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How do we measure acidity using pH, Ka and buffers?

The Bronsted-Lowry theory, strong and weak acids and bases, pH and the calculation of pH for strong and weak acids using Ka and Kw, titration curves and indicators, and the action of buffer solutions.

A CCEA A-Level Chemistry answer on acid-base equilibria, covering the Bronsted-Lowry theory, strong and weak acids and bases, calculating pH using Ka and Kw, titration curves and choice of indicator, and the action and calculation of buffer solutions.

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  1. What this dot point is asking
  2. Bronsted-Lowry theory
  3. pH calculations
  4. Titration curves and indicators
  5. Buffers
  6. Examples in context
  7. Try this

What this dot point is asking

CCEA wants you to use the Bronsted-Lowry theory, distinguish strong and weak acids and bases, calculate pH for strong and weak acids using KaK_a and KwK_w, interpret titration curves and choose indicators, and explain the action of buffer solutions both qualitatively and by calculation.

Bronsted-Lowry theory

In HCl+H2O→H3O++Cl−\text{HCl} + \text{H}_2\text{O} \rightarrow \text{H}_3\text{O}^+ + \text{Cl}^-, the pairs are HCl/Cl−\text{HCl}/\text{Cl}^- and H3O+/H2O\text{H}_3\text{O}^+/\text{H}_2\text{O}. A strong acid fully dissociates; a weak acid only partly dissociates, setting up an equilibrium so that KaK_a is small.

pH calculations

For a strong base such as NaOH\text{NaOH}, find [OH−][\text{OH}^-] from the concentration, then use KwK_w to get [H+][\text{H}^+], then take the log.

Titration curves and indicators

Buffers

The pH of a buffer is found from [H+]=Ka×[acid][salt][\text{H}^+] = K_a \times \dfrac{[\text{acid}]}{[\text{salt}]}. If [acid]=[salt][\text{acid}] = [\text{salt}] then [H+]=Ka[\text{H}^+] = K_a and pH=pKa\text{pH} = \text{p}K_a.

Examples in context

Blood is buffered near pH 7.4\text{pH}\ 7.4 by the carbonic acid and hydrogencarbonate system, H2CO3⇌HCO3−+H+\text{H}_2\text{CO}_3 \rightleftharpoons \text{HCO}_3^- + \text{H}^+, which mirrors the ethanoic acid buffer studied in CCEA. In a CCEA practical, students titrate ethanoic acid against sodium hydroxide and use the pH meter trace to locate the half-neutralisation point, reading pKa\text{p}K_a directly off the curve as the pH where exactly half the acid has been neutralised. The same titration shows phenolphthalein turning pink only within the sharp jump near pH 9\text{pH}\ 9, confirming why methyl orange would change too early and give a false endpoint for this weak-acid/strong-base pairing.

Try this

Q1. Calculate the pH of 0.10 mol dm−30.10\ \text{mol dm}^{-3} hydrochloric acid. [1 mark]

  • Cue. pH=−log⁡10(0.10)=1.0\text{pH} = -\log_{10}(0.10) = 1.0.

Q2. State the two components of an acidic buffer solution and write the expression linking [H+][\text{H}^+] to KaK_a. [2 marks]

  • Cue. A weak acid and its conjugate base (its salt); [H+]=Ka×[acid]/[salt][\text{H}^+] = K_a \times [\text{acid}]/[\text{salt}].

Q3. A buffer contains 0.20 mol dm−30.20\ \text{mol dm}^{-3} ethanoic acid and 0.20 mol dm−30.20\ \text{mol dm}^{-3} sodium ethanoate (Ka=1.74×10−5K_a = 1.74 \times 10^{-5}). Calculate its pH. [2 marks]

  • Cue. [H+]=Ka×(0.20/0.20)=1.74×10−5[\text{H}^+] = K_a \times (0.20/0.20) = 1.74 \times 10^{-5}; pH=4.76\text{pH} = 4.76.

Exam-style practice questions

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

CCEA 20194 marksCalculate the pH of a solution formed when 25.0 cm325.0\ \text{cm}^3 of 0.100 mol dm−30.100\ \text{mol dm}^{-3} ethanoic acid is taken, given that KaK_a for ethanoic acid is 1.74×10−5 mol dm−31.74 \times 10^{-5}\ \text{mol dm}^{-3}.
Show worked answer →

Ethanoic acid is weak, so it only partly dissociates. Use the weak acid approximation Ka=[H+]2/[HA]K_a = [\text{H}^+]^2 / [\text{HA}], treating the equilibrium concentration of acid as the initial concentration.

Rearrange for the hydrogen ion concentration:

[H+]=Ka×[HA]=1.74×10−5×0.100[\text{H}^+] = \sqrt{K_a \times [\text{HA}]} = \sqrt{1.74 \times 10^{-5} \times 0.100}

[H+]=1.74×10−6=1.32×10−3 mol dm−3[\text{H}^+] = \sqrt{1.74 \times 10^{-6}} = 1.32 \times 10^{-3}\ \text{mol dm}^{-3}

Then take the negative log:

pH=−log⁡10(1.32×10−3)=2.88\text{pH} = -\log_{10}(1.32 \times 10^{-3}) = 2.88

Markers reward (1) recognising the acid is weak and using KaK_a not the full concentration, (2) the square-root step, (3) the correct pH to two decimal places. The volume is a distractor since concentration, not amount, sets pH here.

CCEA 20213 marksExplain how a buffer solution made from ethanoic acid and sodium ethanoate resists a change in pH when a small amount of strong acid is added. Include an equation.
Show worked answer →

An acidic buffer contains a large reservoir of weak acid and its conjugate base. The equilibrium present is CH3COOH⇌CH3COO−+H+\text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+.

When acid is added, the extra H+\text{H}^+ ions react with the conjugate base (ethanoate ions):

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

Because the ethanoate reservoir is large, almost all the added hydrogen ions are removed, so [H+][\text{H}^+] and therefore pH barely change. Markers reward (1) naming the reservoir of conjugate base, (2) the equation showing ethanoate mopping up H+\text{H}^+, (3) the statement that pH changes only slightly because the change in [H+][\text{H}^+] is small.

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