In 2019, Wine Australia's national quality audit found that 12% of 480 wine samples submitted by producers had reported total acidity values outside ±0.5 g/L of what independent laboratory titration found — errors large enough to mislabel acidity on the bottle. The root cause in every case: either a poorly standardised NaOH solution or non-concordant titres being averaged. This lesson covers the technique decisions that prevent both.
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
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?
Core Content
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.
Primary standard criteria: high purity, high molar mass, stable in air, non-hygroscopic, readily soluble. Na₂CO₃ (M = 106.0) standardises HCl; H₂C₂O₄·2H₂O (M = 126.1) standardises NaOH. NaOH fails: absorbs CO₂ and H₂O from air. HCl fails: concentration cannot be verified by weighing. Standard solution prep: weigh → dissolve → quantitative transfer to volumetric flask → make to mark.
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Why can NaOH NOT be used as a primary standard?
We just saw what a primary standard is and why NaOH cannot be one. That raises a question: Once you have your standard solution prepared, what specific technique is required for each piece of titration equipment — and why does each step exist? This card answers it → rinse burette with titrant, pipette with analyte, flask with distilled water only; read at bottom of meniscus at eye level; endpoint ≠ equivalence point.
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 |
Burette: rinse with titrant; read bottom of meniscus at eye level; record to ±0.05 mL; remove air bubbles from tip. Pipette: rinse with analyte; do not blow out last drop. Conical flask: rinse with distilled water ONLY — never analyte; white tile underneath. Endpoint (indicator colour change) ≠ equivalence point (stoichiometric completion) — coincide only with correct indicator choice.
Add the highlighted point to your notes before the check below.
A student rinses the conical flask with the analyte solution before adding the pipetted volume. What is the effect on the titre?
We just saw the technique requirements for each piece of apparatus. That raises a question: Once you have a concordant titre, what is the step-by-step calculation to find the unknown concentration? This card answers it → four steps: balanced equation + mole ratio → n(known) = cV → n(unknown) via ratio → c(unknown) = n/V; the mole ratio at Step 3 is where most marks are lost.
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) |
Four steps: (1) balanced equation + mole ratio → (2) n(known) = c × V (V in litres) → (3) n(unknown) = n(known) × ratio (from equation) → (4) c(unknown) = n/V. Concordant titres: within ±0.10 mL; average concordant only. Mole ratio always from balanced equation — never assume 1:1 (H₂SO₄ + 2NaOH: n(H₂SO₄) = n(NaOH)/2).
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For the titration H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O, if n(NaOH) = 2.40 × 10⁻³ mol, what is n(H₂SO₄)?
We just saw the four-step concentration calculation applicable to any acid-base titration. That raises a question: Which specific household substance titrations appear in the HSC prescribed investigation, and what are the indicator choices and calculation routes for each? This card answers it → vinegar: CH₃COOH + NaOH (1:1), phenolphthalein; antacid: NaHCO₃ + HCl (1:1), methyl orange.
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%.
Vinegar: CH₃COOH + NaOH (1:1); phenolphthalein (EP pH ≈ 8.7; range 8.3–10.0); % = (n × 60.06 / mass of sample) × 100%. Antacid: NaHCO₃ + HCl (1:1); methyl orange (EP pH acidic; range 3.1–4.4); % = (n × 84.01 / mass of tablet) × 100%. For liquids: mass = V × density.
Add the highlighted point to your notes before the check below.
Why is phenolphthalein (range 8.3–10.0) the correct indicator for the vinegar + NaOH titration?
We just saw the household substance titrations — vinegar and antacid. That raises a question: What are the specific HSC reporting and error-analysis standards that must appear in every practical response? This card answers it → concordant titres (±0.10 mL, average concordant only); error analysis (name + direction of effect + specific improvement); "human error" without specification earns zero marks.
A student titrates NaOH against HCl and records: 15.35, 15.60, 14.90, 15.40 mL. They average all four: 15.31 mL. The winemaker from the Think First would reject this immediately — 14.90 is non-concordant (differs by 0.50 mL from the others), and averaging it inflates the titre and underestimates [HCl]. In the 2019 Wine Australia audit, this exact error — averaging non-concordant titres — was the most common source of reportable inaccuracy. Here are the HSC standards that prevent it.
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.
Record burette readings to ±0.05 mL; record initial AND final for each titre. Concordant = within ±0.10 mL; average concordant only; exclude non-concordant titres. Error analysis: name → direction (titre too large/small → c too high/low) → specific improvement. "Human error" without specification earns zero HSC marks. Rough titration not included in concordance calculation.
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Titres recorded: 22.45, 22.50, 22.40, 23.10 mL. Which titres are concordant and should be averaged?
✏️ Worked Examples
A student prepares a standard solution by dissolving 1.325 g of anhydrous Na₂CO₃ (M = 106.0 g/mol) in distilled water and making up to exactly 250.0 mL in a volumetric flask. (a) Calculate the concentration of this Na₂CO₃ standard solution. (b) The student uses this solution to determine the concentration of an HCl solution by pipetting 25.00 mL of Na₂CO₃ into a conical flask and titrating with HCl. Three titres: 22.45 mL, 22.40 mL, 22.50 mL. Calculate c(HCl). Equation: Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂.
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) 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.
A student analyses commercial white vinegar. They pipette 10.00 mL of vinegar into a conical flask, add phenolphthalein, and titrate with 0.5000 mol/L NaOH. Four titres: 18.60 mL, 18.55 mL, 18.55 mL, 18.90 mL. Density of vinegar = 1.005 g/mL. M(CH₃COOH) = 60.06 g/mol. (a) Identify concordant titres and calculate the average. (b) Calculate c(CH₃COOH) in mol/L. (c) Calculate the percentage by mass of acetic acid. (d) Compare to the label "5.0% acidity" and suggest one specific source of error.
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
(a) Concordant titres:
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.
Average = (18.60 + 18.55 + 18.55) / 3 = 18.57 mL = 0.01857 L
(b) c(CH₃COOH):
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) = 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) = 9.285 × 10⁻³ × 60.06 = 0.5577 g
mass(vinegar) = 10.00 × 1.005 = 10.05 g
% = (0.5577 / 10.05) × 100% = 5.55%
(d) Comparison and error:
Experimental 5.55% vs label 5.0% — approximately 11% above stated value. Specific error: if the conical flask was rinsed with vinegar (rather than distilled water) before adding the pipetted volume, additional acetic acid remains on the walls — extra moles beyond the 10.00 mL pipetted. This increases n(CH₃COOH), making the titre too large → c overestimated → % overestimated. Improvement: rinse flask with distilled water only.
ANSWERS: (a) 18.90 excluded; average = 18.57 mL. (b) c = 0.9285 mol/L. (c) % = 5.55%. (d) Flask rinsed with analyte → titre too large → % overestimated; fix: distilled water rinse only.
A student determines the concentration of a NaOH solution using oxalic acid dihydrate (H₂C₂O₄·2H₂O, M = 126.1 g/mol) as the primary standard. They weigh out 0.6320 g, dissolve in distilled water, and make up to 100.0 mL in a volumetric flask. They pipette 20.00 mL of this standard into a conical flask and titrate with NaOH. Three titres: 19.85 mL, 19.80 mL, 20.15 mL. Equation: H₂C₂O₄ + 2NaOH → Na₂C₂O₄ + 2H₂O. (a) Calculate c(H₂C₂O₄). (b) Identify concordant titres and calculate c(NaOH). (c) The student forgot to rinse the burette with NaOH before filling. Explain the effect on titre and calculated c(NaOH). (d) Describe correct burette reading technique to eliminate parallax error.
GIVEN: mass = 0.6320 g; M = 126.1 g/mol; V(flask) = 0.1000 L; V(pipette) = 0.02000 L; titres = 19.85, 19.80, 20.15 mL
(a) c(H₂C₂O₄):
n = 0.6320 / 126.1 = 5.012 × 10⁻³ mol
c = 5.012 × 10⁻³ / 0.1000 = 0.05012 mol/L
(b) Concordant titres:
20.15 differs from 19.80 by 0.35 mL > 0.10 mL → exclude 20.15 mL.
Average = (19.85 + 19.80) / 2 = 19.83 mL = 0.01983 L
n(H₂C₂O₄) in 20.00 mL = 0.05012 × 0.02000 = 1.002 × 10⁻³ mol
n(NaOH) = 2 × 1.002 × 10⁻³ = 2.005 × 10⁻³ mol
c(NaOH) = 2.005 × 10⁻³ / 0.01983 = 0.1011 mol/L
(c) Effect of not rinsing burette:
Residual water dilutes the NaOH → effective [NaOH] lower than nominal. More volume of diluted NaOH needed to neutralise fixed moles of H₂C₂O₄ → titre too large. In calculation: c(NaOH) = n/V(titre). n is fixed by stoichiometry; V(titre) is too large → c(NaOH) underestimated. Improvement: rinse burette twice with 5–10 mL NaOH before filling.
(d) Correct burette reading technique:
(1) Position eye at exactly the same horizontal level as the liquid meniscus — not above or below. (2) Read the graduation mark at the bottom of the meniscus for colourless solutions. (3) Record to ±0.05 mL. Parallax error: if eye is above the meniscus, the reading is too high; if below, too low. If parallax differs between initial and final readings, the errors do not cancel — the titre contains a net systematic error.
ANSWERS: (a) c = 0.05012 mol/L. (b) 20.15 excluded; average = 19.83 mL; c(NaOH) = 0.1011 mol/L. (c) Water dilutes NaOH → titre too large → c(NaOH) underestimated. (d) Eye level with meniscus; read bottom of meniscus; prevents unequal parallax 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
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Activities
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, monoprotic). (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 appropriate for this titration and justify 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 = 14.38 mL.
Student 3: Fills the conical flask with the analyte from the pipette, then rinses the flask interior with the analyte solution "to make sure no analyte was lost on the glass."
Check Your Understanding
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 into a conical flask that had previously been rinsed with distilled water (not dried). They add phenolphthalein and titrate. How does the residual distilled water affect the result?
5. A student weighs 0.4820 g of H₂C₂O₄·2H₂O (M = 126.1 g/mol), dissolves to 100.0 mL, and titrates 20.00 mL aliquots against NaOH. Average titre = 15.12 mL. Equation: H₂C₂O₄ + 2NaOH → Na₂C₂O₄ + 2H₂O. What is c(NaOH)?
UnderstandBand 3(4 marks) 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.
ApplyBand 4–5(5 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 appropriate but phenolphthalein (range 8.3–10.0) is not.
EvaluateBand 5–6(6 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: 16.45 mL, 16.50 mL, 16.40 mL, 17.10 mL. Density of wine = 0.990 g/mL. Tartaric acid (M = 150.1 g/mol; diprotic: H₂T + 2NaOH → Na₂T + 2H₂O). (a) Identify concordant titres and calculate the average. (b) Calculate c(tartaric acid) in mol/L. (c) Calculate g/L of tartaric acid. (d) A burette tip air bubble likely affected one titre — identify which, explain the direction of error, and state how it was resolved.
Q1: C — NaOH absorbs CO₂ (forming Na₂CO₃) and water vapour — its effective [OH⁻] decreases unpredictably over time. The correct procedure is standardisation against H₂C₂O₄·2H₂O. Option A is wrong — NaOH concentration is unreliable within days. Option D gives only approximate results.
Q2: B — H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. 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. c = 1.130 × 10⁻³/0.02500 = 0.04520 mol/L. Option A inverts the ratio. Option C fails to convert mL to L.
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.
Q4: A — Distilled water in the flask does not change moles of analyte. Titration is based on moles, not concentration. The titre is unaffected. This is why rinsing with distilled water (not analyte) is correct technique.
Q5: D — c(H₂C₂O₄) = (0.4820/126.1)/0.1000 = 0.03822 mol/L. n in 20 mL = 0.03822 × 0.02000 = 7.644 × 10⁻⁴ mol. Ratio 1:2 → n(NaOH) = 1.529 × 10⁻³ mol. c(NaOH) = 1.529 × 10⁻³/0.01512 = 0.1011 mol/L.
(a) NaOH fails the stability criterion: (1) reacts with atmospheric CO₂: NaOH + CO₂ → Na₂CO₃ + H₂O — Na₂CO₃ is a weaker base, reducing effective [OH⁻]; (2) hygroscopic — absorbs H₂O, making accurate weighing impossible. Both make the label concentration unreliable.
(b) Primary standard: oxalic acid dihydrate (H₂C₂O₄·2H₂O, M = 126.1 g/mol). Accurately weigh; dissolve and make to known volume in volumetric flask; titrate against NaOH to determine actual concentration. Balanced equation: H₂C₂O₄(aq) + 2NaOH(aq) → Na₂C₂O₄(aq) + 2H₂O(l).
(a) n(HCl) = 0.1000 × 0.01965 = 1.965 × 10⁻³ mol; n(NaHCO₃) = 1.965 × 10⁻³ mol (1:1); c(NaHCO₃) = 1.965 × 10⁻³/0.02500 = 0.07860 mol/L.
(b) n total = 0.07860 × 0.2500 = 0.01965 mol; mass = 0.01965 × 84.01 = 1.651 g; % = (1.651/2.404) × 100% = 68.7%.
(c) NaHCO₃ + HCl titration (weak base + strong acid) has EP pH ≈ 3.7–4.0 (acidic — CO₂(aq) in solution). Methyl orange (3.1–4.4) encompasses this EP pH → appropriate. Phenolphthalein (8.3–10.0) would not change colour until far past the equivalence point → unsuitable; would significantly underestimate NaHCO₃ content.
(a) 17.10 mL non-concordant (0.70 mL above 16.40 mL > 0.10 mL); concordant: 16.45, 16.50, 16.40; average = 16.45 mL = 0.01645 L.
(b) n(NaOH) = 0.5000 × 0.01645 = 8.225 × 10⁻³ mol; H₂T:NaOH = 1:2; n(H₂T) = 4.113 × 10⁻³ mol; c(H₂T) = 0.4113 mol/L.
(c) g/L = 0.4113 × 150.1 = 61.7 g/L.
(d) The 17.10 mL titre most likely contained the air bubble. An air bubble in the burette tip creates a void — when released, the recorded volume is larger than actual liquid delivered → titre too large. Resolved by identifying 17.10 mL as non-concordant and excluding it from the average.
Student 1: Mole ratio backwards. 2HCl + Na₂CO₃ → products; HCl:Na₂CO₃ = 2:1, so n(Na₂CO₃) = n(HCl)/2 = 1.075 × 10⁻³ mol. Calculated n(Na₂CO₃) was doubled → c overestimated by ×2.
Student 2: 14.90 differs from 14.15 by 0.75 mL — non-concordant; must be excluded. Concordant: 14.15, 14.20, 14.25. Correct average = 14.20 mL. Including 14.90 inflated average by 0.18 mL → overestimated concentration.
Student 3: Rinsing the flask with analyte adds extra moles beyond pipetted volume → more titrant needed → titre too large → c(analyte) overestimated. Improvement: rinse with distilled water only.
Return to your Think First response about the winemaker's titration. Recall the 2019 Wine Australia audit: 12% of samples had acidity errors traced to poorly standardised NaOH and non-concordant titres. Can you now explain every decision the winemaker made in precise technical terms?
Review
A student is given an unknown concentration of H₂SO₄ and a standard 0.1000 mol/L NaOH solution. Describe the complete procedure, calculation, and error analysis for determining c(H₂SO₄) by titration. Include: (a) preparation and rinsing of equipment; (b) the full four-step calculation showing all working; (c) concordant titre analysis; (d) two specific sources of error with direction of effect and improvement; (e) explanation of indicator choice. (10 marks)
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