Titration Calculations and Curve Shapes Revision Notes | A Level AQA Chemistry

1. Why Titration Curves Tell You More Than a Single Number

A titration gives you a precise volume at the end point — but the shape of the pH curve tells you the whole story of what is happening in the flask at every stage. Whether the acid and base are strong or weak completely changes that shape, and AQA expects you to sketch, label, and explain all four combinations.

There are two skills running in parallel here: the calculation skill (using moles to find unknown concentrations or volumes) and the curve interpretation skill (reading and explaining pH at key points). Both appear on the same paper, so this lesson treats them together.

2. Titration Calculations — The Moles Method

Every titration calculation comes back to one relationship: $n = c \times V$, where $n$ is moles (mol), $c$ is concentration (mol dm$^{-3}$), and $V$ is volume in dm$^{3}$. Because volumes in titrations are usually given in cm$^{3}$, you must convert: divide by 1000 to get dm$^{3}$.

For a monoprotic acid reacting with a monoprotic base, the mole ratio at the equivalence point is always 1:1. That means moles of acid = moles of base at the equivalence point, which is the bridge between the two solutions in the calculation.

Worked example: Titration Calculation: Finding Concentration of a Base

25.0 cm³ of sodium hydroxide solution is titrated against 0.100 mol dm⁻³ hydrochloric acid. The mean titre is 22.4 cm³. Calculate the concentration of the NaOH solution. [3 marks]

1. $V_{\text{HCl}} = \frac{22.4}{1000} = 0.0224 \text{ dm}^3$ — Always convert cm³ to dm³ before using n = cV. This is the most common unit error in titration questions.
2. $n_{\text{HCl}} = 0.100 \times 0.0224 = 0.00224 \text{ mol}$ — [M1] Correct method: n = c × V with consistent units.
3. $\text{HCl}(aq) + \text{NaOH}(aq) \rightarrow \text{NaCl}(aq) + \text{H}_2\text{O}(l)$ $n_{\text{NaOH}} = n_{\text{HCl}} = 0.00224 \text{ mol}$ — HCl and NaOH are both monoprotic, so the ratio is 1:1. State the equation to justify this step.
4. $V_{\text{NaOH}} = \frac{25.0}{1000} = 0.0250 \text{ dm}^3$ — Use the volume of the solution whose concentration you are finding.
5. $c_{\text{NaOH}} = \frac{n}{V} = \frac{0.00224}{0.0250} = 0.0896 \text{ mol dm}^{-3}$ — [A1] Final answer to 3 significant figures, matching the data given. Include units.

Answer: $c_{\text{NaOH}} = 0.0896 \text{ mol dm}^{-3}$

3. The Shape of pH Curves — What to Look For

Before sketching any curve, fix three reference points in your mind: the initial pH (before any base is added), the equivalence point (where moles of acid = moles of base), and the final pH (large excess of base). The steep, near-vertical section of the curve always surrounds the equivalence point.

The key variable that changes the shape is whether the acid and base are strong or weak. A strong acid or base is fully dissociated in solution; a weak acid or base is only partially dissociated. This distinction controls the starting pH, the pH at equivalence, and whether a buffer region appears before the equivalence point.

A weak acid, by contrast, only partially dissociates. For ethanoic acid, the equilibrium $\text{CH}_3\text{COOH}(aq) \rightleftharpoons \text{CH}_3\text{COO}^-(aq) + \text{H}^+(aq)$ lies well to the left, so the pH of a 0.1 mol dm$^{-3}$ solution is around 2.9 rather than 1.0. This higher starting pH is the first visible difference on the curve.

3a. Strong Acid–Strong Base (e.g. HCl and NaOH)

Picture adding NaOH from a burette into a flask of HCl. At the start, the pH is low — around 1 for a 0.1 mol dm$^{-3}$ solution. As NaOH is added, pH rises slowly at first, then shoots up almost vertically through the equivalence point, then levels off at a high pH in the excess base region.

The equivalence point for a strong acid–strong base titration is at pH 7. The salt formed (NaCl) does not hydrolyse in water, so the solution is neutral. The steep section runs roughly from pH 3 to pH 11, which means both methyl orange (pH range 3.1–4.4) and phenolphthalein (pH range 8.3–10.0) are suitable indicators — both lie within the steep section.

3b. Weak Acid–Strong Base (e.g. CH₃COOH and NaOH)

This curve has a distinctly different shape in the early stages. The starting pH is higher (around 2.9 for 0.1 mol dm$^{-3}$ ethanoic acid) because the weak acid only partially dissociates. As NaOH is added, a buffer region develops — a relatively flat section where pH changes very slowly. This happens because the flask now contains both the weak acid and its conjugate base (ethanoate ions), which resist pH change.

At the half-equivalence point — when exactly half the acid has been neutralised — pH = p$K_a$ of the weak acid. For ethanoic acid ($K_a = 1.74 \times 10^{-5}$ mol dm$^{-3}$), p$K_a \approx 4.76$, so the pH at the half-equivalence point is approximately 4.76. This is a useful landmark on the curve and a common AQA calculation target.

The equivalence point is above pH 7 (around pH 8–9) because the salt formed is sodium ethanoate, and the ethanoate ion is the conjugate base of a weak acid. It partially hydrolyses water: $\text{CH}_3\text{COO}^-(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{CH}_3\text{COOH}(aq) + \text{OH}^-(aq)$, producing a slightly alkaline solution.

The steep section for a weak acid–strong base curve is shorter and shifted upward compared to the strong–strong curve. It typically runs from around pH 7 to pH 11. This means methyl orange is not suitable (its colour change at pH 3.1–4.4 falls in the buffer region, not the steep section), but phenolphthalein (pH 8.3–10.0) is suitable because its range lies within the steep section.

3c. Strong Acid–Weak Base (e.g. HCl and NH₃)

Now flip the situation: the flask contains a strong acid and the burette contains a weak base (ammonia solution). The starting pH is low (around 1 for 0.1 mol dm$^{-3}$ HCl). As ammonia is added, pH rises, but the equivalence point is now below pH 7 — around pH 5–6. The ammonium chloride salt formed hydrolyses to give a slightly acidic solution: $\text{NH}_4^+(aq) \rightleftharpoons \text{NH}_3(aq) + \text{H}^+(aq)$.

After the equivalence point, excess ammonia causes the pH to rise more slowly, as the solution becomes increasingly basic. The steep section is shorter than for strong–strong and is shifted downward, running roughly from pH 3 to pH 7. Methyl orange (pH 3.1–4.4) is suitable here; phenolphthalein is not, because its range (pH 8.3–10.0) lies above the steep section.

3d. Weak Acid–Weak Base (e.g. CH₃COOH and NH₃)

This combination produces the most unusual curve. The starting pH is relatively high (around 2.9), and as the weak base is added, the pH rises gradually throughout. There is no sharp, steep section around the equivalence point — instead, the curve shows a gradual, almost linear rise with no distinct steep section or clear inflection point. The equivalence point pH depends on the relative strengths of the acid and base.

Because there is no steep section, no simple indicator is suitable for a weak acid–weak base titration. The colour change of any indicator would be too gradual to identify a precise end point. This is an important AQA exam point: you must be able to justify why no indicator works here.

4. Sketching pH Curves — The Examiner's Checklist

AQA sketch questions award marks for specific features, not for artistic accuracy. When you draw a pH curve, you need to hit four marks: the correct starting pH, the correct shape of the curve before the equivalence point (including any buffer region), the position and pH of the equivalence point, and the correct final pH in excess base.

• Strong acid start: pH ≈ 1 for 0.1 mol dm⁻³; weak acid start: pH ≈ 2.9–3.0 for 0.1 mol dm⁻³
• Weak acid curves: show a buffer region (gradual, flat-ish rise) before the equivalence point
• Mark the equivalence point clearly — label it and state the approximate pH
• Strong–strong equivalence: pH 7; weak acid–strong base: pH above 7; strong acid–weak base: pH below 7
• After the equivalence point, the curve levels off as excess base dominates
• The steep section must be near-vertical and centred on the equivalence point

One mark-winning habit: always label the x-axis as 'Volume of base added / cm³' and the y-axis as 'pH'. Mark the equivalence point volume explicitly. Examiners reward precision in labelling, not just the shape.

5. The Half-Equivalence Point and pKa

This is a calculation feature unique to weak acid–strong base curves. At the half-equivalence point, exactly half the weak acid has been converted to its conjugate base, so $[\text{HA}] = [\text{A}^-]$. Substituting into the Henderson–Hasselbalch equation gives $\text{pH} = \text{p}K_a + \log_{10}(1) = \text{p}K_a$. This means you can read p$K_a$ directly from the curve at the half-equivalence volume.

Worked example: Reading pKa from a pH Curve

A 25.0 cm³ sample of a weak acid is titrated with 0.100 mol dm⁻³ NaOH. The equivalence point is reached at 20.0 cm³ of NaOH. The pH at 10.0 cm³ of NaOH added is 4.74. Determine the pKa of the weak acid.

1. $V_{\text{equivalence}} = 20.0 \text{ cm}^3$ — This is where moles of NaOH = moles of weak acid.
2. $V_{\text{half-equivalence}} = \frac{20.0}{2} = 10.0 \text{ cm}^3$ — At this volume, exactly half the weak acid has been neutralised, so [HA] = [A⁻].
3. $\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) = \text{p}K_a + \log_{10}(1) = \text{p}K_a$ — Because [HA] = [A⁻], the log term equals zero and drops out.
4. $\text{p}K_a = \text{pH at half-equivalence} = 4.74$ — Read directly from the curve at 10.0 cm³. This is a one-line answer once you know the method.

Answer: $\text{p}K_a = 4.74$

6. Comparing All Four Curve Types

Now that you have seen each curve individually, the exam will ask you to compare them. The key differences to articulate are: starting pH, presence or absence of a buffer region, the pH at the equivalence point, and the length and position of the steep section. These four features map directly onto the four mark points in a typical 'compare' question.

When comparing strong–strong with weak acid–strong base, for instance: both have a steep section, but the weak acid curve starts higher, has a buffer region before the equivalence point, and reaches equivalence above pH 7. The strong–strong curve has no buffer region, starts at a lower pH, and reaches equivalence at exactly pH 7.

7. Exam Technique and Common Mistakes

Titration questions on AQA papers follow predictable patterns. Calculation questions almost always require you to show the mole ratio step explicitly — even if it is 1:1, write it out. Examiners award a method mark for the ratio, and skipping it loses you that mark even if your final answer is correct.

For curve-sketching questions, the mark scheme typically awards one mark per feature: starting pH, shape before equivalence (including buffer region if appropriate), equivalence point position and pH, and the levelling-off after equivalence. Practise drawing all four curve types from memory, labelling each feature as you go.