OCR OCR 2019-style practice: A solution contains 12 g of sodium hydroxide dissolved in 500\ cm^3 of solution. Calculate the concentration in g/dm^3, and then in mol/dm^3. The relative formula mass of sodium hydroxide is 40.

A C5.1 concentration calculation. Reward: first convert the volume to cubic decimetres: 500\ cm^3 = 0.5\ dm^3. Concentration in g/dm3 = massvolume = 120.5 = 24\ g/dm^3. To convert to mol/dm3, divide by the relative formula mass: 2440 = 0.6\ mol/dm^3. Markers credit converting the volume to dm3, the concentration in g/dm3 (24), and the concentration in mol/dm3 (0.6). A common error is to forget to convert cm3 to dm3 (dividing by 1000), which makes the answer 1000 times too small or large.

OCR OCR 2022-style practice: In a titration, 25.0\ cm^3 of sodium hydroxide solution is neutralised by 20.0\ cm^3 of hydrochloric acid of concentration 0.10\ mol/dm^3. The equation is NaOH + HCl → NaCl + H_2O. Calculate the concentration of the sodium hydroxide solution in mol/dm^3.

A Higher tier titration calculation. Reward: moles of HCl = concentration × volume in dm^3 = 0.10 × 20.01000 = 0.0020 mol (1 mark). The equation shows a 1:1 ratio of HCl to NaOH, so moles of NaOH = 0.0020 mol (1 mark). The volume of NaOH is 25.01000 = 0.025\ dm^3 (1 mark). Concentration of NaOH = molesvolume = 0.00200.025 = 0.080\ mol/dm^3 (1 mark for method, 1 mark for answer). Markers reward moles of acid, the mole ratio, the volume conversion, and the final concentration of 0.08 mol/dm3. A common error is not converting the volumes to dm3.

EnglandChemistrySyllabus dot point

How do we measure concentration, and how does a titration find the concentration of an acid or alkali?

Concentration in grams and moles per cubic decimetre, calculating concentration, the method and apparatus of a titration, and calculating an unknown concentration from titration results.

A focused answer to OCR Gateway GCSE Chemistry A topic C5.1 on concentration and titrations, covering concentration in grams and moles per cubic decimetre, the titration method and apparatus, and calculating an unknown concentration from titration results.

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

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What this dot point is asking
Concentration
The titration method
Calculating an unknown concentration

What this dot point is asking

OCR wants you to express concentration in grams and moles per cubic decimetre ( $\text{g/dm}^3$ and $\text{mol/dm}^3$ ), calculate concentration, describe the method and apparatus of a titration, and calculate an unknown concentration from titration results. This is a major piece of quantitative chemistry and a key practical.

Concentration

For example, $24$ g of sodium hydroxide in $1\ \text{dm}^3$ is $24\ \text{g/dm}^3$ , which (dividing by $M_r = 40$ ) is $0.6\ \text{mol/dm}^3$ .

The titration method

A single indicator such as phenolphthalein or methyl orange is used (not universal indicator) because it gives a sharp colour change at the end point.

Calculating an unknown concentration

To find an unknown concentration from a titration:

Calculate the moles of the solution you know ( $\text{moles} = \text{concentration} \times \text{volume in dm}^3$ ).
Use the mole ratio from the balanced equation to find the moles of the other solution.
Divide those moles by that solution's volume in dm $^3$ to get its concentration.

Be careful to convert all volumes from cm $^3$ to dm $^3$ first.

Finding the concentration of an acid by titration

In a titration, $25.0\ \text{cm}^3$ of sodium hydroxide of concentration $0.10\ \text{mol/dm}^3$ is exactly neutralised by $20.0\ \text{cm}^3$ of hydrochloric acid. The equation is $\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}$ . Find the concentration of the acid in mol/dm $^3$ .

step 1: Find the moles of the known solution

Moles of NaOH $= \text{concentration} \times \text{volume in dm}^3 = 0.10 \times \dfrac{25.0}{1000} = 0.0025$ mol.

step 2: Use the mole ratio

The equation shows a $1:1$ ratio of NaOH to HCl, so moles of HCl $= 0.0025$ mol.

step 3: Convert the acid volume to dm $^3$

Volume of HCl $= \dfrac{20.0}{1000} = 0.020\ \text{dm}^3$ .

step 4: Calculate the acid concentration

Concentration of HCl $= \dfrac{\text{moles}}{\text{volume}} = \dfrac{0.0025}{0.020} = 0.125\ \text{mol/dm}^3$ .

Exam-style practice questions

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

OCR 20193 marksA solution contains 12 g of sodium hydroxide dissolved in

500\ \text{cm}^3

of solution. Calculate the concentration in

\text{g/dm}^3

, and then in

\text{mol/dm}^3

. The relative formula mass of sodium hydroxide is 40.

Show worked answer →

A C5.1 concentration calculation. Reward: first convert the volume to cubic decimetres: $500\ \text{cm}^3 = 0.5\ \text{dm}^3$ . Concentration in g/dm3 $= \dfrac{\text{mass}}{\text{volume}} = \dfrac{12}{0.5} = 24\ \text{g/dm}^3$ . To convert to mol/dm3, divide by the relative formula mass: $\dfrac{24}{40} = 0.6\ \text{mol/dm}^3$ . Markers credit converting the volume to dm3, the concentration in g/dm3 (24), and the concentration in mol/dm3 (0.6). A common error is to forget to convert cm3 to dm3 (dividing by 1000), which makes the answer 1000 times too small or large.

OCR 20225 marksIn a titration,

25.0\ \text{cm}^3

of sodium hydroxide solution is neutralised by

20.0\ \text{cm}^3

of hydrochloric acid of concentration

0.10\ \text{mol/dm}^3

. The equation is

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

. Calculate the concentration of the sodium hydroxide solution in

\text{mol/dm}^3

Show worked answer →

A Higher tier titration calculation. Reward: moles of HCl $= \text{concentration} \times \text{volume in dm}^3 = 0.10 \times \dfrac{20.0}{1000} = 0.0020$ mol (1 mark). The equation shows a $1:1$ ratio of HCl to NaOH, so moles of NaOH $= 0.0020$ mol (1 mark). The volume of NaOH is $\dfrac{25.0}{1000} = 0.025\ \text{dm}^3$ (1 mark). Concentration of NaOH $= \dfrac{\text{moles}}{\text{volume}} = \dfrac{0.0020}{0.025} = 0.080\ \text{mol/dm}^3$ (1 mark for method, 1 mark for answer). Markers reward moles of acid, the mole ratio, the volume conversion, and the final concentration of 0.08 mol/dm3. A common error is not converting the volumes to dm3.

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Sources & how we know this

OCR GCSE (9-1) Gateway Science Suite Chemistry A (J248) specification — OCR (2016)