Back Titration Calculations Explained

Some substances refuse to be titrated directly. Chalk doesn’t dissolve in water. An antacid tablet reacts slowly and fizzes. Ammonia is a gas that escapes from solution. For awkward analytes like these, chemists use an indirect trick: add a known excess of a reagent, let it react completely, then titrate whatever is left over. The amount that reacted with the analyte is found by subtraction. This is a back titration, and once you’ve seen the pattern, the calculations are very manageable.

When to use a back titration

• The analyte is an insoluble solid (calcium carbonate, magnesium oxide, antacid tablets).
• The reaction with the analyte is slow (so a direct titration would have no clear end point).
• The analyte is volatile (ammonia, which would escape during a direct titration).
• The direct end point is hard to see (for example, a coloured or cloudy sample).
• A reaction needs heating to go to completion (such as hydrolysing aspirin).

The logic in one line

Moles of reagent that reacted with the analyte = moles of reagent added − moles of reagent left over

The “left over” amount comes from the titration.

The step-by-step method

1. Write both equations: the reaction of the analyte with the excess reagent, and the titration reaction.
2. Calculate moles of reagent added at the start (c × V).
3. Calculate moles of titrant used in the titration (c × V).
4. Convert to moles of excess reagent left over using the titration equation’s ratio.
5. Subtract to find moles of reagent that reacted with the analyte.
6. Convert to moles of analyte using the first equation’s ratio.
7. Find the mass, percentage or concentration of the analyte.

Watch for scaling: if the reaction mixture was made up to a volume and only a portion was titrated, scale the leftover amount up to the whole mixture before subtracting.

Worked example 1: limestone

A 1.000 g sample of limestone is added to 50.0 cm³ of 0.500 mol/dm³ HCl. After the reaction, the excess acid needs 17.20 cm³ of 0.500 mol/dm³ NaOH. Calculate the percentage of CaCO₃ in the limestone.

Equations: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂ HCl + NaOH → NaCl + H₂O

1. HCl added = 0.500 × 0.0500 = 0.0250 mol
2. NaOH used = 0.500 × 0.01720 = 8.60 × 10⁻³ mol
3. HCl left over = 8.60 × 10⁻³ mol (1 : 1)
4. HCl reacted with CaCO₃ = 0.0250 − 0.00860 = 0.0164 mol
5. CaCO₃ = 0.0164 ÷ 2 = 8.20 × 10⁻³ mol
6. Mass CaCO₃ = 8.20 × 10⁻³ × 100.09 = 0.821 g
7. % CaCO₃ = 0.821 ÷ 1.000 × 100 = 82.1%

Sense check: the mass of CaCO₃ (0.821 g) is less than the sample mass (1.000 g), as it must be. The rest of the limestone is insoluble impurities such as sand and clay.

Worked example 2: antacid tablet with a portion titrated

An antacid tablet containing magnesium hydroxide is dissolved in 100.0 cm³ of 0.200 mol/dm³ HCl. The solution is made up to 250.0 cm³. A 25.00 cm³ portion needs 12.50 cm³ of 0.0500 mol/dm³ NaOH. Calculate the mass of Mg(OH)₂ in the tablet.

Equations: Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O HCl + NaOH → NaCl + H₂O

1. HCl added = 0.200 × 0.1000 = 0.0200 mol
2. NaOH used = 0.0500 × 0.01250 = 6.25 × 10⁻⁴ mol = HCl in 25.00 cm³ portion
3. HCl left in all 250.0 cm³ = 6.25 × 10⁻⁴ × 10 = 6.25 × 10⁻³ mol
4. HCl reacted = 0.0200 − 0.00625 = 0.01375 mol
5. Mg(OH)₂ = 0.01375 ÷ 2 = 6.875 × 10⁻³ mol
6. Mass = 6.875 × 10⁻³ × 58.32 = 0.401 g

The scaling step (× 10) is where most marks are lost. See also titrating an antacid tablet.

Worked example 3: nitrogen in a fertiliser

0.600 g of an ammonium sulfate fertiliser is warmed with 50.0 cm³ of 0.500 mol/dm³ NaOH until all the ammonia has been driven off. The remaining NaOH needs 32.45 cm³ of 0.500 mol/dm³ HCl. Calculate the percentage of nitrogen in the fertiliser, and its purity.

Equations: NH₄⁺ + OH⁻ → NH₃ + H₂O NaOH + HCl → NaCl + H₂O

1. NaOH added = 0.500 × 0.0500 = 0.0250 mol
2. HCl used = 0.500 × 0.03245 = 0.01623 mol = NaOH left over
3. NaOH reacted with NH₄⁺ = 0.0250 − 0.01623 = 0.00877 mol
4. NH₄⁺ = N = 0.00877 mol
5. Mass N = 0.00877 × 14.01 = 0.1229 g
6. % N = 0.1229 ÷ 0.600 × 100 = 20.5%

Pure ammonium sulfate, (NH₄)₂SO₄ (M = 132.14 g/mol), contains 2 × 14.01 ÷ 132.14 × 100 = 21.2% nitrogen. So the fertiliser is about 20.5 ÷ 21.2 × 100 = 97% pure ammonium sulfate.

This method works because ammonia is volatile: it’s driven off by heating, and a direct titration of the ammonium salt would be inaccurate.

Worked example 4: aspirin

Aspirin (acetylsalicylic acid, C₉H₈O₄, M = 180.16 g/mol) can be analysed by back titration because hydrolysing it with hot alkali uses 2 moles of NaOH per mole of aspirin: one to neutralise the carboxylic acid group and one to hydrolyse the ester group. Heating is needed for the hydrolysis to go to completion, so a direct titration wouldn’t work.

A 0.500 g aspirin tablet is heated with 50.0 cm³ of 0.200 mol/dm³ NaOH. After cooling, the excess NaOH needs 26.40 cm³ of 0.100 mol/dm³ H₂SO₄. Calculate the mass of aspirin in the tablet and its percentage by mass.

Equations: C₉H₈O₄ + 2NaOH → products H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

1. NaOH added = 0.200 × 0.0500 = 0.0100 mol
2. H₂SO₄ used = 0.100 × 0.02640 = 2.64 × 10⁻³ mol
3. NaOH left over = 2 × 2.64 × 10⁻³ = 5.28 × 10⁻³ mol
4. NaOH reacted with aspirin = 0.0100 − 0.00528 = 4.72 × 10⁻³ mol
5. Aspirin = 4.72 × 10⁻³ ÷ 2 = 2.36 × 10⁻³ mol
6. Mass = 2.36 × 10⁻³ × 180.16 = 0.425 g
7. Percentage = 0.425 ÷ 0.500 × 100 = 85%

The rest of the tablet is binders and fillers, which is typical.

Worked example 5: chloride by Volhard’s method

50.00 cm³ of 0.1000 mol/dm³ AgNO₃ is added to 25.00 cm³ of NaCl solution. The excess Ag⁺ needs 18.75 cm³ of 0.0800 mol/dm³ KSCN. Find [Cl⁻].

• Ag⁺ added = 5.000 × 10⁻³ mol
• Excess Ag⁺ = SCN⁻ = 1.500 × 10⁻³ mol
• Ag⁺ reacted = 3.500 × 10⁻³ mol = Cl⁻
• [Cl⁻] = 3.500 × 10⁻³ ÷ 0.02500 = 0.1400 mol/dm³

See precipitation titrations.

Common mistakes

1. Forgetting to subtract: using the titre moles as if they were the analyte moles.
2. Missing the scale-up when only a portion of the mixture was titrated.
3. Using the wrong ratio, especially for H₂SO₄ (2 NaOH per H₂SO₄), CaCO₃ and Mg(OH)₂ (2 HCl each) or aspirin (2 NaOH).
4. Not sense-checking: an analyte mass greater than the sample mass means an error.
5. Mixing up which reagent is in excess: the excess reagent is the one added first, not the titrant.

Key takeaways

• A back titration adds a known excess of reagent, then titrates what’s left.
• Reagent reacted = reagent added − reagent left over.
• Use two equations and apply each ratio carefully; scale up if only a portion was titrated.
• Back titrations suit insoluble, slow-reacting or volatile analytes.
• Always check that the answer is physically possible. For direct titrations, see titration calculations.