Chemistry • Year 11 • Module 2 • Lesson 10
Volumetric Analysis & Titration
Build HSC Band 5–6 extended-response technique on multi-step titration calculations, back-titration design, and evaluating experimental data in real industrial and pharmaceutical contexts.
1. Multi-step calculation — purity of a vinegar sample (Band 4–5)
8 marks Band 4–5
Scenario. A food scientist is testing a commercial apple cider vinegar sample. She dissolves 10.00 mL of vinegar in water and makes it up to 100.0 mL in a volumetric flask. A 20.00 mL aliquot is titrated against 0.05000 mol L−1 NaOH using phenolphthalein. Three accurate titres are: 24.60, 24.55, 24.58 mL. The equation is: CH3COOH + NaOH → CH3COONa + H2O. Molar mass of CH3COOH = 60.05 g mol−1. Density of the original vinegar = 1.004 g mL−1.
(a) Calculate the average concordant titre and apply all 4 steps to find n(CH3COOH) in the 20.00 mL aliquot. Show full working for each step. (4 marks)
(b) Calculate the total mass of CH3COOH in the original 10.00 mL of vinegar. Show the scale-up step explicitly. (2 marks)
(c) Calculate the percentage by mass of acetic acid in the vinegar. (1 mark)
(d) Australian Food Standards require vinegar to contain at least 4.0% w/w acetic acid. State whether this sample meets the standard. Identify one assumption you made in (b) that could affect the accuracy of this conclusion. (1 mark)
2. Experimental design — back-titration for impure limestone (Band 5–6)
8 marks Band 5–6
Research question. A geology student has a sample of impure limestone (~1.0 g) and wants to determine its percentage purity (CaCO3 content). CaCO3 does not dissolve readily in water but reacts readily with excess HCl:
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
The student has: an accurate mass of limestone (m0 g); 1.000 mol L−1 HCl (excess, 50.00 mL measured precisely); 0.1000 mol L−1 NaOH (standard); phenolphthalein; a 250 mL volumetric flask; a 25.00 mL pipette; and a burette. The CO2 is allowed to escape before titration. MM(CaCO3) = 100.09 g mol−1.
Q2. Design a complete back-titration procedure and show how to calculate % purity. In your response you must:
- Describe each step of the procedure in the correct order, including any rinsing steps and why they matter.
- State the purpose of using excess HCl in this design, not a precise stoichiometric amount.
- Write the balanced equation for the back-titration step (NaOH reacting with excess HCl).
- Show, algebraically, how n(CaCO3) is determined from the titration data.
- Write the formula for % purity in terms of the measured quantities.
- Identify one source of systematic error specific to this back-titration design and suggest how to minimise it.
3. Data + scenario — evaluating a student’s titration results (Band 5–6)
7 marks Band 5–6
Scenario. Two students, Alex and Jordan, each independently titrated 25.00 mL of an unknown HCl solution against 0.1000 mol L−1 NaOH. Their results are shown below. The equation is HCl + NaOH → NaCl + H2O.
| Student | Run | Titre (mL) | Note |
|---|---|---|---|
| Alex | Rough | 21.60 | Excluded |
| 1 | 20.40 | ||
| 2 | 20.35 | ||
| 3 | 20.80 | ||
| Jordan | Rough | 22.10 | Excluded |
| 1 | 20.45 | ||
| 2 | 20.48 | ||
| 3 | 20.50 |
Note: Alex forgot to rinse the burette with titrant before Run 1. The burette had been rinsed with distilled water.
Q3. Analyse and evaluate both students’ data to determine whose result is more reliable, and assess the impact of Alex’s rinsing error. In your response you must:
- Identify which of Alex’s titres are concordant, calculate the average, and calculate c(HCl) from Alex’s data.
- Calculate c(HCl) from Jordan’s data (all three titres are concordant).
- Explain, with reference to the concentration formula, why rinsing with water (not titrant) causes a systematic error in the direction you identify.
- Compare the two calculated concentrations and assess which is more reliable and why.
- Suggest one additional procedural improvement that would reduce random error for the next run.
Q1 — Vinegar purity calculation
(a) Concordance check: all three titres 24.55, 24.58, 24.60 mL; range = 0.05 mL ≤ 0.10 mL → all concordant. Average = (24.60 + 24.55 + 24.58) ÷ 3 = 24.58 mL = 0.02458 L.
Step 1: n(NaOH) = 0.05000 × 0.02458 = 1.229 × 10−3 mol
Step 2: 1:1 ratio (CH3COOH : NaOH) → n(CH3COOH) in aliquot = 1.229 × 10−3 mol
Steps 3/4 deferred to (b) below.
(b) Scale-up: n(CH3COOH) in 100.0 mL total = 1.229 × 10−3 × (100.0 ÷ 20.00) = 6.145 × 10−3 mol.
(This came from 10.00 mL of vinegar.)
m(CH3COOH) = 6.145 × 10−3 × 60.05 = 0.3690 g = 369.0 mg
(c) Mass of 10.00 mL vinegar = 10.00 × 1.004 = 10.04 g.
% w/w = (0.3690 ÷ 10.04) × 100 = 3.67%
(d) 3.67% < 4.0% minimum → the sample does not meet the Australian Food Standards requirement. Assumption: all acidity comes from acetic acid only (no other acids present); if other acids (e.g. malic acid) are present, the actual acetic acid % is lower than 3.67%, making the failure more significant.
Marking criteria: 1 mark per step (average titre; n(NaOH); scale-up; mass); 1 mark conclusion + assumption.
Q2 — Back-titration design (8 marks)
Step 1: Accurately weigh ~1.0 g of limestone sample on a 4-decimal-place balance; record mass as m0. [1]
Step 2: Pipette exactly 50.00 mL of 1.000 mol L−1 HCl into a 250 mL conical flask. Add the weighed limestone. Allow all fizzing (CO2) to cease — ensure all CaCO3 has reacted. Warming gently may accelerate the reaction. [1]
Step 3: Transfer quantitatively to the 250 mL volumetric flask and make up to the mark with deionised water. Mix thoroughly. [1]
Step 4: Rinse the pipette with the solution. Pipette 25.00 mL aliquot into a conical flask. Rinse the burette with 0.1000 mol L−1 NaOH before filling. Add 2–3 drops phenolphthalein. Titrate until a persistent pale pink persists. Repeat to concordance. [1]
Purpose of excess HCl: CaCO3 is a solid with limited and variable reactivity; using excess HCl guarantees all the CaCO3 dissolves completely. A stoichiometric amount might leave unreacted CaCO3, giving an underestimate of purity. [1]
Back-titration equation: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
Algebraic derivation:
n(HCl)initial = 1.000 × 0.05000 = 0.05000 mol
n(NaOH) in aliquot = c(NaOH) × Vtitre
n(HCl)excess in aliquot = n(NaOH) × 1/1 (1:1 ratio)
n(HCl)excess total = n(HCl)excess in aliquot × (250.0 ÷ 25.00)
n(HCl)reacted with CaCO3 = n(HCl)initial − n(HCl)excess total
n(CaCO3) = n(HCl)reacted ÷ 2 (from 1:2 ratio)
[1]
% purity formula: % purity = [n(CaCO3) × 100.09 ÷ m0] × 100% [1]
Systematic error: CO2 remaining dissolved in solution before making up to 250 mL would consume some NaOH during back-titration (CO2 + 2NaOH → Na2CO3 + H2O), overestimating excess HCl and underestimating purity. Minimise by boiling gently to expel all CO2 before transfer to the volumetric flask. [1]
Q3 — Evaluating student data (7 marks)
Alex’s data: Titres 20.40, 20.35, 20.80 mL. Concordance: |20.40 − 20.35| = 0.05 mL → concordant; |20.80 − 20.40| = 0.40 mL → NOT concordant. Average of concordant titres: (20.40 + 20.35) ÷ 2 = 20.38 mL.
c(HCl) from Alex = n(NaOH) ÷ V(HCl) = (0.1000 × 0.02038) ÷ 0.02500 = 0.0815 mol L−1. [1+1]
Jordan’s data: Titres 20.45, 20.48, 20.50 mL; range = 0.05 mL → all concordant. Average = (20.45 + 20.48 + 20.50) ÷ 3 = 20.48 mL.
c(HCl) = (0.1000 × 0.02048) ÷ 0.02500 = 0.0819 mol L−1. [1]
Water-rinsing systematic error (Alex): Water left in the burette dilutes the NaOH titrant below 0.1000 mol L−1. With lower [NaOH], a larger volume of NaOH must be added to neutralise the same moles of HCl. This overestimates the titre V for Run 1, making n(NaOH) apparently larger, and therefore making c(HCl) appear higher than the true value. This is a systematic error that shifts all results from a rinsed-with-water burette in one direction (higher titre, higher apparent c(HCl)). [2]
Comparison + reliability: Jordan’s data is more reliable because (1) all three accurate titres are concordant (smaller spread), (2) the burette was correctly prepared (no rinsing error). Alex’s Run 3 titre (20.80 mL) is non-concordant and may reflect the progressively diluted titrant in Run 1 affecting subsequent runs via back-contamination or inconsistent endpoint detection. Jordan’s three concordant titres averaged gives high precision. [1]
One procedural improvement: Perform at least three concordant runs (Jordan already did this) and additionally: clamp the burette vertically, read the meniscus at eye-level to eliminate parallax, or use a burette with a white backing stripe. Any valid improvement that targets random error is acceptable. [1]