Titration Calculations

Higher tier only: This whole lesson is Higher tier. Foundation candidates carry out the titration practical (previous lesson) but are not asked for these calculations.

A titration is only half the story: once you have a mean titre you can use it to calculate an unknown concentration. This is one of the most rewarding calculations at GCSE because it ties together everything from Topic C3 — moles, balanced equations and mole ratios — with the concentration ideas of C5. This lesson, part of Topic C5 of OCR Gateway Science A, sets out a reliable four-step method and works through examples with both a 1:1 ratio and a non-1:1 ratio, finishing by converting an answer into g/dm³.

By the end of this lesson you should be able to use titration results to calculate an unknown concentration in mol/dm³, apply the correct mole ratio from a balanced equation, and convert the answer into g/dm³.

These calculations reward a methodical, step-by-step approach more than almost any other at GCSE. The arithmetic is straightforward, but there are several places to slip — forgetting to convert a volume, misreading the mole ratio, or rounding too early — so laying the working out clearly is the surest way to full marks.

The Four-Step Method

Every titration calculation follows the same four steps. Learn them as a routine and the marks follow.

Moles of the known solution $= \text{concentration} \times \text{volume (dm}^3)$ .
Mole ratio from the balanced equation — how many moles of the unknown react with each mole of the known.
Moles of the unknown = moles of the known $\times$ the ratio.
Concentration of the unknown $= \dfrac{\text{moles}}{\text{volume (dm}^3)}$ .

Recall from Topic C3 that $\text{moles} = \dfrac{\text{mass}}{M_r}$ and that the mole ratio comes only from the big numbers in the balanced equation — exactly the ideas reused here. As always, every volume must be in dm³ (divide a cm³ value by 1000).

Before starting, make sure you have a balanced equation for the neutralisation, because the whole calculation depends on reading the correct mole ratio from it. The acid–alkali reactions you will meet are: $\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}$ (ratio $1:1$ ); $\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$ (ratio $1:2$ ); and $\text{HNO}_3 + \text{NaOH} \rightarrow \text{NaNO}_3 + \text{H}_2\text{O}$ (ratio $1:1$ ). Notice that sulfuric acid is the one to watch — because it provides two hydrogen ions per molecule, one mole of it neutralises two moles of a single-hydroxide alkali such as sodium hydroxide. Spotting the ratio correctly at step 2 is the part of the calculation that most often separates a full-mark answer from a near-miss.

Exam Tip: Write the four steps as labelled lines every time: moles of known → ratio → moles of unknown → concentration. Showing the working earns the method marks even if your final arithmetic slips.

The reason this method works is worth understanding rather than just memorising. A titration measures the exact volume at which the acid and alkali have just reacted — the end point. At that moment the moles of acid and the moles of alkali present are linked by the balanced equation. So if you know everything about one solution (its concentration and the volume used), you can work out its moles, then use the equation's ratio to find the moles of the other solution, and finally — because you also measured its volume — calculate its concentration. The whole calculation is really just a chain of the ideas you already met in Topic C3 (moles from concentration and volume; mole ratios from equations), applied to two solutions at once. Once you see it as that chain, every titration calculation looks the same, whatever the chemicals involved.

A common worry is which solution is the "known" and which is the "unknown". The known is simply the one whose concentration the question gives you; the unknown is the one you are asked to find. It does not matter whether that is the acid or the alkali — the four steps run in exactly the same order. Start from whichever solution you have full information about (concentration and volume), and finish with the solution you measured a volume for but whose concentration is missing.

Worked Example 1 — A 1 : 1 Ratio

In a titration, $25.0 \text{ cm}^3$ of sodium hydroxide solution is exactly neutralised by $20.0 \text{ cm}^3$ of hydrochloric acid of concentration $0.100 \text{ mol/dm}^3$ . Calculate the concentration of the sodium hydroxide in mol/dm³.

The balanced equation is:

$\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}$

Step 1 — moles of the known (HCl). Convert the volume: $20.0 \text{ cm}^3 = 0.0200 \text{ dm}^3$ .

$\text{moles of HCl} = \text{conc} \times \text{volume} = 0.100 \times 0.0200 = 0.00200 \text{ mol}$

Step 2 — mole ratio. The equation shows $\text{HCl} : \text{NaOH} = 1 : 1$ .

Step 3 — moles of the unknown (NaOH) $= 0.00200 \text{ mol}$ (same as HCl, because the ratio is $1:1$ ).

Step 4 — concentration of NaOH. Convert its volume: $25.0 \text{ cm}^3 = 0.0250 \text{ dm}^3$ .

$\text{concentration} = \dfrac{\text{moles}}{\text{volume}} = \dfrac{0.00200}{0.0250} = 0.0800 \text{ mol/dm}^3$

Answer: the sodium hydroxide is $0.0800 \text{ mol/dm}^3$ .

Worked Example 2 — A Non-1 : 1 Ratio

$25.0 \text{ cm}^3$ of sulfuric acid of concentration $0.0500 \text{ mol/dm}^3$ is exactly neutralised by $20.0 \text{ cm}^3$ of sodium hydroxide solution. Calculate the concentration of the sodium hydroxide in mol/dm³.

The balanced equation is:

$\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$

Step 1 — moles of the known ( $\text{H}_2\text{SO}_4$ ). Convert the volume: $25.0 \text{ cm}^3 = 0.0250 \text{ dm}^3$ .

$\text{moles of H}_2\text{SO}_4 = 0.0500 \times 0.0250 = 0.00125 \text{ mol}$

Step 2 — mole ratio. The equation shows $\text{H}_2\text{SO}_4 : \text{NaOH} = 1 : 2$ — each mole of acid reacts with two moles of alkali.

Step 3 — moles of the unknown (NaOH) $= 0.00125 \times 2 = 0.00250 \text{ mol}$ .

Step 4 — concentration of NaOH. Convert its volume: $20.0 \text{ cm}^3 = 0.0200 \text{ dm}^3$ .

$\text{concentration} = \dfrac{0.00250}{0.0200} = 0.125 \text{ mol/dm}^3$

Answer: the sodium hydroxide is $0.125 \text{ mol/dm}^3$ .

Exam Tip: The mole ratio is everything in step 2. With $\text{H}_2\text{SO}_4 + 2\text{NaOH}$ the ratio is $1:2$ , so you multiply the acid's moles by 2 to get the alkali's moles. Ignoring the ratio (treating it as $1:1$ ) is the single biggest source of lost marks here.

Titration Calculations

Titration Calculations

The Four-Step Method

Worked Example 1 — A 1 : 1 Ratio

Worked Example 2 — A Non-1 : 1 Ratio

Worked Example 3 — Converting the Answer to g/dm³

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