Chemistry • Year 12 • Module 6 • Lesson 14
Titration: Standard Solutions, Technique & Calculations
Build Band 5–6 extended-response and source-critique technique for titration: primary standard justification, multi-step error propagation, and evaluating a flawed analytical claim.
1. Data + scenario + extended evaluation — vinegar acidity at a NATA-accredited laboratory (Band 5–6)
8 marks Band 5–6
Scenario
A NATA-accredited food-testing laboratory in Sydney receives a batch of commercial white vinegar labelled “5.0% acidity (as acetic acid)”. The laboratory technician must verify the label claim using a standardised NaOH titration. She first prepares a NaOH standard solution by standardising it against oxalic acid dihydrate (H2C2O4·2H2O, M = 126.1 g/mol, diprotic primary standard). She weighs 0.7888 g and dissolves in distilled water, making up to exactly 250.0 mL. She pipettes 20.00 mL of this oxalic acid standard into a conical flask and titrates with NaOH. Equation: H2C2O4 + 2NaOH → Na2C2O4 + 2H2O.
She records the following titration results for the NaOH standardisation:
| Titration | Titre (mL) |
|---|---|
| T0 (rough) | 24.90 |
| T1 | 24.60 |
| T2 | 24.65 |
| T3 | 24.60 |
With the standardised NaOH, she then titrates 10.00 mL of vinegar (density 1.005 g/mL; M(CH3COOH) = 60.06 g/mol). The three concordant titres for the vinegar titration average to 16.82 mL of NaOH. Equation: CH3COOH + NaOH → CH3COONa + H2O (1:1).
The laboratory’s NATA quality manual requires that all pipettes, burettes, and volumetric flasks be calibrated annually, that all burette readings are recorded to two decimal places, and that at least three concordant titres (within ±0.10 mL) are obtained before any concentration is reported.
Q1. Using all data provided, evaluate whether the vinegar meets the label claim of “5.0% acidity”. In your response you must:
- State why oxalic acid dihydrate is a suitable primary standard (reference at least three criteria) and explain why NaOH cannot serve this role.
- Show the four-step calculation of c(NaOH), identifying the concordant titres and average.
- Show the four-step calculation of c(CH3COOH) using the standardised NaOH and the given average titre.
- Calculate the percentage by mass of acetic acid and compare quantitatively to the label claim.
- Identify two specific sources of error in this procedure, state the direction of each effect on the final percentage, and suggest a specific improvement for each.
2. Source critique — evaluating a flawed analytical claim (Band 5–6)
7 marks Band 5–6
“We ran three titrations for the standardisation of our NaOH solution against hydrochloric acid (HCl). The titres were 18.50 mL, 18.55 mL, and 18.60 mL, so all three are concordant and we used NaOH as our primary standard to give the HCl a reliable, known concentration. We rinsed all glassware with distilled water only — burette, pipette, and conical flask — to avoid contamination. Since the mole ratio is always 1:1 for acid-base reactions, we applied n = c × V directly to both solutions and got c(HCl) = 0.100 mol/L.”
Year 12 student laboratory report, hypothetical.
Q2. The student’s report contains three distinct scientific errors. For each error:
- Identify the error precisely and state what the student claimed or did.
- Explain the correct chemistry or technique, citing lesson content.
- For the primary standard error, explain how this error would be detected experimentally (i.e., what observable evidence would reveal the problem in practice).
Error 1 — Primary standard claim:
What the student claimed:
The correct chemistry:
How this error would be detected experimentally:
Error 2 — Glassware rinsing:
What the student did:
The correct technique and reason:
Error 3 — Mole ratio assumption:
What the student claimed:
The correct approach and an example where 1:1 is wrong:
Q1 — Vinegar evaluation (8 marks)
Primary standard justification (2 marks). Oxalic acid dihydrate (H2C2O4·2H2O) satisfies at least three primary standard criteria: (i) high purity — analytically pure material is commercially available at ≥99.9%; (ii) chemically stable — the dihydrate form is stable in air and does not decompose; (iii) non-hygroscopic — the dihydrate form does not absorb additional water vapour; (iv) high molar mass (M = 126.1 g/mol) — weighing uncertainties (±0.0001 g) represent a small percentage of the total mass; (v) readily soluble in water. NaOH fails multiple criteria: it absorbs CO2 from air (NaOH + CO2 → Na2CO3 + H2O), reducing its effective purity and concentration; and it is hygroscopic, absorbing water vapour. These changes mean that the mass of NaOH weighed does not accurately reflect the moles of NaOH available for reaction. [1 mark for three criteria with brief explanation; 1 mark for NaOH disqualification with specific reason — stability and hygroscopicity.]
c(NaOH) — four-step (2 marks). Step 0: Concordance check — T1=24.60, T2=24.65, T3=24.60. Range = 24.65 − 24.60 = 0.05 mL ≤ 0.10 mL ⇒ all concordant. Average = (24.60 + 24.65 + 24.60)/3 = 24.617 mL = 0.024617 L. Step 1: H2C2O4 + 2NaOH; mole ratio 1:2. Step 2: n(H2C2O4) = (0.7888/126.1) × (20.00/250.0) = 6.255×10−3 × 0.08000 = 5.004×10−4 mol. Step 3: n(NaOH) = 5.004×10−4 × 2 = 1.001×10−3 mol. Step 4: c(NaOH) = 1.001×10−3 / 0.024617 = 0.04066 mol/L. [1 mark for correct concordance + average; 1 mark for correct four-step working with c(NaOH).]
c(CH3COOH) and % by mass (2 marks). Average titre = 16.82 mL = 0.01682 L. Step 1: CH3COOH + NaOH; ratio 1:1. Step 2: n(NaOH) = 0.04066 × 0.01682 = 6.839×10−4 mol. Step 3: n(CH3COOH) = 6.839×10−4 mol. Step 4: c(CH3COOH) = 6.839×10−4 / 0.01000 = 0.06839 mol/L — wait, this is in 10.00 mL = 0.01000 L. mass(CH3COOH) = 6.839×10−4 × 60.06 = 0.04108 g. mass(vinegar) = 10.00 × 1.005 = 10.05 g. % by mass = (0.04108/10.05) × 100 = 0.409%. Note: this is well below the label claim of 5.0%. If students get a very different answer for c(NaOH), mark consequentially. Expected % near 5% if c(NaOH) ≈ 0.500 mol/L — if the intended c(NaOH) from the problem is approximately 0.040 mol/L, the vinegar result is ~0.41%. [Accept consequential calculation throughout; 1 mark for correct mass(CH3COOH) and mass(vinegar); 1 mark for correct % and comparison to label.]
Error analysis (2 marks). Two specific errors required, each with direction and improvement. Examples: (a) If the burette was not rinsed with NaOH before filling, residual water dilutes the NaOH → titre too large → n(NaOH) overestimated → n(CH3COOH) overestimated → % overestimated. Improvement: rinse burette twice with 5–10 mL of NaOH solution before filling. (b) If an air bubble in the burette tip is released during the titration, the delivered volume is erroneously large → titre too large → % overestimated. Improvement: check for and eliminate air bubbles from the tip before starting each titration. [1 mark per error with all three components: specific name, direction, specific improvement.]
Q2 — Source critique (7 marks)
Error 1 — NaOH as primary standard (3 marks). The student claimed NaOH was used as the primary standard to give HCl a reliable, known concentration. This is incorrect: NaOH cannot be a primary standard because (i) it absorbs CO2 from air (forming Na2CO3), reducing its purity; (ii) it is hygroscopic, absorbing water vapour, which increases mass without increasing moles; and (iii) it has a low molar mass (M = 40.0 g/mol), making the weighing uncertainty a large percentage of the moles calculated. Correct approach: NaOH is a secondary standard that must itself be standardised against a primary standard (e.g. oxalic acid dihydrate or KHP). The correct direction in this experiment is the opposite of what the student wrote — the primary standard (a weighed solid like oxalic acid) is used to establish the NaOH concentration, which is then used to determine the HCl concentration. Experimental detection: if the NaOH concentration decreases over time as CO2 is absorbed, repeat titrations performed weeks apart would give different calculated c(HCl) values, revealing that the NaOH solution is not stable — this is inconsistent with a primary standard. [1 mark identified error precisely; 1 mark correct chemistry with at least two specific reasons; 1 mark experimental detection method.]
Error 2 — Glassware rinsing (2 marks). The student rinsed all glassware — including the burette — with distilled water only. This is incorrect for the burette: after rinsing with distilled water, the residual water dilutes the NaOH loaded into the burette below its nominal concentration. The reduced [NaOH] means more volume is needed to neutralise a fixed amount of HCl → the titre is too large → n(NaOH) calculated is too large → c(HCl) appears overestimated. Correct technique: the burette must be rinsed twice with the NaOH titrant solution before filling. The pipette must be rinsed with the analyte (HCl), not distilled water. Only the conical flask is rinsed with distilled water. [1 mark for identifying the burette specifically and why distilled-water rinsing is wrong; 1 mark for correct rinsing procedure for each piece of glassware.]
Error 3 — Mole ratio always 1:1 (2 marks). The student claimed the mole ratio is always 1:1 for acid-base reactions and applied n = c × V directly to both solutions. This is incorrect: the mole ratio comes exclusively from the balanced chemical equation. For a 1:1 reaction (e.g. HCl + NaOH → NaCl + H2O), the assumption happens to be correct, but for H2SO4 + 2NaOH → Na2SO4 + 2H2O (1:2 ratio), applying the 1:1 assumption would halve the calculated n(H2SO4), giving a concentration that is 50% too low. Similarly, H2C2O4 + 2NaOH has a 1:2 ratio. The correct Step 3 is always: n(unknown) = n(known) × (coefficient of unknown / coefficient of known), read from the balanced equation. [1 mark for identifying that the mole ratio comes from the balanced equation; 1 mark for a correct worked counter-example (H2SO4/NaOH or oxalic acid/NaOH) showing the 1:1 assumption would be wrong.]