Titration Calculations

Titrations are a quantitative analytical technique used to determine the concentration of an unknown solution by reacting it with a solution of known concentration. This lesson covers acid-base titrations, back titrations, and multi-step titration problems.

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The Titration Method

Apparatus

• Burette: Delivers variable, measured volumes of solution (to ± 0.05 cm³). The solution in the burette is called the titrant.
• Pipette: Delivers a fixed, precise volume (usually 25.0 cm³, to ± 0.06 cm³). The solution measured by the pipette is called the analyte.
• Conical flask: Where the reaction takes place.
• Indicator: Shows the end point (the point at which the indicator changes colour, signalling that the reaction is complete).

Common Indicators

Indicator	Colour in acid	Colour in alkali	Used for
Methyl orange	Red	Yellow	Strong acid + weak base
Phenolphthalein	Colourless	Pink	Weak acid + strong base
Either	—	—	Strong acid + strong base

Key Point: Neither indicator is suitable for weak acid + weak base titrations because the pH change at the equivalence point is too gradual.

Concordant Titres

You must repeat the titration until you obtain concordant titres — at least two titre values that are within 0.10 cm³ of each other. The mean titre is calculated from the concordant values only (discard any anomalous results, including the rough titration).

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Standard Titration Calculations

Method

1. Write the balanced equation for the reaction.
2. Calculate the moles of the substance whose concentration and volume are both known (usually the pipetted solution or the titrant).
3. Use the mole ratio to find moles of the unknown substance.
4. Calculate the concentration (or mass) of the unknown.

Worked Example 1: Simple Acid-Base Titration

25.0 cm³ of sodium hydroxide solution was titrated against 0.100 mol dm⁻³ hydrochloric acid. The mean titre was 22.5 cm³. Calculate the concentration of the sodium hydroxide.

NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)

Step 1: n(HCl) = c × V = 0.100 × 22.5/1000 = 2.25 × 10⁻³ mol

Step 2: Mole ratio NaOH : HCl = 1 : 1 n(NaOH) = 2.25 × 10⁻³ mol

Step 3: c(NaOH) = n / V = 2.25 × 10⁻³ / 0.0250 = 0.0900 mol dm⁻³

Worked Example 2: Diprotic Acid

25.0 cm³ of 0.120 mol dm⁻³ sodium carbonate solution was titrated against hydrochloric acid. The mean titre was 30.0 cm³. Calculate the concentration of the HCl.

Na₂CO₃(aq) + 2HCl(aq) → 2NaCl(aq) + H₂O(l) + CO₂(g)

n(Na₂CO₃) = 0.120 × 0.0250 = 3.00 × 10⁻³ mol

Mole ratio Na₂CO₃ : HCl = 1 : 2

n(HCl) = 2 × 3.00 × 10⁻³ = 6.00 × 10⁻³ mol

c(HCl) = 6.00 × 10⁻³ / 0.0300 = 0.200 mol dm⁻³

Worked Example 3: Sulfuric Acid and NaOH

25.0 cm³ of sulfuric acid was neutralised by 20.0 cm³ of 0.150 mol dm⁻³ NaOH. Calculate the concentration of the sulfuric acid.

H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)

n(NaOH) = 0.150 × 0.0200 = 3.00 × 10⁻³ mol

Mole ratio H₂SO₄ : NaOH = 1 : 2

n(H₂SO₄) = 3.00 × 10⁻³ / 2 = 1.50 × 10⁻³ mol

c(H₂SO₄) = 1.50 × 10⁻³ / 0.0250 = 0.0600 mol dm⁻³

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Finding Molar Mass or Identity from Titration Data

Worked Example 4: Finding Molar Mass

1.325 g of a monobasic acid HA was dissolved in water to make 250 cm³ of solution. 25.0 cm³ of this solution required 18.5 cm³ of 0.108 mol dm⁻³ NaOH for complete neutralisation. Calculate the molar mass of HA.

HA(aq) + NaOH(aq) → NaA(aq) + H₂O(l)

n(NaOH) = 0.108 × 18.5/1000 = 2.00 × 10⁻³ mol

Mole ratio HA : NaOH = 1 : 1

n(HA) in 25.0 cm³ = 2.00 × 10⁻³ mol

n(HA) in 250 cm³ = 2.00 × 10⁻³ × (250/25.0) = 0.0200 mol

M(HA) = m / n = 1.325 / 0.0200 = 66.3 g mol⁻¹

Exam Tip: When the solute is dissolved in a volumetric flask and then a pipette is used to take a portion (aliquot), remember to scale up the moles to the total volume. This is a very common calculation step that students forget.

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Back Titrations

A back titration is used when:

• The substance being analysed is insoluble in water (e.g., calcium carbonate).
• The substance reacts slowly with the titrant.
• Direct titration would give an unclear end point.

Method

1. React the analyte with a known excess of reagent A.
2. Titrate the unreacted excess of reagent A with reagent B.
3. Calculate moles of A originally added, subtract moles of A that reacted with B, to find moles of A that reacted with the analyte.

Worked Example 5: Back Titration