What are buffer solutions?

To calculate a buffer pH, rearrange the K_a expression to [H^+] = K_a × [HA][A^-]. At the half-equivalence point of a weak-acid titration, [HA] = [A^-], so [H^+] = K_a and pH = pK_a, a useful way to find K_a from a curve.

Calculate the pH of 0.010\ mol dm^-3 hydrochloric acid. [1 mark]

Pearson Edexcel Edexcel 2019-style practice: Calculate the pH of 0.150\ mol dm^-3 ethanoic acid. (K_a = 1.74 × 10^-5\ mol dm^-3.) State one assumption you make.

Use the weak-acid approximation and take logs. For a weak acid, K_a = [H^+]^2[HA] assuming [H^+] = [A^-] and that dissociation is small so [HA] stays at 0.150 (1, assumption). [H^+] = K_a × [HA] = 1.74 × 10^-5 × 0.150 = 2.61 × 10^-6 = 1.62 × 10^-3\ mol dm^-3 (2). pH = -_10(1.62 × 10^-3) = 2.79 (1). Markers reward the correct rearrangement, value of [H^+], and pH to 2 decimal places, plus a stated assumption.

Pearson Edexcel Edexcel 2021-style practice: A buffer is made by dissolving 0.20\ mol of sodium ethanoate in 1.0\ dm^3 of 0.10\ mol dm^-3 ethanoic acid. (K_a = 1.74 × 10^-5.) (a) Calculate the pH. (b) Explain how the buffer resists a rise in pH when a little alkali is added.

Use the buffer expression, then explain the equilibrium response. (a) K_a = [H^+][A^-][HA], so [H^+] = K_a [HA][A^-] = 1.74 × 10^-5 × 0.100.20 = 8.7 × 10^-6\ mol dm^-3 (2). pH = -_10(8.7 × 10^-6) = 5.06 (1). (b) Added OH^- reacts with the weak acid CH_3COOH (the reservoir): CH_3COOH + OH^- → CH_3COO^- + H_2O (1), so [H^+] and hence pH change only slightly (1).

EnglandChemistrySyllabus dot point

How do we measure and control the acidity of a solution?

The Bronsted-Lowry theory, the pH scale, strong and weak acids, Ka and Kw, titration curves, indicator choice and the action of buffer solutions.

An Edexcel 9CH0 Topic 17 answer covering the Bronsted-Lowry theory, the pH scale, strong and weak acids, Ka and Kw, titration curves, indicators and buffers.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-02

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

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What this topic is asking

Edexcel Topic 17 wants you to apply the Bronsted-Lowry theory, calculate pH for strong and weak acids and bases using $K_a$ , $K_w$ and $\text{p}K_a$ , interpret titration curves, choose suitable indicators, and explain and calculate the action of buffer solutions.

The answer

Bronsted-Lowry theory and pH

Strong and weak acids

For a strong acid such as $\text{HCl}$ (full dissociation), $[\text{H}^+]$ equals the acid concentration, so the pH follows directly. For a weak acid (partial dissociation), use the acid dissociation constant:

The ionic product of water is $K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14}\ \text{mol}^2\,\text{dm}^{-6}$ at $298\ \text{K}$ . For a strong base, find $[\text{OH}^-]$ from the concentration, then use $[\text{H}^+] = K_w / [\text{OH}^-]$ to get the pH.

Titration curves and indicators

Buffer solutions

To calculate a buffer pH, rearrange the $K_a$ expression to $[\text{H}^+] = K_a \times \dfrac{[\text{HA}]}{[\text{A}^-]}$ . At the half-equivalence point of a weak-acid titration, $[\text{HA}] = [\text{A}^-]$ , so $[\text{H}^+] = K_a$ and $\text{pH} = \text{p}K_a$ , a useful way to find $K_a$ from a curve.

Calculating the pH change when alkali is added to a buffer

A buffer contains $0.10\ \text{mol}$ ethanoic acid and $0.10\ \text{mol}$ sodium ethanoate in $1.0\ \text{dm}^3$ . ( $K_a = 1.74 \times 10^{-5}$ .) Find the pH before and after adding $0.01\ \text{mol}$ of $\text{NaOH}$ .

step 1: pH of the original buffer

$[\text{H}^+] = K_a \dfrac{[\text{HA}]}{[\text{A}^-]} = 1.74 \times 10^{-5} \times \dfrac{0.10}{0.10} = 1.74 \times 10^{-5}$ . $\text{pH} = -\log_{10}(1.74 \times 10^{-5}) = 4.76$ .

step 2: React the added alkali

$\text{OH}^-$ converts $0.01\ \text{mol}$ acid into ethanoate: acid $= 0.10 - 0.01 = 0.09\ \text{mol}$ ; ethanoate $= 0.10 + 0.01 = 0.11\ \text{mol}$ .

step 3: Recalculate $[\text{H}^+]$

$[\text{H}^+] = 1.74 \times 10^{-5} \times \dfrac{0.09}{0.11} = 1.42 \times 10^{-5}$ .

step 4: New pH and conclusion

$\text{pH} = -\log_{10}(1.42 \times 10^{-5}) = 4.85$ . The pH rose by only $0.09$ , showing the buffer resists change; the same alkali added to pure water would have changed the pH far more.

Examples in context

Example 1. Blood as a buffer. Human blood is buffered at about pH $7.4$ by the carbonic acid / hydrogencarbonate system, $\text{H}_2\text{CO}_3 \rightleftharpoons \text{H}^+ + \text{HCO}_3^-$ . Added acid is removed by $\text{HCO}_3^-$ and added base by $\text{H}_2\text{CO}_3$ , while the lungs and kidneys adjust the components. This is exactly the acidic-buffer action of Topic 17, and a deviation of even $0.4$ pH units can be life-threatening, showing why buffering is biologically vital.

Example 2. Choosing the indicator for an ethanoic acid titration. When standardising sodium hydroxide against ethanoic acid (a weak acid with a strong base), the equivalence point lies above pH $7$ because the salt formed is slightly alkaline. Phenolphthalein (range $8.3$ to $10$ ) changes colour within the steep section, giving a sharp end point, whereas methyl orange would change far too early. Picking the indicator to match the curve is a routine but heavily marked skill.

Try this

Q1. Calculate the pH of $0.010\ \text{mol dm}^{-3}$ hydrochloric acid. [1 mark]

Cue. Strong acid, $[H^+] = 0.010$ ; $\text{pH} = -\log_{10}(0.010) = 2.00$ .

Q2. Explain how an acidic buffer resists a rise in pH when a small amount of alkali is added. [2 marks]

Cue. The weak acid reacts with the added $OH^-$ , removing it and keeping $[H^+]$ approximately constant.

Exam-style practice questions

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

Edexcel 20194 marksCalculate the pH of

0.150\ \text{mol dm}^{-3}

ethanoic acid. (

K_a = 1.74 \times 10^{-5}\ \text{mol dm}^{-3}

.) State one assumption you make.

Show worked answer →

Use the weak-acid approximation and take logs.

For a weak acid, $K_a = \dfrac{[\text{H}^+]^2}{[\text{HA}]}$ assuming $[\text{H}^+] = [\text{A}^-]$ and that dissociation is small so $[\text{HA}]$ stays at $0.150$ (1, assumption).

$[\text{H}^+] = \sqrt{K_a \times [\text{HA}]} = \sqrt{1.74 \times 10^{-5} \times 0.150} = \sqrt{2.61 \times 10^{-6}} = 1.62 \times 10^{-3}\ \text{mol dm}^{-3}$ (2).

$\text{pH} = -\log_{10}(1.62 \times 10^{-3}) = 2.79$ (1).

Markers reward the correct rearrangement, value of $[\text{H}^+]$ , and pH to 2 decimal places, plus a stated assumption.

Edexcel 20215 marksA buffer is made by dissolving

0.20\ \text{mol}

of sodium ethanoate in

1.0\ \text{dm}^3

0.10\ \text{mol dm}^{-3}

ethanoic acid. (

K_a = 1.74 \times 10^{-5}

.) (a) Calculate the pH. (b) Explain how the buffer resists a rise in pH when a little alkali is added.

Show worked answer →

Use the buffer expression, then explain the equilibrium response.

(a) $K_a = [\text{H}^+]\dfrac{[\text{A}^-]}{[\text{HA}]}$ , so $[\text{H}^+] = K_a \dfrac{[\text{HA}]}{[\text{A}^-]} = 1.74 \times 10^{-5} \times \dfrac{0.10}{0.20} = 8.7 \times 10^{-6}\ \text{mol dm}^{-3}$ (2). $\text{pH} = -\log_{10}(8.7 \times 10^{-6}) = 5.06$ (1).

(b) Added $\text{OH}^-$ reacts with the weak acid $\text{CH}_3\text{COOH}$ (the reservoir): $\text{CH}_3\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O}$ (1), so $[\text{H}^+]$ and hence pH change only slightly (1).

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Sources & how we know this

Pearson Edexcel A-Level Chemistry (9CH0) specification — Pearson Edexcel (2015)