# AHL: Advanced stoichiometry and titration calculations

> IB Chemistry HL · R2: How much / how fast / how far?
> Source: https://www.owlsprep.com/study/ib-chemistry-hl-u5-ahl-advanced-stoichiometry-and-titration/

This sub-topic covers advanced titration calculations including back titrations, polyprotic acid titrations, percentage purity calculations, and limiting reactant problems common in IB Chemistry HL Paper 2.

**Prerequisites:** [Basic stoichiometry and mole concept](https://www.owlsprep.com/study/ib-chemistry-hl-u1-mole-concept-stoichiometry/); [Fundamental acid-base titration theory](https://www.owlsprep.com/study/ib-chemistry-hl-u8-acid-base-titration-theory/)

## Learning objectives

- Calculate reacting quantities for indirect back titration experiments
- Determine unknown concentrations for polyprotic acid titrations
- Calculate percentage purity of impure samples from titration data
- Solve multi-step limiting reactant titration problems

## Back Titration Calculations

**Back Titration** — An indirect titration technique used for insoluble, weak, or impure analytes. A known excess of standard reagent is added to the analyte, then the unreacted excess is titrated to find how much reagent reacted with the analyte.

*Example:* Used to find the mass of calcium carbonate in impure eggshell

Back titrations are used when direct titration is not possible, for example when the analyte is insoluble in water or a weak acid/base that gives a unclear endpoint. The key step is calculating moles of unreacted excess reagent, then subtracting from the initial moles of excess to find moles that reacted with the analyte.

**Worked example:** 2.50 g of impure calcium carbonate is reacted with 50.0 cm³ of 2.00 mol dm⁻³ hydrochloric acid. The excess HCl requires 36.5 cm³ of 1.50 mol dm⁻³ sodium hydroxide for neutralization. Calculate the moles of CaCO₃ in the sample.

1. Calculate initial moles of HCl added:

   $$n(HCl)_{initial} = c \times V = 2.00 \times \frac{50.0}{1000} = 0.100 \text{ mol}$$
2. Calculate moles of excess HCl from the NaOH titration. The reaction is 1:1, so moles of NaOH = moles of excess HCl:

   $$n(NaOH) = 1.50 \times \frac{36.5}{1000} = 0.05475 \text{ mol} = n(HCl)_{excess}$$
3. Calculate moles of HCl that reacted with CaCO₃:

   $$n(HCl)_{reacted} = 0.100 - 0.05475 = 0.04525 \text{ mol}$$
4. Use the 1:2 mole ratio from the balanced reaction $CaCO_3 + 2HCl \rightarrow CaCl_2 + CO_2 + H_2O$ to find moles of CaCO₃:

   $$n(CaCO_3) = \frac{0.04525}{2} = 0.0226 \text{ mol (3 sig figs)}$$

> **Exam tip:** Always write the balanced reaction between the analyte and excess reagent, the mole ratio is almost never 1:1 for back titrations.

## Polyprotic Acid Titration Calculations

**Polyprotic Acid** — An acid that can donate more than one proton (H⁺) to a strong base in neutralization, with each proton reacting stepwise at different pH values.

*Notation:* Diprotic = 2 acidic protons, Triprotic = 3 acidic protons

*Example:* Sulfuric acid ($H_2SO_4$) is diprotic, phosphoric acid ($H_3PO_4$) is triprotic

For titration calculations, the key detail is identifying how many protons are neutralized at the measured equivalence point. For full neutralization of an n-protic acid, the mole ratio of acid to strong base is always 1:n.

**Worked example:** A 25.0 cm³ sample of sulfuric acid is titrated with 0.100 mol dm⁻³ NaOH. 32.4 cm³ of NaOH is required to reach the second (final) equivalence point. Calculate the concentration of H₂SO₄.

1. Write the balanced full neutralization reaction:

   $$H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$$
2. Calculate moles of NaOH used:

   $$n(NaOH) = 0.100 \times \frac{32.4}{1000} = 0.00324 \text{ mol}$$
3. Use the 1:2 mole ratio of H₂SO₄ to NaOH:

   $$n(H_2SO_4) = \frac{0.00324}{2} = 0.00162 \text{ mol}$$
4. Calculate concentration of H₂SO₄:

   $$c(H_2SO_4) = \frac{n}{V} = \frac{0.00162}{0.0250} = 0.0648 \text{ mol dm}^{-3}$$

> **tip**
>
> If the question asks for the first equivalence point of a diprotic acid, only one proton reacts, so the mole ratio is 1:1, not 1:2. Always confirm which equivalence point is given in the question.

## Percentage Purity Calculations

Percentage purity is a very common exam question that asks for the percentage of pure target compound in an impure sample, calculated from titration data. The general formula is:

$$\text{Percentage purity} = \frac{\text{Mass of pure compound}}{\text{Total mass of impure sample}} \times 100\%$$

**Worked example:** From the earlier back titration example, 2.50 g of impure CaCO₃ gave 0.0226 mol of pure CaCO₃. Calculate the percentage purity ($M_r(CaCO_3) = 100.18$ g mol⁻¹).

1. Calculate the mass of pure CaCO₃:

   $$\text{Mass} = n \times M_r = 0.0226 \times 100.18 = 2.26 \text{ g}$$
2. Substitute into the percentage purity formula:

   $$\text{Percentage purity} = \frac{2.26}{2.50} \times 100 = 90.4\%$$

> **Exam tip:** Always report percentage purity to the same number of significant figures as the least precise given data, which is almost always 3 sig figs in IB exams.

## Mixture Analysis Titration Problems

Advanced titration problems often ask to find the composition of a mixture of two reactive compounds. These problems require setting up algebraic equations for total moles and total mass of products to solve for the unknown quantities of each component in the mixture.

**Worked example:** 1.80 g of a mixture of NaCl and NaBr is reacted with excess AgNO₃ to precipitate all halide ions as AgCl and AgBr. The total mass of the dry precipitate is 3.60 g. Calculate the percentage by mass of NaCl in the original mixture.

1. Let $x$ = mass of NaCl, so $(1.80 - x)$ = mass of NaBr. Moles of each halide equals moles of each silver precipitate:

   $$n(NaCl) = n(AgCl) = \frac{x}{58.44}, \quad n(NaBr) = n(AgBr) = \frac{1.80 - x}{102.89}$$
2. Set up the total mass equation, with $M(AgCl) = 143.32$ g mol⁻¹, $M(AgBr) = 187.77$ g mol⁻¹:

   $$\left(\frac{x}{58.44} \times 143.32\right) + \left(\frac{1.80 - x}{102.89} \times 187.77\right) = 3.60$$
3. Simplify and solve for $x$:

   $$2.452x + 1.825(1.80 - x) = 3.60 \rightarrow 0.627x = 0.315 \rightarrow x = 0.502 \text{ g}$$
4. Calculate percentage by mass of NaCl:

   $$\% NaCl = \frac{0.502}{1.80} \times 100 = 27.9\%$$

## Common pitfalls

- **Wrong:** Forgetting to account for the stoichiometric ratio when calculating analyte moles in back titration
  - Why it fails: Most analytes react with multiple moles of excess reagent, so skipping the ratio step gives a result double the correct value
  - Correct: Always write the full balanced reaction between the analyte and excess reagent before calculating moles of analyte
- **Wrong:** Using the 1:n ratio for polyprotic acids when only one proton is neutralized
  - Why it fails: Questions can ask for the first equivalence point of a polyprotic acid, where only one proton reacts, leading to half the correct concentration
  - Correct: Always check the question to confirm which equivalence point is being titrated before selecting the mole ratio
- **Wrong:** Dividing pure mass by pure mass or using pure mass as the denominator for percentage purity
  - Why it fails: Students forget the sample is impure, and mix up which mass corresponds to the whole sample
  - Correct: Remember percentage purity = (mass of pure / mass of total impure sample) × 100%, always confirm which mass is given
- **Wrong:** Using volume in cm³ directly in concentration calculations without converting to dm³
  - Why it fails: Concentration is given in mol dm⁻³, so units mismatch leads to a result that is 1000 times larger than the correct value
  - Correct: Always divide titration volume in cm³ by 1000 to convert to dm³ before substituting into $c = n/V$

## Cheatsheet

| Calculation Type | Key Rule/Formula | Exam Tip |
| --- | --- | --- |
| Back Titration | $n(analyte) = \frac{n(excess_{initial}) - n(excess_{unreacted})}{stoichiometric ratio}$ | Write the balanced reaction first |
| Polyprotic Acid | $n(base) = n(acid) \times \text{number of neutralized protons}$ | Confirm which equivalence point is given |
| Percentage Purity | $\% Purity = \frac{mass_{pure}}{mass_{impure}} \times 100\%$ | Match sig figs to given data |
| General Titration | $n = cV$, $V(dm^3) = \frac{V(cm^3)}{1000}$ | Always check mole ratios from balanced equations |

## What's next

Advanced titration stoichiometry is the foundation for almost all quantitative questions in IB Chemistry HL, from enthalpy change calculations to equilibrium constant determinations. Mastery of these techniques is critical for scoring full marks on multi-part Paper 2 questions, where these problems are typically worth 6-8 marks. The stoichiometric reasoning you developed here transfers directly to all other quantitative chemistry topics you will encounter in the rest of the syllabus. After completing this sub-topic, you can build on your knowledge with the following topics.

- [AHL: Rate laws and reaction order](https://www.owlsprep.com/study/ib-chemistry-hl-u5-ahl-rate-laws-and-reaction/)
- [AHL: Activation energy and Arrhenius equation](https://www.owlsprep.com/study/ib-chemistry-hl-u5-ahl-activation-energy-and-arrhenius/)
- [AHL: Extended Le Chatelier's principle](https://www.owlsprep.com/study/ib-chemistry-hl-u5-ahl-extended-le-chatelier-s/)

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