Chemistry • Year 12 • Module 6 • Lesson 14
Titration: Standard Solutions, Technique & Calculations
Apply the four-step titration method to real data, interpret a concordant-results table, and reason about technique errors.
1. Interpret titration results — concordant results and four-step calculation
A student standardises a NaOH solution by titrating it against a standard solution of potassium hydrogen phthalate (KHP, KHC8H4O4, M = 204.2 g/mol). KHP is a primary standard that reacts with NaOH in a 1:1 mole ratio: KHC8H4O4(aq) + NaOH(aq) → KNaC8H4O4(aq) + H2O(l).
The student dissolves 2.042 g of KHP and makes up to exactly 100.0 mL in a volumetric flask. They pipette 20.00 mL portions into the conical flask and titrate with the NaOH. The results are recorded below.
| Titration | Initial burette reading (mL) | Final burette reading (mL) | Titre (mL) | Concordant? |
|---|---|---|---|---|
| Rough (T0) | 0.00 | 19.70 | 19.70 | Excluded |
| Trial 1 (T1) | 0.10 | 19.85 | ||
| Trial 2 (T2) | 0.05 | 20.25 | ||
| Trial 3 (T3) | 0.15 | 19.95 |
1.1 Calculate each titre (T1, T2, T3) and write them in the table above. 3 marks
1.2 Identify which titres are concordant (within ±0.10 mL of each other) and mark the table. Calculate the average concordant titre. Show all working. 3 marks
1.3 Calculate the concentration of the KHP standard solution. 2 marks
1.4 Using the four-step method, calculate the concentration of the NaOH solution. Show each step labelled. 4 marks
2. Interpret a titration results graph — concordant results visualised
The graph below shows the four titres recorded by another student during the standardisation of HCl against a Na2CO3 standard. The horizontal dashed lines mark the concordance band (±0.10 mL around the median of the acceptable titres). 7 marks
Figure 2.1. Titration results for standardisation of HCl against Na2CO3 standard. Concordance band illustrated (±0.10 mL). Stylised data.
2.1 How many of the four recorded titres (T0–T3) are concordant? Identify them by name and state the range of the concordant titres. 2 marks
2.2 Explain why T0 is excluded from the concordance check even though no concordance rule requires it. 2 marks
2.3 If all four titres (T0–T3) were averaged instead of only the concordant ones, by approximately how much would the average titre differ from the correct concordant average? State whether this would make the calculated concentration of HCl too high or too low. 3 marks
3. Cause-and-effect chains — technique errors
Complete the effect chain for each technique error below. Each arrow represents the logical consequence of the step before it. 8 marks (2 each chain)
Chain 3.1 — Burette not rinsed with NaOH before filling:
Chain 3.2 — Conical flask rinsed with the vinegar analyte before pipetting:
4. Case study — acidity analysis at the Australian Wine Research Institute (AWRI)
The Australian Wine Research Institute (AWRI) in Adelaide provides standardised protocols for measuring titratable acidity (TA) in Australian wine. A winemaker analyses a 5.000 mL sample of Barossa Valley red wine. The wine’s total acidity is expressed as grams of tartaric acid (H2C4H4O6, M = 150.1 g/mol, diprotic) per litre. The standardised NaOH titrant has concentration 0.3330 mol/L. The balanced equation for the neutralisation is: H2C4H4O6(aq) + 2NaOH(aq) → Na2C4H4O6(aq) + 2H2O(l). Three titres: 8.40 mL, 8.45 mL, 8.40 mL. 8 marks
4.1 Confirm the three titres are concordant and calculate the average titre. 2 marks
4.2 Using the four-step method, calculate n(tartaric acid) in the 5.000 mL wine sample. 3 marks
4.3 Calculate the concentration of tartaric acid in the wine, expressed as g/L. Show the unit conversion. 3 marks
Q1 — KHP titration data
1.1 Titres: T1 = 19.85 − 0.10 = 19.75 mL; T2 = 20.25 − 0.05 = 20.20 mL; T3 = 19.95 − 0.15 = 19.80 mL.
1.2 Concordance: T1 = 19.75, T2 = 20.20, T3 = 19.80. Difference T2 − T1 = 0.45 mL > 0.10 mL ⇒ T2 is non-concordant. T3 − T1 = 0.05 mL ≤ 0.10 mL ⇒ T1 and T3 are concordant. Average concordant titre = (19.75 + 19.80) / 2 = 19.78 mL.
1.3 c(KHP): n(KHP) = mass / M = 2.042 / 204.2 = 1.000 × 10−2 mol. c(KHP) = 1.000 × 10−2 / 0.1000 = 0.1000 mol/L.
1.4 c(NaOH): Step 1: KHP + NaOH → products; ratio 1:1. Step 2: n(KHP) = 0.1000 × 0.02000 = 2.000 × 10−3 mol. Step 3: n(NaOH) = 2.000 × 10−3 mol. Step 4: c(NaOH) = 2.000 × 10−3 / 0.01978 = 0.1011 mol/L.
Q2 — Graph interpretation
2.1 Two titres are concordant: T1 (22.40 mL) and T2 (22.45 mL). Range = 22.45 − 22.40 = 0.05 mL ≤ 0.10 mL. T3 (22.80 mL) differs from T1 by 0.40 mL — non-concordant.
2.2 T0 is the rough titre, performed to approximately locate the endpoint before recording careful results. The student approaches the endpoint quickly in the rough titration and is likely to overshoot, so T0 is always excluded regardless of its numerical value. It is used only as a guide for where to slow the addition in subsequent titrations.
2.3 Concordant average = (22.40 + 22.45) / 2 = 22.425 mL. If all four were averaged: (22.90 + 22.40 + 22.45 + 22.80) / 4 = 22.638 mL. Difference = 22.638 − 22.425 = +0.21 mL. A larger average titre means more n(NaOH) is calculated, so the calculated c(HCl) (from Step 4: c = n/V) would be too low (an overestimate of n would not occur; V(titre) appears in the denominator for calculating c of the unknown species in the flask — see marking note). Marking note: c(unknown in flask) = n(unknown) / V(flask), where n(unknown) comes from n(standard used from burette). Larger V(titre) ⇒ more n(standard) ⇒ more n(unknown) in flask ⇒ calculated c(HCl) appears overestimated if HCl is in burette, but if HCl is in the flask and NaOH is in the burette, a larger V(titre) means more NaOH ⇒ more n(acid) assigned ⇒ overestimate. Accept a clear directional argument with correct reasoning.
Q3 — Cause-and-effect chains
Chain 3.1 — Burette not rinsed with NaOH: Residual water dilutes NaOH in burette → effective [NaOH] is lower than nominal → more volume of dilute NaOH needed to neutralise fixed moles of acid → titre is too large → calculated n(NaOH) is too large → n(unknown) is too large → c(unknown) is overestimated (too high).
Chain 3.2 — Flask rinsed with vinegar: Extra CH3COOH remains on flask walls → more moles of acid present than the pipetted volume alone → more NaOH needed to reach the endpoint → titre is too large → n(NaOH) calculated is too large → n(CH3COOH) assigned is too large → c(CH3COOH) is overestimated (too high).
Q4 — AWRI wine analysis
4.1 Titres: 8.40, 8.45, 8.40. Range = 8.45 − 8.40 = 0.05 mL ≤ 0.10 mL ⇒ all concordant. Average = (8.40 + 8.45 + 8.40) / 3 = 8.417 mL.
4.2 Step 1: H2C4H4O6 + 2NaOH → … Mole ratio 1:2. Step 2: n(NaOH) = 0.3330 × 0.008417 = 2.803 × 10−3 mol. Step 3: n(tartaric acid) = 2.803 × 10−3 / 2 = 1.401 × 10−3 mol.
4.3 mass(tartaric acid) in 5.000 mL sample = 1.401 × 10−3 × 150.1 = 0.2103 g. This is in 5.000 mL. Scale to 1000 mL: 0.2103 × (1000/5.000) = 42.06 g/L ≈ 42.1 g/L titratable acidity as tartaric acid (4 sig. fig.).