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Titration Calculations Worksheet

General chemistryGrades 9–12 · GCSE / A-level · IB · AP · CBSE11 items · answer key included

A printable titration calculations worksheet with answers for GCSE, A-level, IB, AP and CBSE chemistry. Students complete a burette results table, choose concordant titres, calculate the mean titre, then use moles = concentration × volume to find an unknown concentration. Further acid–base titration questions cover a strong acid–strong base titration, a diprotic sulfuric acid (H₂SO₄ + 2NaOH) calculation, a volume-needed problem and a back-to-mass percentage purity question. Every answer in the key is fully worked and numerically checked.

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Titration Calculations Worksheet

Name Class Date

Part A — Results table and mean titre

25.00 cm³ of sodium hydroxide solution was pipetted into a conical flask and titrated with 0.100 mol dm⁻³ hydrochloric acid using phenolphthalein. Complete the table, then answer the questions.

1.
RoughRun 1Run 2Run 3
Final burette reading / cm³24.6024.2524.1024.35
Initial burette reading / cm³0.000.100.000.20
Titre / cm³24.6024.1524.1024.15
Concordant? (✓)✓✓✓
2.
Calculate the mean titre using concordant results only.
(24.15 + 24.10 + 24.15) ÷ 3 = 24.13 cm3 (rough titre 24.60 cm3 excluded).
3.
Write the balanced equation and calculate the moles of HCl in the mean titre.
NaOH + HCl → NaCl + H2O
n(HCl) = 0.100 × 24.13/1000 = 2.413 × 10−3 mol
4.
Calculate the concentration of the sodium hydroxide in mol dm−3.
1 : 1 ratio → n(NaOH) = 2.413 × 10−3 mol
[NaOH] = 2.413 × 10−3 ÷ 0.02500 = 0.0965 mol dm−3
5.
Explain why the rough titre is larger than the accurate titres and is not used in the mean.
In the rough run acid is added quickly, so the end point is overshot; it is only used to find the approximate end point so later runs can be done dropwise.

Part B — Titration calculations

Show your working: moles of known → mole ratio → moles of unknown → concentration or mass. Give answers to 3 significant figures.

6.
Strong acid–strong base. What volume of 0.120 mol dm−3 HNO3 exactly neutralises 20.0 cm3 of 0.150 mol dm−3 KOH?
KOH + HNO3 → KNO3 + H2O
n(KOH) = 0.150 × 0.0200 = 3.00 × 10−3 mol = n(HNO3)
V = 3.00 × 10−3 ÷ 0.120 = 0.0250 dm3 = 25.0 cm3
7.
Diprotic acid. 25.0 cm3 of 0.200 mol dm−3 NaOH is neutralised by 21.30 cm3 of sulfuric acid. Calculate the concentration of the H2SO4 in mol dm−3.
H2SO4 + 2NaOH → Na2SO4 + 2H2O
n(NaOH) = 0.200 × 0.0250 = 5.00 × 10−3 mol
n(H2SO4) = ½ × 5.00 × 10−3 = 2.50 × 10−3 mol
[H2SO4] = 2.50 × 10−3 ÷ 0.02130 = 0.117 mol dm−3
8.
Convert your answer to B2 into a mass concentration in g dm−3 (Mr H2SO4 = 98.08).
0.11737 mol dm−3 × 98.08 g mol−1 = 11.5 g dm−3
9.
Purity. 1.00 g of impure NaOH was dissolved and made up to 250.0 cm3 in a volumetric flask. 25.0 cm3 portions needed a mean titre of 22.40 cm3 of 0.100 mol dm−3 HCl. Calculate the percentage purity of the NaOH (Mr = 40.00).
n(HCl) = 0.100 × 0.02240 = 2.240 × 10−3 mol = n(NaOH) in 25.0 cm3
In 250.0 cm3: 2.240 × 10−3 × 10 = 0.02240 mol
Mass = 0.02240 × 40.00 = 0.896 g
Purity = 0.896 ÷ 1.00 × 100 = 89.6 %
10.
Which indicator would you use for a strong acid–strong base titration, and what is the colour change when acid is added to alkali?
Phenolphthalein (pink → colourless) or methyl orange (yellow → orange/red). Both work because the pH changes sharply from about 3 to 11 at the equivalence point.
11.
Explain why the burette is rinsed with the acid, but the conical flask is rinsed only with distilled water.
Rinsing the burette with acid stops water diluting the acid (which would increase the titre). Water left in the flask does not change the number of moles of alkali pipetted in, so the result is unaffected.

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Answer key: Titration Calculations Worksheet

Part A — Results table and mean titre

1.
RoughRun 1Run 2Run 3
Final burette reading / cm³24.6024.2524.1024.35
Initial burette reading / cm³0.000.100.000.20
Titre / cm³24.6024.1524.1024.15
Concordant? (✓)✓✓✓
2.
(24.15 + 24.10 + 24.15) ÷ 3 = 24.13 cm3 (rough titre 24.60 cm3 excluded).
3.
NaOH + HCl → NaCl + H2O
n(HCl) = 0.100 × 24.13/1000 = 2.413 × 10−3 mol
4.
1 : 1 ratio → n(NaOH) = 2.413 × 10−3 mol
[NaOH] = 2.413 × 10−3 ÷ 0.02500 = 0.0965 mol dm−3
5.
In the rough run acid is added quickly, so the end point is overshot; it is only used to find the approximate end point so later runs can be done dropwise.

Part B — Titration calculations

6.
KOH + HNO3 → KNO3 + H2O
n(KOH) = 0.150 × 0.0200 = 3.00 × 10−3 mol = n(HNO3)
V = 3.00 × 10−3 ÷ 0.120 = 0.0250 dm3 = 25.0 cm3
7.
H2SO4 + 2NaOH → Na2SO4 + 2H2O
n(NaOH) = 0.200 × 0.0250 = 5.00 × 10−3 mol
n(H2SO4) = ½ × 5.00 × 10−3 = 2.50 × 10−3 mol
[H2SO4] = 2.50 × 10−3 ÷ 0.02130 = 0.117 mol dm−3
8.
0.11737 mol dm−3 × 98.08 g mol−1 = 11.5 g dm−3
9.
n(HCl) = 0.100 × 0.02240 = 2.240 × 10−3 mol = n(NaOH) in 25.0 cm3
In 250.0 cm3: 2.240 × 10−3 × 10 = 0.02240 mol
Mass = 0.02240 × 40.00 = 0.896 g
Purity = 0.896 ÷ 1.00 × 100 = 89.6 %
10.
Phenolphthalein (pink → colourless) or methyl orange (yellow → orange/red). Both work because the pH changes sharply from about 3 to 11 at the equivalence point.
11.
Rinsing the burette with acid stops water diluting the acid (which would increase the titre). Water left in the flask does not change the number of moles of alkali pipetted in, so the result is unaffected.

How to use this worksheet

  1. Use Part A straight after a practical titration so students process real-looking burette data before their own.
  2. Remind students that burette readings are recorded to 2 d.p. (ending in 0 or 5) and that the rough titre is never used in the mean.
  3. Part B escalates from a 1 : 1 calculation to the 1 : 2 H₂SO₄ ratio and a dilution-factor purity question — set Q4–Q5 as stretch.
  4. Print the answer key for self-marking; each step (moles, ratio, concentration) is shown on its own line.
  5. Set the title, student lines and answer option on the left, then press Download PDF and choose “Save as PDF” (or print directly). Your choices are remembered on this device.

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Frequently asked questions

What are concordant titres?

Concordant titres are accurate titre values that agree within 0.10 cm³ of each other. Only concordant results are used to calculate the mean titre; the rough (trial) titre is excluded.

How do you calculate concentration from a titration?

Find moles of the known solution (moles = concentration × volume in dm³), use the balanced equation's mole ratio to get moles of the unknown, then divide by the unknown's volume in dm³ to get its concentration in mol dm⁻³.

Why is the ratio 1 : 2 for sulfuric acid and sodium hydroxide?

H₂SO₄ is diprotic: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Each mole of sulfuric acid neutralises two moles of sodium hydroxide, so moles of H₂SO₄ = ½ × moles of NaOH.

Are the answers included?

Yes. The answer key shows the completed results table, the mean titre and every calculation step, with values computed and checked.

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