Winemakers have used acid-base titration to measure wine acidity for over a century — the same procedure you will perform in the HSC prescribed investigation, the same four-step calculation, and the same sources of error that professional oenologists manage every day in production laboratories.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
A winemaker takes a 10.00 mL sample of her shiraz and titrates it against a standard 0.5000 mol/L NaOH solution. She adds NaOH drop by drop from a burette, swirling constantly. At exactly 16.45 mL of NaOH added, the solution turns faint pink and stays pink for more than 30 seconds. She records this as her endpoint, calculates the total acidity, and adjusts the blend accordingly.
Before reading on: What specific technique decisions did the winemaker make in this procedure that affect the accuracy of her result? Why did she stop at a "faint pink that stays pink for 30 seconds" rather than a deeper colour? And why does she need to know the NaOH concentration precisely before she begins?
By the end of this lesson you will be able to:
Every titration calculation depends on knowing the exact concentration of one solution before the experiment begins — and the reason this starting concentration must be established from a primary standard rather than taken on trust from a label is that most reagents, including NaOH, change concentration over time in ways that make the label unreliable.
A standard solution is a solution of precisely known concentration used as the reference reagent in a titration. To prepare one, a primary standard must be used — a substance that can be accurately weighed and dissolved to give a reliably known concentration. The five criteria for a primary standard:
Anhydrous sodium carbonate (Na₂CO₃, M = 106.0 g/mol) — used to standardise HCl. Stable (when freshly dried), high molar mass, non-hygroscopic. Reaction: Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂.
Oxalic acid dihydrate (H₂C₂O₄·2H₂O, M = 126.1 g/mol) — used to standardise NaOH. The dihydrate form is stable and non-hygroscopic (unlike anhydrous oxalic acid). Reaction: H₂C₂O₄ + 2NaOH → Na₂C₂O₄ + 2H₂O.
NaOH cannot be a primary standard — it absorbs CO₂ from air (NaOH + CO₂ → Na₂CO₃ + H₂O) and absorbs water vapour, both of which change its effective concentration. NaOH solutions must always be standardised against a primary standard before use in accurate titrations. Similarly, HCl cannot be a primary standard — concentrated HCl is a gas dissolved in water at an approximate concentration; the exact concentration cannot be verified by weighing.
Preparing a standard solution: (1) Accurately weigh primary standard on an analytical balance (4 decimal places). (2) Dissolve in a small volume of distilled water in a beaker. (3) Quantitatively transfer to a volumetric flask of exact volume (e.g. 250.0 mL) using a wash bottle to rinse all traces into the flask. (4) Make up to the calibration mark with distilled water and invert repeatedly to mix.
Every specific technique requirement in a titration exists for a quantitative reason — and being able to explain why each step is performed, not just describe what is done, is what distinguishes a Band 6 practical report from a Band 3 description.
Burette technique: The burette is a precision glass tube graduated to ±0.05 mL (0.01 mL graduations; read to nearest 0.05 mL). Before use: rinse twice with small volumes of the titrant solution — residual water would dilute the titrant and reduce its effective concentration. Fill above the zero mark, then open the stopcock to fill the tip and remove air bubbles — air bubbles in the tip produce an erroneously large delivered volume when released. Read the burette at the bottom of the meniscus, with the eye level with the liquid surface to eliminate parallax error. Record readings to ±0.05 mL (two decimal places: e.g. 16.45 mL). Titre = final reading − initial reading.
Pipette technique: A volumetric pipette delivers a single, fixed, accurately known volume (e.g. 10.00 mL or 25.00 mL). Before use: rinse twice with the analyte solution (not distilled water). Fill by suction using a pipette filler (never by mouth), drain to the calibration mark, touch the tip against the inside of the receiving flask to drain the last drop — do not blow out the remaining small volume (the calibrated volume assumes the last drop remains in the tip).
Conical flask technique: The conical flask holds the analyte. Use a conical flask rather than a beaker because: the tapered neck allows vigorous swirling without spilling; swirling mixes the reactants at the endpoint more effectively than stirring; the narrow opening reduces evaporation. The flask can be rinsed with distilled water during the titration — this does not change the moles of analyte. A white tile placed under the flask improves visibility of the indicator colour change.
Endpoint vs equivalence point: The equivalence point is the theoretical point at which moles of acid exactly equal moles of base (stoichiometric completion). The endpoint is the observed colour change of the indicator. A correct indicator choice makes these coincide; a wrong indicator produces a systematic error (endpoint ≠ equivalence point).
| Equipment | Key technique requirement | Reason | Error if not followed |
|---|---|---|---|
| Burette | Read at bottom of meniscus, eye level | Eliminates parallax error | Systematic error in all titre readings |
| Burette | Rinse with titrant before filling | Prevents dilution of titrant | Titrant more dilute → titre too large → c(unknown) overestimated |
| Burette | Fill tip; remove air bubbles | Volume delivered = volume calculated | Air bubble release → erroneously large titre |
| Pipette | Rinse with analyte solution | Prevents dilution of analyte | Fewer moles of analyte → titre too small → c(unknown) underestimated |
| Pipette | Do not blow out last drop | Last drop not in calibrated volume | Delivers more than stated volume → too many moles of analyte |
| Conical flask | Rinse walls with distilled water during titration | Moles of analyte unchanged | No error — this is safe practice |
| White tile | Place under flask | Improves endpoint visibility | Endpoint may be missed or judged too late |
Every titration concentration calculation follows the same four-step logic regardless of which acid and base are involved — learning the steps in order, rather than as isolated formulas, means the method transfers to any acid-base combination in the HSC.
Step 1: Write the balanced equation and identify the mole ratio. Step 2: Calculate moles of the known (standard) solution using n = c × V (V in litres). Step 3: Apply the mole ratio to find moles of the unknown: n(unknown) = n(known) × (coefficient of unknown / coefficient of known). Step 4: Calculate c(unknown) = n(unknown) / V(unknown) (V in litres).
The most common error occurs at Step 3. For H₂SO₄ + 2NaOH: if NaOH is in the burette and H₂SO₄ is in the flask, n(H₂SO₄) = n(NaOH)/2 — because 2 moles of NaOH react per mole of H₂SO₄. Inverting this ratio — writing n(H₂SO₄) = 2 × n(NaOH) — doubles the answer. Always write the balanced equation and read the mole ratio from it explicitly.
| Titration | Balanced equation | Mole ratio (acid:base) | Step 3 rule |
|---|---|---|---|
| HCl + NaOH | HCl + NaOH → NaCl + H₂O | 1:1 | n(HCl) = n(NaOH) |
| H₂SO₄ + NaOH | H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O | 1:2 | n(H₂SO₄) = n(NaOH)/2 |
| HCl + Na₂CO₃ | 2HCl + Na₂CO₃ → 2NaCl + H₂O + CO₂ | 2:1 | n(HCl) = 2 × n(Na₂CO₃) |
| H₂C₂O₄ + NaOH | H₂C₂O₄ + 2NaOH → Na₂C₂O₄ + 2H₂O | 1:2 | n(H₂C₂O₄) = n(NaOH)/2 |
| CH₃COOH + NaOH | CH₃COOH + NaOH → CH₃COONa + H₂O | 1:1 | n(CH₃COOH) = n(NaOH) |
The HSC prescribed investigation requires chemical analysis of a common household substance — both vinegar (acetic acid concentration) and antacid tablets (sodium hydrogen carbonate or calcium carbonate content) are standard examples that connect titration technique directly to consumer chemistry.
Vinegar analysis (% acetic acid by mass): Vinegar is a dilute solution of acetic acid (CH₃COOH) in water. Commercial vinegar is labelled "5% acidity" — approximately 5 g of acetic acid per 100 mL. To verify this: pipette 10.00 mL of vinegar into a conical flask; add 2–3 drops of phenolphthalein (appropriate because CH₃COOH + NaOH is weak acid + strong base — equivalence point pH ≈ 8.7, within phenolphthalein's range 8.3–10.0); titrate with standard NaOH until the first permanent faint pink colour persists for 30 seconds. Apply four-step method: write equation (CH₃COOH + NaOH → CH₃COONa + H₂O; 1:1); calculate n(NaOH) = c × V; n(CH₃COOH) = n(NaOH); mass(CH₃COOH) = n × 60.06 g/mol; % = (mass/mass of vinegar sample) × 100%.
Antacid analysis (NaHCO₃ or CaCO₃ content): Dissolve a known mass of antacid tablet in distilled water; titrate with standard HCl using methyl orange as indicator (NaHCO₃ + HCl reaction gives EP at acidic pH — within methyl orange's range 3.1–4.4). Equation: NaHCO₃ + HCl → NaCl + H₂O + CO₂ (1:1). Calculate n(HCl), then n(NaHCO₃) = n(HCl), then mass(NaHCO₃) = n × 84.01 g/mol, then % = (mass/mass of tablet) × 100%.
The HSC chemistry prescribed investigation for Module 6 is a titration — the marking criteria assess not just whether the numerical answer is correct but whether the procedure, data recording, and error analysis meet the standards of quantitative analytical chemistry.
Pre-titration requirements: Prepare or verify the standard solution concentration; rinse all equipment appropriately; check for air bubbles in burette tip; perform a rough titration to approximately locate the endpoint before recording concordant titres.
Recording titration data: Record all burette readings to two decimal places (±0.05 mL precision); record both initial and final readings for each titre; perform at least three titrations; identify concordant titres (within ±0.10 mL) and average only these; state clearly which titres are concordant and which are excluded as non-concordant.
Error analysis — HSC requirements: For every source of error identified: (1) name the specific error; (2) state the direction of its effect (does it make the titre too large or too small? Does it make the calculated concentration too high or too low?); (3) suggest a specific, actionable improvement. A named error without these two follow-on points earns partial marks. "Human error" and "parallax error" without specification earn no marks.
"NaOH cannot be a primary standard because it is a strong base." Basicity is irrelevant to primary standard criteria. NaOH fails because it is chemically unstable in air — absorbs CO₂ and H₂O, altering its concentration. The reason must reference instability, not strength.
"The conical flask should be rinsed with the analyte before adding the pipetted volume." Wrong — only rinse with distilled water. Rinsing with analyte leaves extra moles of analyte on the walls, increasing the titre and overestimating concentration.
"All titres should be averaged to reduce random error." Only concordant titres (within ±0.10 mL of each other) are averaged. Non-concordant titres indicate a technique error and must be excluded.
"Endpoint = equivalence point." The equivalence point is the stoichiometric point of complete neutralisation. The endpoint is the observed indicator colour change — these coincide only if the indicator is correctly chosen. A wrong indicator choice separates them, introducing systematic error.
"The mole ratio in Step 3 is always 1:1." The mole ratio comes from the balanced equation. H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O: the ratio is 1:2. Always write the balanced equation first and read the ratio explicitly — never assume 1:1.
GIVEN: mass(Na₂CO₃) = 1.325 g; M = 106.0 g/mol; V(flask) = 250.0 mL = 0.2500 L; V(pipette) = 25.00 mL = 0.02500 L; titres = 22.45, 22.40, 22.50 mL
FIND: (a) c(Na₂CO₃); (b) c(HCl)
(a) ANSWER — Standard solution concentration:
n(Na₂CO₃) = mass / M = 1.325 / 106.0 = 1.250 × 10⁻² mol
c(Na₂CO₃) = n / V = 1.250 × 10⁻² / 0.2500 = 0.05000 mol/L
(b) Concordant titres and average:
Check: 22.45, 22.40, 22.50 — largest difference = 22.50 − 22.40 = 0.10 mL ≤ 0.10 mL ✓ — all three concordant.
Average titre = (22.45 + 22.40 + 22.50) / 3 = 22.45 mL = 0.02245 L
Step 2 — moles of Na₂CO₃: n(Na₂CO₃) = 0.05000 × 0.02500 = 1.250 × 10⁻³ mol
Step 3 — mole ratio: Na₂CO₃ : HCl = 1 : 2 → n(HCl) = 2 × 1.250 × 10⁻³ = 2.500 × 10⁻³ mol
Step 4 — c(HCl): c = n / V = 2.500 × 10⁻³ / 0.02245 = 0.1114 mol/L
ANSWERS: (a) c(Na₂CO₃) = 0.05000 mol/L. (b) All three titres concordant; average = 22.45 mL; n(HCl) = 2.500 × 10⁻³ mol; c(HCl) = 0.1114 mol/L.
GIVEN: V(vinegar) = 10.00 mL = 0.01000 L; c(NaOH) = 0.5000 mol/L; titres = 18.60, 18.55, 18.55, 18.90 mL; density = 1.005 g/mL; M = 60.06 g/mol
FIND: concordant average; c(CH₃COOH); % by mass; error analysis
(a) Concordant titres:
Check: 18.90 mL differs from 18.55 by 0.35 mL and from 18.60 by 0.30 mL — both > 0.10 mL → exclude 18.90 mL (non-concordant).
Concordant: 18.60, 18.55, 18.55. Average = (18.60 + 18.55 + 18.55) / 3 = 55.70 / 3 = 18.57 mL = 0.01857 L
(b) c(CH₃COOH) — four-step method:
Step 1: CH₃COOH + NaOH → CH₃COONa + H₂O. Mole ratio 1:1.
Step 2: n(NaOH) = 0.5000 × 0.01857 = 9.285 × 10⁻³ mol
Step 3: n(CH₃COOH) = n(NaOH) × (1/1) = 9.285 × 10⁻³ mol
Step 4: c(CH₃COOH) = 9.285 × 10⁻³ / 0.01000 = 0.9285 mol/L
(c) Percentage by mass:
mass(CH₃COOH) in 10.00 mL = n × M = 9.285 × 10⁻³ × 60.06 = 0.5577 g
mass(vinegar sample) = V × density = 10.00 × 1.005 = 10.05 g
% by mass = (0.5577 / 10.05) × 100% = 5.55%
(d) Comparison and error analysis:
Experimental result = 5.55%; label = 5.0%. The experimental value is 0.55 percentage points above the label — approximately 11% above the stated value.
Specific source of error: If the conical flask was rinsed with the vinegar sample rather than distilled water before adding the pipetted volume, additional acetic acid would remain on the flask walls — adding extra moles of CH₃COOH beyond the 10.00 mL measured by the pipette. This would require more NaOH to reach the endpoint, making the titre too large. A larger titre → more n(NaOH) calculated → more n(CH₃COOH) assigned → overestimated concentration and percentage acidity.
Improvement: Rinse the conical flask with distilled water only before adding the pipetted volume; never rinse with the analyte solution.
ANSWERS: (a) 18.90 mL excluded; average = 18.57 mL. (b) n(NaOH) = 9.285 × 10⁻³ mol; n(CH₃COOH) = 9.285 × 10⁻³ mol; c = 0.9285 mol/L. (c) mass(CH₃COOH) = 0.5577 g; mass(vinegar) = 10.05 g; % = 5.55%. (d) Experimental 5.55% is above labelled 5.0%; possible cause: flask rinsed with analyte → extra CH₃COOH → titre too large → overestimated %; improvement: rinse flask with distilled water only.
GIVEN: mass = 0.6320 g; M = 126.1 g/mol; V(flask) = 100.0 mL = 0.1000 L; V(pipette) = 20.00 mL = 0.02000 L; titres = 19.85, 19.80, 20.15 mL
FIND: c(H₂C₂O₄); concordant average; c(NaOH); error direction; parallax technique
(a) ANSWER — c(H₂C₂O₄):
n(H₂C₂O₄·2H₂O) = mass / M = 0.6320 / 126.1 = 5.012 × 10⁻³ mol
Note: the water of crystallisation is included in M but does not change the moles of H₂C₂O₄ — each formula unit provides one diprotic acid molecule.
c(H₂C₂O₄) = 5.012 × 10⁻³ / 0.1000 = 0.05012 mol/L
(b) Concordant titres and c(NaOH):
Check: 20.15 differs from 19.85 by 0.30 mL and from 19.80 by 0.35 mL — both > 0.10 mL → exclude 20.15 mL.
Concordant: 19.85 and 19.80. Average = (19.85 + 19.80) / 2 = 19.83 mL = 0.01983 L
Step 2: n(H₂C₂O₄) in 20.00 mL = 0.05012 × 0.02000 = 1.002 × 10⁻³ mol
Step 3: H₂C₂O₄ : NaOH = 1 : 2 → n(NaOH) = 2 × 1.002 × 10⁻³ = 2.005 × 10⁻³ mol
Step 4: c(NaOH) = 2.005 × 10⁻³ / 0.01983 = 0.1011 mol/L
(c) Effect of not rinsing burette with NaOH:
If the burette contained residual water from a previous rinse, the NaOH loaded into the burette mixes with this water — diluting the NaOH below its nominal concentration. The effective [NaOH] in the burette is lower than 0.1011 mol/L.
Since the fixed moles of H₂C₂O₄ in the flask must still be neutralised, more volume of the diluted NaOH is required to deliver the same moles → the titre is too large.
In the calculation: c(NaOH) = n(NaOH) / V(titre). The n(NaOH) calculated is fixed by stoichiometry (from n(H₂C₂O₄)), but V(titre) is too large → c(NaOH) = n / V(too large) → calculated c(NaOH) is underestimated (too small).
Improvement: Rinse the burette twice with 5–10 mL of the NaOH solution before filling to the zero mark.
(d) Correct burette reading technique:
To eliminate parallax error: (1) position the eye at exactly the same horizontal level as the liquid meniscus — not above or below; (2) read the graduation mark that aligns with the bottom of the meniscus for colourless solutions; (3) record to ±0.05 mL (two decimal places).
Parallax error arises when the eye is above or below the liquid level — the line of sight intersects a different graduation mark than the one aligned with the meniscus. If the eye is above the meniscus, the graduation read is higher than the true level — producing a reading that is too high. If the eye is below, the reading is too low.
Critically: if the parallax error differs between the initial and final readings (e.g. eye level at initial but above level at final), the errors do not cancel — the titre (final − initial) contains a net systematic error that propagates into every calculated concentration value.
ANSWERS: (a) n = 5.012 × 10⁻³ mol; c(H₂C₂O₄) = 0.05012 mol/L. (b) 20.15 mL excluded (non-concordant); average = 19.83 mL; n(NaOH) = 2.005 × 10⁻³ mol; c(NaOH) = 0.1011 mol/L. (c) Water dilutes NaOH → effective [NaOH] lower → more volume needed → titre too large → c(NaOH) = n/V(large) → underestimated c(NaOH); improvement: rinse burette twice with titrant. (d) Eye level with meniscus; read bottom of meniscus; prevents unequal parallax error in initial and final readings that would produce a systematic titre error.
📷 Image slot — Titration apparatus burette, conical flask with white tile, correct eye-level meniscus reading position
📷 Image slot — Volumetric flask preparation: step-by-step transfer of primary standard from beaker to 250.0 mL flask
Write these by hand for retention.
Apply the four-step method to each scenario. Show all working including the balanced equation and mole ratio.
Scenario 1: A student prepares a standard solution by dissolving 0.9507 g of anhydrous Na₂CO₃ (M = 106.0 g/mol) in distilled water and making up to 250.0 mL. They pipette 20.00 mL into a conical flask and titrate with H₂SO₄. Four titres: 18.75 mL, 18.70 mL, 18.70 mL, 19.20 mL. (a) Calculate c(Na₂CO₃). (b) Identify concordant titres and calculate c(H₂SO₄). Equation: Na₂CO₃ + H₂SO₄ → Na₂SO₄ + H₂O + CO₂.
Scenario 2: A vitamin C tablet is dissolved in water and made up to 100.0 mL. A 25.00 mL aliquot is titrated against 0.1000 mol/L NaOH. Average titre = 23.45 mL. The tablet label states 1000 mg ascorbic acid (M = 176.1 g/mol, Ka ≈ 8 × 10⁻⁵, monoprotic for titration purposes). (a) Calculate the mass of ascorbic acid in the 100.0 mL solution. (b) Calculate the percentage difference between your experimental result and the label claim. (c) Identify one indicator that would be appropriate for this titration and justify your choice using the equivalence point pH.
Each student made at least one error. Identify the error, state its effect on the calculated concentration, and suggest the specific improvement.
Student 1: Titrating HCl (in burette) against Na₂CO₃ (in flask). They calculate: "n(HCl) = 0.1000 × 0.02150 = 2.150 × 10⁻³ mol; n(Na₂CO₃) = 2 × 2.150 × 10⁻³ = 4.300 × 10⁻³ mol."
Student 2: Records titres of 14.20 mL, 14.15 mL, 14.90 mL, 14.25 mL. They average all four: (14.20 + 14.15 + 14.90 + 14.25)/4 = 57.50/4 = 14.38 mL.
Student 3: Fills the conical flask with the analyte solution from the pipette, then rinses the flask interior with the analyte solution "to make sure no analyte was lost on the glass."
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
1. A student is about to use a NaOH solution labelled "0.1000 mol/L — prepared 3 months ago." Which action is most appropriate before proceeding?
2. A student titrates 25.00 mL of H₂SO₄ with 0.1000 mol/L NaOH and obtains an average titre of 22.60 mL. Which calculation correctly determines c(H₂SO₄)?
3. A student titrating vinegar with NaOH records titres of 16.30 mL, 16.20 mL, 16.25 mL, and 16.75 mL. They average all four to get 16.38 mL. Which statement correctly evaluates this approach?
4. A student pipettes the analyte solution from a pipette into a conical flask that had previously been rinsed with distilled water (not dried). They then add 2 drops of phenolphthalein and titrate. How does the distilled water remaining in the flask affect the result?
5. A student weighs 0.4820 g of H₂C₂O₄·2H₂O (M = 126.1 g/mol) and dissolves it to make 100.0 mL of standard solution. They titrate 20.00 mL aliquots against NaOH and obtain an average titre of 15.12 mL. Equation: H₂C₂O₄ + 2NaOH → Na₂C₂O₄ + 2H₂O. What is c(NaOH)?
Q6. A student wishes to prepare a 0.1000 mol/L NaOH standard solution. A teacher says this is impossible. (a) Explain why NaOH cannot be used as a primary standard. (b) Describe the correct procedure to obtain a NaOH solution of accurately known concentration, naming a suitable primary standard and writing the balanced equation for the standardisation reaction. (4 marks)
Q7. A student dissolves a 2.404 g antacid tablet (containing NaHCO₃) in distilled water and makes up to 250.0 mL. They pipette 25.00 mL aliquots and titrate against 0.1000 mol/L HCl using methyl orange. Average titre = 19.65 mL. Equation: NaHCO₃ + HCl → NaCl + H₂O + CO₂. (a) Calculate the concentration of NaHCO₃ in the 250.0 mL solution. (b) Calculate the percentage by mass of NaHCO₃ in the tablet. M(NaHCO₃) = 84.01 g/mol. (c) Explain why methyl orange (range 3.1–4.4) is an appropriate indicator for this titration but phenolphthalein (range 8.3–10.0) is not. (5 marks)
Q8. A winemaker measures the acidity of a red wine by titrating 10.00 mL samples with 0.5000 mol/L NaOH. Four titres are recorded: 16.45 mL, 16.50 mL, 16.40 mL, 17.10 mL. The density of the wine is 0.990 g/mL and its acidity is expressed as tartaric acid equivalents (M(tartaric acid) = 150.1 g/mol; the reaction is diprotic: H₂T + 2NaOH → Na₂T + 2H₂O). (a) Identify concordant titres and calculate the average. (b) Calculate the concentration of tartaric acid in the wine (mol/L). (c) Calculate the g/L of tartaric acid in the wine. (d) A second winemaker suggests the burette tip may have contained an air bubble during one of the titrations. Identify which titre this most likely affected, explain the direction of error, and state how this was resolved in the calculation. (6 marks)
Return to your Think First response. Can you now explain every decision the winemaker made in precise technical terms?
Q1: C — NaOH absorbs CO₂ (forming Na₂CO₃) and water vapour from air — its effective [OH⁻] decreases unpredictably over time. A 3-month-old NaOH solution cannot be trusted at face value. The correct procedure is standardisation against H₂C₂O₄·2H₂O — a primary standard — to determine actual concentration before use. Option A is wrong — NaOH concentration is unreliable within days of preparation. Option B is irrelevant — dilution does not solve the accuracy problem. Option D gives only approximate results, insufficient for quantitative analytical work.
Q2: B — Balanced equation: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Mole ratio H₂SO₄ : NaOH = 1:2. n(NaOH) = 0.1000 × 0.02260 = 2.260 × 10⁻³ mol. n(H₂SO₄) = 2.260 × 10⁻³/2 = 1.130 × 10⁻³ mol (divide by 2, NOT multiply). c(H₂SO₄) = 1.130 × 10⁻³/0.02500 = 0.04520 mol/L. Option A inverts the ratio (×2 instead of ÷2). Option C fails to convert mL to L — moles are 1000× too large. Option D uses 1:1 ratio, ignoring the diprotic nature of H₂SO₄.
Q3: B — Concordant criterion is ±0.10 mL. 16.75 differs from 16.20 by 0.55 mL — non-concordant; exclude. Correct average = (16.30 + 16.20 + 16.25)/3 = 16.25 mL. Averaging 16.38 overestimates the titre by 0.13 mL — this directly overestimates n(NaOH) and thus overestimates c(acid). Option A is wrong — non-concordant titres introduce systematic errors that averaging does not remove. Option D uses a ±1.0 mL criterion — the correct criterion is ±0.10 mL for burette precision.
Q4: A — Distilled water in the conical flask is irrelevant to the number of moles of analyte. The analyte was delivered by a calibrated pipette (e.g. 10.00 mL). Diluting it with water does not change n(analyte) — and titration is based on moles, not concentration. The titre is unaffected. This is why rinsing the conical flask with distilled water (not analyte solution) is the correct technique. Option B is wrong — concentration of analyte in the flask does not affect titre; Option D is wrong — the flask need not be dried.
Q5: D — c(H₂C₂O₄) = (0.4820/126.1)/0.1000 = 3.822 × 10⁻³/0.1000 = 0.03822 mol/L. n(H₂C₂O₄) in 20.00 mL aliquot = 0.03822 × 0.02000 = 7.644 × 10⁻⁴ mol. Balanced equation: H₂C₂O₄ + 2NaOH → Na₂C₂O₄ + 2H₂O → ratio 1:2 → n(NaOH) = 2 × 7.644 × 10⁻⁴ = 1.529 × 10⁻³ mol. c(NaOH) = 1.529 × 10⁻³/0.01512 = 0.1011 mol/L. Option A uses a 1:1 ratio (wrong). Option B uses a 1:4 ratio (wrong). The answer text in D works through the complete calculation.
(a) Why NaOH cannot be a primary standard (2 marks): NaOH fails the stability criterion in two ways. First, it reacts with atmospheric CO₂: NaOH + CO₂ → Na₂CO₃ + H₂O — the Na₂CO₃ formed is a weaker base, reducing the effective [OH⁻] below the stated concentration. Second, it is hygroscopic — it absorbs water vapour from air, increasing its mass and making accurate weighing impossible. Any concentration calculated from the mass of NaOH weighed out is therefore unreliable.
(b) Correct procedure (2 marks): Primary standard: oxalic acid dihydrate (H₂C₂O₄·2H₂O, M = 126.1 g/mol) — stable, non-hygroscopic, high molar mass. Procedure: accurately weigh a mass of H₂C₂O₄·2H₂O; dissolve in distilled water and make up to a known volume in a volumetric flask to prepare a standard solution; titrate aliquots of this standard against the NaOH solution to determine its actual concentration. Balanced equation: H₂C₂O₄(aq) + 2NaOH(aq) → Na₂C₂O₄(aq) + 2H₂O(l).
(a) c(NaHCO₃) (2 marks): n(HCl) = 0.1000 × 0.01965 = 1.965 × 10⁻³ mol. NaHCO₃ : HCl = 1:1 → n(NaHCO₃) = 1.965 × 10⁻³ mol. c(NaHCO₃) = 1.965 × 10⁻³/0.02500 = 0.07860 mol/L.
(b) % NaHCO₃ by mass (2 marks): n(NaHCO₃) total in 250.0 mL = 0.07860 × 0.2500 = 0.01965 mol. mass(NaHCO₃) = 0.01965 × 84.01 = 1.651 g. % by mass = (1.651/2.404) × 100% = 68.7%.
(c) Indicator justification (1 mark): NaHCO₃ + HCl titration (weak base + strong acid) has an equivalence point at pH < 7 — the solution at equivalence contains CO₂(aq) and H₂O, giving pH approximately 3.7–4.0. Methyl orange (range 3.1–4.4) changes colour within this pH range and is appropriate. Phenolphthalein (range 8.3–10.0) would not change colour until pH 8.3+ — far past the equivalence point. Using phenolphthalein would require excess acid to be added, destroying the NaHCO₃ and reducing the apparent titre, significantly underestimating the NaHCO₃ content.
(a) Concordant titres and average (1 mark): 17.10 mL differs from 16.40 mL by 0.70 mL > 0.10 mL — non-concordant; exclude. Concordant: 16.45, 16.50, 16.40 mL. Average = (16.45 + 16.50 + 16.40)/3 = 49.35/3 = 16.45 mL = 0.01645 L.
(b) c(tartaric acid) (2 marks): n(NaOH) = 0.5000 × 0.01645 = 8.225 × 10⁻³ mol. H₂T + 2NaOH → Na₂T + 2H₂O; ratio H₂T : NaOH = 1:2. n(H₂T) = 8.225 × 10⁻³/2 = 4.113 × 10⁻³ mol. c(H₂T) = 4.113 × 10⁻³/0.01000 = 0.4113 mol/L.
(c) g/L of tartaric acid (1 mark): g/L = c × M = 0.4113 × 150.1 = 61.7 g/L.
(d) Air bubble analysis (2 marks): The 17.10 mL titre is the most likely candidate for an air bubble error. An air bubble in the burette tip creates a void in the liquid path — when the stopcock is opened, the bubble is released along with the titrant, making the volume delivered appear larger than the actual liquid delivered. The recorded titre (17.10 mL) therefore overestimates the volume of NaOH actually dispensed. This error was resolved by identifying 17.10 mL as non-concordant (0.70 mL above the other titres) and excluding it from the average. Only the three concordant titres (16.40, 16.45, 16.50 mL) were averaged, eliminating the air bubble error from the calculation.
Student 1 Error: Mole ratio applied backwards. Equation: 2HCl + Na₂CO₃ → products — HCl : Na₂CO₃ = 2:1, so n(Na₂CO₃) = n(HCl)/2, NOT 2 × n(HCl). Correct: n(Na₂CO₃) = 2.150 × 10⁻³/2 = 1.075 × 10⁻³ mol. Effect: calculated n(Na₂CO₃) is doubled → c(Na₂CO₃) overestimated by a factor of 2. The student should write the balanced equation first and read the mole ratio explicitly before Step 3.
Student 2 Error: 14.90 mL differs from 14.15 mL by 0.75 mL — non-concordant; must be excluded. Concordant titres: 14.15, 14.20, 14.25 (all within 0.10 mL). Correct average = (14.15 + 14.20 + 14.25)/3 = 14.20 mL (not 14.38 mL). Including 14.90 inflated the average by 0.18 mL, overestimating the titre and overestimating the concentration of the unknown.
Student 3 Error: Rinsing the flask with analyte solution adds extra moles of analyte beyond those delivered by the pipette. The effective moles of analyte in the flask exceed the pipetted volume. Effect: more titrant is needed to reach the endpoint → titre too large → calculated concentration of analyte is overestimated. Improvement: rinse the conical flask with distilled water only (not with analyte solution). Distilled water does not change the moles of analyte and is therefore harmless to the titration result.
Salt Hydrolysis