Titration Calculations

Titration calculations combine several stoichiometry skills into one problem. You use the volume and concentration of one solution (the one you know) to find the unknown concentration or volume of the other solution. The balanced equation provides the mole ratio that links them.

The method

1. Write the balanced equation for the reaction.
2. Calculate the moles of the solution whose concentration and volume are both known, using n = c x V. (Remember to convert cm3 to dm3.)
3. Use the mole ratio from the balanced equation to find the moles of the other substance.
4. Use the moles and the remaining known value (either concentration or volume) to find the unknown, using c = n / V or V = n / c.

The formula

n = c x V (to find moles from the known solution)

Then the mole ratio links the two substances.

Then c = n / V or V = n / c (to find the unknown).

Worked examples

Example 1: Finding unknown concentration

Question: 25.0 cm3 of sodium hydroxide solution was neutralised by 20.0 cm3 of 0.1 mol/dm3 hydrochloric acid. Calculate the concentration of the sodium hydroxide. The equation is NaOH + HCl -> NaCl + H2O.

Working:

• Moles of HCl = c x V = 0.1 x (20.0 / 1000) = 0.002 mol
• Mole ratio: 1 NaOH : 1 HCl
• Moles of NaOH = 0.002 mol
• Volume of NaOH = 25.0 / 1000 = 0.025 dm3
• Concentration of NaOH = n / V = 0.002 / 0.025 = 0.08 mol/dm3

Answer: 0.08 mol/dm3

Examiner note: This is the classic titration question. Work through each step in order and show all unit conversions.

Example 2: Finding unknown concentration with a 2:1 ratio

Question: 25.0 cm3 of sodium carbonate solution required 30.0 cm3 of 0.1 mol/dm3 hydrochloric acid for complete reaction. Find the concentration of the sodium carbonate. The equation is Na2CO3 + 2HCl -> 2NaCl + H2O + CO2.

Working:

• Moles of HCl = 0.1 x (30.0 / 1000) = 0.003 mol
• Mole ratio: 1 Na2CO3 : 2 HCl
• Moles of Na2CO3 = 0.003 / 2 = 0.0015 mol
• Volume of Na2CO3 = 25.0 / 1000 = 0.025 dm3
• Concentration of Na2CO3 = 0.0015 / 0.025 = 0.06 mol/dm3

Answer: 0.06 mol/dm3

Examiner note: The mole ratio is 1:2. Divide the moles of HCl by 2 to get moles of Na2CO3. This is where many students lose marks.

Example 3: Finding unknown volume

Question: What volume of 0.2 mol/dm3 sulfuric acid is needed to neutralise 25.0 cm3 of 0.1 mol/dm3 potassium hydroxide? The equation is H2SO4 + 2KOH -> K2SO4 + 2H2O.

Working:

• Moles of KOH = 0.1 x (25.0 / 1000) = 0.0025 mol
• Mole ratio: 1 H2SO4 : 2 KOH
• Moles of H2SO4 = 0.0025 / 2 = 0.00125 mol
• Volume of H2SO4 = n / c = 0.00125 / 0.2 = 0.00625 dm3
• Convert: 0.00625 x 1000 = 6.25 cm3

Answer: 6.25 cm3

Examiner note: The question asks for volume, so rearrange to V = n / c in the final step.

Example 4: Finding concentration in g/dm3

Question: 25.0 cm3 of NaOH solution was titrated against 0.1 mol/dm3 HCl. The mean titre was 22.5 cm3. Find the concentration of NaOH in g/dm3. (Ar: Na = 23, O = 16, H = 1) The equation is NaOH + HCl -> NaCl + H2O.

Working:

• Moles of HCl = 0.1 x (22.5 / 1000) = 0.00225 mol
• Mole ratio: 1:1, so moles of NaOH = 0.00225 mol
• Concentration of NaOH in mol/dm3 = 0.00225 / 0.025 = 0.09 mol/dm3
• Mr of NaOH = 40
• Concentration in g/dm3 = 0.09 x 40 = 3.6 g/dm3

Answer: 3.6 g/dm3

Examiner note: After finding concentration in mol/dm3, multiply by Mr to convert to g/dm3.

Example 5: Using titration data with averaging

Question: A student titrated 25.0 cm3 of potassium hydroxide against 0.05 mol/dm3 sulfuric acid. The titre results were 24.0, 23.5, 23.4, and 23.5 cm3. Calculate the concentration of the potassium hydroxide. The equation is 2KOH + H2SO4 -> K2SO4 + 2H2O.

Working:

• Concordant results: 23.5, 23.4, and 23.5 cm3 (discard 24.0 as anomalous)
• Mean titre = (23.5 + 23.4 + 23.5) / 3 = 23.47 cm3
• Moles of H2SO4 = 0.05 x (23.47 / 1000) = 0.001174 mol
• Mole ratio: 2 KOH : 1 H2SO4
• Moles of KOH = 0.001174 x 2 = 0.002347 mol
• Volume of KOH = 25.0 / 1000 = 0.025 dm3
• Concentration of KOH = 0.002347 / 0.025 = 0.0939 mol/dm3

Answer: 0.094 mol/dm3 (3 s.f.)

Examiner note: Discard the anomalous result (24.0 cm3) and average only the concordant values before calculating.

Common mistakes

1. Forgetting to convert cm3 to dm3. This is the most common error. Divide by 1000 every time you see cm3.
2. Getting the mole ratio backwards. If the ratio is 1 A : 2 B and you know moles of B, divide by 2. If you know moles of A, multiply by 2. Think carefully about which direction you are converting.
3. Using the wrong titre value. Use the mean of the concordant results only, not the mean of all results.
4. Confusing which solution is which. Label clearly: which is in the burette and which is in the conical flask. The titre is the volume from the burette.
5. Not showing the mole ratio step. Examiners award a mark for using the ratio. Even when it is 1:1, write it explicitly.

When this appears

Titration calculations are Extended tier content and appear on Paper 4. They build on concentration calculations and volume from concentration. See the stoichiometry topic for the broader context.