=Ka×[HA]=1.74×105×0.150=2.61×106=1.616×103[\text{H}^+] = \sqrt{K_a \times [\text{HA}]} = \sqrt{1.74 \times 10^{-5} \times 0.150} = \sqrt{2.61 \times 10^{-6}} = 1.616 \times 10^{-3} mol dm-3.

pH=log10[H+]=log10(1.616×103)=2.79\text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(1.616 \times 10^{-3}) = 2.79.

Markers reward the weak-acid approximation, the square root for [H+][\text{H}^+], and the pH to two decimal places.

WJEC 20193 marksExplain how a buffer made from ethanoic acid and sodium ethanoate resists a change in pH when a small amount of acid is added.
Show worked answer →

The buffer contains a reservoir of the weak acid CH3COOH\text{CH}_3\text{COOH} and its conjugate base CH3COO\text{CH}_3\text{COO}^-.

When acid (H+\text{H}^+) is added, the conjugate base removes it: CH3COO+H+CH3COOH\text{CH}_3\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{COOH}.

Because the added H+\text{H}^+ is converted to undissociated acid, the free [H+][\text{H}^+] and so the pH change very little.

Markers reward identifying the conjugate base reservoir, the equation removing added H+\text{H}^+, and the small resulting pH change.

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