The problems are on the titration worksheet. Try them on your own first, then compare with the answers below.
Quick answer: A titration finds an unknown concentration by measuring the exact volume of one solution that reacts with a known volume of another. These worked answers correct the 20 problems of the Chemistry 9-10 worksheet on titration, from name the apparatus to an unknown acid: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows moles of the known solution first, then the mole ratio, then the concentration of the unknown. Check the volume units and the dilution factor in every problem.
Level: Chemistry grades 9-10 (GCSE · IGCSE · O Level · Class 9-10) · Curriculum: GCSE C4 RP
Problem 1 Name the apparatus
1. A is the burette; it holds the solution that is added drop by drop, usually the acid.2. B is the tap, which controls the flow.
3. The white background makes the colour change at the end point easier to see.
Problem 2 Reading burette results
1. The bottom of the meniscus is at 23.85 cm³ (to the nearest 0.05).2. 23.85 − 0.40 = 23.45 cm³.
3. Looking from above or below gives a parallax error that changes the reading.
Problem 3 Which indicator?
1. Pink.2. Pink to colourless.
3. The indicator is itself a weak acid and would react with some of the alkali, so too much would alter the volume of acid needed.
Problem 4 Moles from c and V
1. 0.200 × 0.0250 = 0.00500 mol.2. 0.100 × 0.0150 = 0.00150 mol.
3. 0.0500 × 0.0225 = 0.001125 mol.
Problem 5 Choosing the mean
1. The rough titration was done quickly, so its value is less accurate; it only shows where to slow down.2. The largest minus the smallest is 22.50 − 22.40 = 0.10 cm³, so they are concordant.
3. (22.40 + 22.50 + 22.45) ÷ 3 = 22.45 cm³.
Problem 6 Technique steps
1. Water left inside would dilute the acid, so a larger volume would be needed.2. A bubble leaves during the run and the reading would show too large a volume of acid.
3. Swirling mixes the solutions so the acid reacts at once and the end point is sharp.
Problem 7 Alkali in the flask
1. 0.100 × 0.02000 = 0.00200 mol.2. Ratio 1 : 1, so c = 0.00200 ÷ 0.0250 = 0.0800 mol/dm³.
3. 0.0800 × 40 = 3.20 g/dm³.
Problem 8 How strong is the vinegar?
1. n(NaOH) = 0.100 × 0.02150 = 0.002150 mol = n(acid). c = 0.002150 ÷ 0.0250 = 0.0860 mol/dm³.2. The dilution was 250 ÷ 25 = 10 times, so c = 0.860 mol/dm³.
3. M(CH₃COOH) = 60 g/mol; 0.860 × 60 = 51.6 g/dm³.
Problem 9 Testing limewater
1. n(HCl) = 0.0200 × 0.01550 = 0.000310 mol.2. Ratio 1 : 2, so n(Ca(OH)₂) = 0.000155 mol and c = 0.000155 ÷ 0.0250 = 0.00620 mol/dm³.
3. M = 74; 0.00620 × 74 = 0.459 g/dm³.
Problem 10 Drain cleaner
1. n(HCl) = 0.100 × 0.02000 = 0.00200 mol = n(NaOH). c = 0.00200 ÷ 0.0250 = 0.0800 mol/dm³.2. 500 ÷ 10.0 = 50.
3. 0.0800 × 50 = 4.00 mol/dm³, and 4.00 × 40 = 160 g/dm³.
Problem 11 Sodium carbonate and acid
1. 0.100 × 0.02200 = 0.00220 mol.2. Ratio Na₂CO₃ : HCl is 1 : 2, so n = 0.00110 mol and c = 0.00110 ÷ 0.0250 = 0.0440 mol/dm³.
3. M = 106; 0.0440 × 106 = 4.66 g/dm³.
Problem 12 Analyse the results table
1. Rough 25.40; run 1 24.30; run 2 48.50 − 24.30 = 24.20; run 3 24.45 − 0.20 = 24.25.2. Runs 1, 2 and 3 lie within 0.10 cm³ (24.30 − 24.20), so the mean is (24.30 + 24.20 + 24.25) ÷ 3 = 24.25 cm³.
3. n(acid) = 0.100 × 0.02425 = 0.002425 mol; c(alkali) = 0.002425 ÷ 0.0250 = 0.0970 mol/dm³.
Problem 13 Read the pH curve
1. The steep part is at 24 cm³.2. The curve passes through pH 7.
3. n(NaOH) = 0.150 × 0.0200 = 0.00300 mol; the ratio is 1 : 1; c(acid) = 0.00300 ÷ 0.0240 = 0.125 mol/dm³.
Problem 14 Household ammonia
1. n(HCl) = 0.100 × 0.02300 = 0.00230 mol = n(NH₃); c = 0.00230 ÷ 0.0250 = 0.0920 mol/dm³.2. Dilution is 250 ÷ 10.0 = 25 times: 0.0920 × 25 = 2.30 mol/dm³; × 17 = 39.1 g/dm³.
3. The salt NH₄Cl makes the end point slightly acidic (about pH 5), where methyl orange changes; phenolphthalein changes in alkaline solution and would change colour too early.
Problem 15 Purity of sodium hydroxide
1. n(HCl) = 0.200 × 0.02250 = 0.00450 mol, so n(NaOH) = 0.00450 mol.2. In 250 cm³: 0.0450 mol × 40 = 1.80 g.
3. 1.80 ÷ 2.00 × 100 = 90.0%.
Problem 16 Making a standard solution
1. n = 0.0500 × 0.250 = 0.0125 mol; M = 106; m = 1.325 g (1.33 g).2. Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂; ratio 1 : 2.
3. n(Na₂CO₃) = 0.0500 × 0.0250 = 0.00125 mol; n(HCl) = 0.00250 mol; c = 0.00250 ÷ 0.02080 = 0.120 mol/dm³.
Problem 17 Acid in lemon juice
1. 0.100 × 0.02400 = 0.00240 mol.2. Ratio acid : NaOH is 1 : 3, so n(acid) = 0.000800 mol and c = 0.000800 ÷ 0.0250 = 0.0320 mol/dm³.
3. The dilution factor is 10: 0.320 mol/dm³. M(C₆H₈O₇) = 192, so 0.320 × 192 = 61.4 g/dm³.
Problem 18 Which errors matter?
1. Too high: the acid is diluted, so a larger titre is needed.2. Unaffected: water does not change the moles of alkali placed in the flask.
3. Too high: the volume read from the burette includes the volume of the bubble.
4. Too high: excess acid was added, so the titre is larger than the true value.
Problem 19 Checking a label
1. n(NaOH) = 0.0500 × 0.02050 = 0.001025 mol = n(HCl); c = 0.001025 ÷ 0.0250 = 0.0410 mol/dm³.2. Dilution is 250 ÷ 10.0 = 25: 0.0410 × 25 = 1.025 mol/dm³.
3. (1.025 − 1.00) ÷ 1.00 × 100 = 2.5%, so the label is within ±5%.
Problem 20 An unknown acid
1. n(NaOH) = 0.100 × 0.02000 = 0.00200 mol = n(HX) in 25.0 cm³: 0.00200 mol.2. 0.00200 × 4 = 0.00800 mol.
3. M = 0.600 ÷ 0.00800 = 75.0 g/mol.
4. A diprotic acid would use 2 mol of NaOH per mole, so the moles of acid would be half and the molar mass would be twice as large (150 g/mol).
