Year 12 Chemistry Module 6, Acid/Base Reactions ⏱ ~45 min Lesson 14 of 19 IQ3

Titration Calculations, Standard Solutions, Concordant Titres & Errors

Commercial wine producers report total acidity from titration, and a result that drifts from what an independent laboratory finds usually traces back to the same two causes: a poorly standardised NaOH solution, and averaging non-concordant titres. Both are avoidable, and both are exactly what this lesson is abou

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Today's hook, When a wine's reported acidity disagrees with an independent laboratory, the cause is almost always a poorly standardised NaOH solution or averaged non-concordant titres. How do standard solutions, primary standards, and concordant titre averaging prevent these errors?
0/5QUESTS
1
You’re here

Get oriented

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

Think First, The Winemaker's Titration

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?

Safety
Titration is a supervised practical with a risk assessment. Standard acids and bases (HCl, NaOH) are corrosive, so wear safety glasses and gloves and clean up spills immediately. Always fill a pipette using a pipette filler or bulb, never by mouth. Fill and clamp the burette at or below eye level (use a funnel, then remove it), and check there is no air bubble in the burette tip. Phenolphthalein and other indicator solutions are made up in ethanol, which is flammable, so keep them away from flames. Run waste acid and base to the labelled waste container, not the sink, unless your teacher directs otherwise.
📐 Titration Formulas, Standard Solutions & Calculations
n = c × V    (V must be in litres)
n, moles (mol) c, concentration (mol/L) V, volume in litres (convert mL ÷ 1000)
Four-step titration calculation method:
Step 1, Write balanced equation and identify mole ratio Step 2, n(known) = c(known) × V(known)  [V in litres] Step 3, n(unknown) = n(known) × (coefficient of unknown / coefficient of known) Step 4, c(unknown) = n(unknown) / V(unknown)  [V in litres]
Standard solution concentration:
n(primary standard) = mass / M c(standard) = n / V(volumetric flask)  [V in litres] Primary standard criteria: high purity, high molar mass, chemically stable, non-hygroscopic, readily soluble Na₂CO₃ (M = 106.0 g/mol) standardises HCl. H₂C₂O₄·2H₂O (M = 126.1 g/mol) standardises NaOH.
Percentage by mass of active ingredient:
mass(active) = n(active) × M(active)  [g] mass(sample) = volume(sample) × density  [g] % by mass = (mass(active) / mass(sample)) × 100% For liquids (vinegar): use density to convert volume to mass. For tablets: use mass of tablet directly.
Learning Intentions
Know

Key facts

  • A standard solution has accurately known concentration
  • Primary standards must be: pure, stable, non-hygroscopic, soluble, high molar mass
  • The four-step titration calculation: n = cV → mole ratio → n(unknown) → c(unknown)
Understand

Concepts

  • Why NaOH and HCl cannot be primary standards
  • The difference between equivalence point and endpoint
  • How technique errors affect accuracy and precision in titration
Can do

Skills

  • Select appropriate primary standards for acid-base titrations
  • Perform titration calculations using the four-step method
  • Describe correct burette, pipette, and conical flask technique
Key Terms
Titration
A volumetric technique to determine the concentration of a solution by reacting it with a standard solution of known concentration.
Standard solution
A solution of accurately known concentration; prepared by dissolving a precisely weighed primary standard.
Primary standard
A highly pure, stable, non-hygroscopic substance used to prepare or calibrate standard solutions (e.g., Na₂CO₃, oxalic acid).
Equivalence point
The point in a titration where stoichiometrically equivalent amounts of acid and base have reacted.
Titre
The volume of standard solution delivered from the burette to reach the endpoint; used in concentration calculations.
Concordant titres
Titres within ±0.10 mL of each other; only concordant results are averaged in the calculation.
2

Standard solutions and primary standards

Cross-lesson links: Titration calculations here apply the neutralisation equations from L02–L04 and the Henderson-Hasselbalch equation from L13. Standard solution preparation using primary standards is the quantitative prerequisite for titration curves (L15–L16). Wine acidity analysis reappears in industrial food analysis (L19).
1
Standard Solutions and Primary Standards

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:

  1. High purity (≥ 99.9%), so that mass accurately reflects moles
  2. High molar mass so that small weighing errors (±0.0001 g on an analytical balance) produce negligible percentage errors in moles
  3. Chemical stability must not decompose, react with air, or change composition while being weighed or stored
  4. Non-hygroscopic must not absorb water vapour from air, which would increase mass and give incorrect moles
  5. Readily soluble in water to give a stable, homogeneous solution

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.

High purity

Anhydrous Na₂CO₃:
H₂C₂O₄·2H₂O:
NaOH (NOT a primary standard): ✗ (absorbs CO₂ and H₂O)

High molar mass

Anhydrous Na₂CO₃: ✓ (106.0 g/mol)
H₂C₂O₄·2H₂O: ✓ (126.1 g/mol)
NaOH (NOT a primary standard): ✗ (M = 40.0 g/mol, small weighing errors are significant)

Stable in air

Anhydrous Na₂CO₃: ✓ (when freshly dried)
H₂C₂O₄·2H₂O:
NaOH (NOT a primary standard): ✗ (reacts with CO₂)

Non-hygroscopic

Anhydrous Na₂CO₃:
H₂C₂O₄·2H₂O:
NaOH (NOT a primary standard): ✗ (absorbs H₂O from air)

Readily soluble

Anhydrous Na₂CO₃:
H₂C₂O₄·2H₂O:
NaOH (NOT a primary standard): ✓ (irrelevant, fails other criteria)

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.

Must Do
When explaining why a substance is or is not suitable as a primary standard, address each relevant criterion specifically, purity, molar mass, stability, hygroscopicity, and solubility. "NaOH is not a primary standard because it absorbs CO₂ from air, reducing its purity and making its effective concentration unreliable, it must be standardised against a primary standard before use" is the minimum complete response.
Common Error
"NaOH cannot be a primary standard because it is a strong base." Basicity is not a criterion for primary standard suitability. The issue is physical and chemical stability, NaOH absorbs CO₂ and H₂O from air. Similarly, students sometimes list HCl as a primary standard, it cannot be, because its concentration cannot be verified by mass.
Exam Tip
In acid-base calculations, always write the balanced equation first, identify the conjugate pair, and state any assumptions before substituting into the Ka expression.

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.

Pause, copy the highlighted definition into your book before moving on.

Why can NaOH NOT be used as a primary standard?

3

Burette, pipette and flask technique

2
Burette, Pipette, and Conical Flask Technique

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).

EquipmentKey technique requirementReasonError if not followed
BuretteRead at bottom of meniscus, eye levelEliminates parallax errorSystematic error in all titre readings
BuretteRinse with titrant before fillingPrevents dilution of titrantTitrant more dilute → titre too large → c(unknown) overestimated
BuretteFill tip; remove air bubblesVolume delivered = volume calculatedAir bubble release → erroneously large titre
PipetteRinse with analyte solutionPrevents dilution of analyteFewer moles of analyte → titre too small → c(unknown) underestimated
PipetteDo not blow out last dropLast drop not in calibrated volumeDelivers more than stated volume → too many moles of analyte
Conical flaskRinse walls with distilled water during titrationMoles of analyte unchangedNo error, this is safe practice
White tilePlace under flaskImproves endpoint visibilityEndpoint may be missed or judged too late
Must Do
Distinguish endpoint from equivalence point in every practical question. The equivalence point is stoichiometric (calculated); the endpoint is observed (indicator colour change). In HSC: "the endpoint was recorded when the solution turned faint pink and remained pink for 30 seconds, this is taken as the equivalence point, assuming phenolphthalein has been correctly selected for this titration type."
Common Error
Students rinse the conical flask with the analyte solution before adding the pipetted volume. This is wrong, it leaves excess analyte on the flask walls, increasing the moles of analyte and giving a titre that is too large. Only the burette and pipette are rinsed with their respective solutions. The conical flask is rinsed with distilled water only.

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?

2b
Investigation, Finding the Concentration of an Unknown by Titration

Syllabus: conduct practical investigations to analyse the concentration of an unknown acid or base by titration

We just saw the technique for each piece of volumetric glassware. That raises a question: how do those techniques combine into a result you can defend to three significant figures? This card answers it → standardise, titrate to concordance, then report with an uncertainty.

Inquiry question. What is the concentration of the supplied unknown sodium hydroxide solution, and what is the uncertainty in that value?

Safety
Teacher supervision and a written risk assessment. Safety glasses and gloves throughout. NaOH solutions are corrosive and are particularly damaging to eyes, rinse any splash for 15 minutes and report it. Never pipette by mouth, use a pipette filler. Fill the burette below eye level, using a funnel, and remove the funnel before titrating. Rinse all glassware into the sink with plenty of water at the end.

Equipment. 50 mL burette and stand, 20 or 25 mL volumetric pipette and filler, 250 mL conical flasks, 250 mL volumetric flask, funnel, white tile, wash bottle of distilled water, primary standard (anhydrous sodium carbonate or potassium hydrogen phthalate), analytical balance, approximately 0.1 mol L⁻¹ HCl, the unknown NaOH, methyl orange and phenolphthalein.

1
Prepare the primary standard. Accurately weigh the primary standard by difference, dissolve it fully, and make up to the mark in a 250 mL volumetric flask. Record the mass to the precision of the balance, this measurement propagates into every result that follows.
2
Standardise the HCl against your primary standard. The HCl is not a primary standard, so its stated concentration is approximate until you have measured it.
3
Condition the glassware. Rinse the burette with the HCl it will hold, and the pipette with the NaOH it will deliver. Rinse the conical flask with distilled water only, rinsing it with the solution would add unmeasured extra moles.
4
Pipette an accurate aliquot of the unknown NaOH into a conical flask. Add 2–3 drops of indicator, no more, the indicator is itself a weak acid.
5
Rough titration. Run the standardised HCl in quickly to find the approximate endpoint. This one is not used in the average.
6
Accurate titrations. Repeat, adding dropwise within 1 mL of the rough endpoint and swirling constantly, until the indicator changes on a single drop. Wash the flask walls down with distilled water before the endpoint, this does not change the moles present.
7
Continue until you have three concordant titres within 0.10 mL. Titrating three times is not the same as achieving concordance, keep going if they disagree.
8
Average the concordant titres only, then calculate c(NaOH) using n = cV and the mole ratio from the balanced equation.
Titration results. Ring the concordant titres and average only those
TitrationInitial burette / mLFinal burette / mLTitre / mLConcordant?
Roughnot used
1
2
3
4 (if needed)

Analysis and evaluation.

  • Write the balanced equation and state the mole ratio you used. Show the full calculation from mean titre to c(NaOH).
  • Combine the uncertainties: two burette readings at ±0.05 mL each, the pipette tolerance, and the volumetric flask tolerance. Quote c(NaOH) with an absolute uncertainty and a sensible number of significant figures.
  • Which single piece of glassware contributes most to your uncertainty? Support the answer with the percentage uncertainty of each.
  • Justify your indicator choice from the equivalence-point pH for this titration, not from habit.
  • Suppose the conical flask had been rinsed with the NaOH before the aliquot was added. Would your calculated c(NaOH) be too high or too low? Explain in terms of moles, not volumes.

Conclusion. Report c(NaOH) with its uncertainty and state the evidence for its reliability, which is the concordance of your titres, not the number of times you titrated.

Concordance is the reliability measure: three titres within 0.10 mL is a claim about precision that a reader can check. “I repeated it three times” is not, and averaging a non-concordant titre in silently degrades the result.
Common Error: Using the nominal bottle concentration of the HCl instead of the standardised value. Every result then inherits that error, and it is invisible in the arithmetic because the numbers still divide correctly.

Sequence: weigh primary standard by difference → make to the mark → standardise the HCl → condition burette and pipette with their own solution, flask with water only → rough titre → accurate titres to three concordant within 0.10 mL → average the concordant only → n = cV and the mole ratio → report with combined uncertainty.

Pause, write the highlighted sequence into your book.

4

The four-step calculation

3
Calculating Unknown Concentration, The Four-Step Method

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.

TitrationBalanced equationMole ratio (acid:base)Step 3 rule
HCl + NaOHHCl + NaOH → NaCl + H₂O1:1n(HCl) = n(NaOH)
H₂SO₄ + NaOHH₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O1:2n(H₂SO₄) = n(NaOH)/2
HCl + Na₂CO₃2HCl + Na₂CO₃ → 2NaCl + H₂O + CO₂2:1n(HCl) = 2 × n(Na₂CO₃)
H₂C₂O₄ + NaOHH₂C₂O₄ + 2NaOH → Na₂C₂O₄ + 2H₂O1:2n(H₂C₂O₄) = n(NaOH)/2
CH₃COOH + NaOHCH₃COOH + NaOH → CH₃COONa + H₂O1:1n(CH₃COOH) = n(NaOH)
Must Do
Record at least three concordant titres (within ±0.10 mL of each other) and average only the concordant results. The first rough titre (used to locate the endpoint approximately) is excluded from the concordance calculation. If titres of 22.45, 22.40, 22.50, and 23.10 mL are recorded, 23.10 is non-concordant (differs by >0.10 mL from the others) and is excluded. Correct average = (22.45 + 22.40 + 22.50)/3 = 22.45 mL.
Common Error
Using the volume in mL directly in n = c × V. Always convert: 16.45 mL = 0.01645 L. n = 0.5000 × 16.45 = 8.225 mol is clearly wrong; n = 0.5000 × 0.01645 = 8.225 × 10⁻³ mol is correct. The sanity check: moles in a typical titration are in the range 10⁻³ to 10⁻⁴ mol. Any answer larger than 0.01 mol is almost certainly a unit conversion error.

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).

Pause, write the highlighted definition into your book before moving on.

For the titration H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O, if n(NaOH) = 2.40 × 10⁻³ mol, what is n(H₂SO₄)?

5

Household analysis and reporting standards

4
Analysing Household Substances, Vinegar and Antacids

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%.

White vinegar

Active component: CH₃COOH (~5%)
Titration type: Weak acid + strong base
Indicator: Phenolphthalein
Endpoint colour change: Colourless → faint pink (persistent)

Antacid tablet

Active component: NaHCO₃ or CaCO₃
Titration type: Base + strong acid
Indicator: Methyl orange
Endpoint colour change: Yellow → orange-red

Vitamin C tablet

Active component: Ascorbic acid (Ka ≈ 8 × 10⁻⁵)
Titration type: Weak acid + strong base
Indicator: Phenolphthalein
Endpoint colour change: Colourless → faint pink
Must Do
For the vinegar titration in HSC practical reports, calculate the percentage acidity and compare to the labelled value (~5%). Any discrepancy should be attributed to specific sources of error, not to the acid being "weaker" or "stronger" than expected. Acetic acid has a fixed Ka regardless of brand; concentration variation explains discrepancies, not strength variation.
Common Error
Using methyl orange (range 3.1–4.4) for the vinegar titration. For CH₃COOH + NaOH, the equivalence point pH ≈ 8.7, this is completely outside methyl orange's range. Methyl orange would change colour at pH ~4, long before the equivalence point is reached, significantly underestimating the acidity. The correct indicator is phenolphthalein (range 8.3–10.0).

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?

5
Performing and Reporting a Titration to HSC Standards

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]. Averaging non-concordant titres is one of the most common sources of reportable inaccuracy in commercial titration work. 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.

What is expected
  • Correct formula, balanced, with state symbols
  • Readings to 0.05 mL; concordant titres identified; average calculated
  • All four steps shown with units at each step
  • Named indicator; equivalence point pH stated; indicator pH range cited
  • Named specific error; direction of effect; specific improvement
Common failure
  • Missing state symbols; incorrect stoichiometry
  • Average includes non-concordant titre; readings given to 0.1 mL only
  • Volumes not converted to litres; mole ratio skipped; answer has no units
  • "Used phenolphthalein because it changes colour", no pH justification
  • "Human error" or "parallax error" without specifying what was misread
Insight
The direction of error analysis works systematically. Ask: did this error increase or decrease the titre? If the titre is too large (more NaOH delivered than needed): n(NaOH) calculated is too large → n(unknown) is too large (Step 3) → c(unknown) is too large (overestimated). If the titre is too small: c(unknown) is underestimated. Trace the error through the four steps to determine the final direction. This analysis earns Band 6 error marks.
Common Error
Averaging all titres including non-concordant ones. If titres are 14.20, 14.25, and 14.85 mL, concordant are 14.20 and 14.25 (within ±0.10 mL); average = 14.23 mL. Including 14.85 gives average 14.43 mL, a 0.20 mL systematic error that propagates through the entire calculation. Always exclude non-concordant titres before averaging.

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.

Pause, copy the highlighted definition into your book before moving on.

Titres recorded: 22.45, 22.50, 22.40, 23.10 mL. Which titres are concordant and should be averaged?

!
Misconceptions to Fix
✗ "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.
✗ "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.
6

Worked examples and practice

Worked Example 1, Standard Solution Preparation and Basic Titration Calculation
ApplyBand 3–4

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.

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)

2

(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

3

(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

4

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.

Worked Example 2, Vinegar Analysis: Percentage Acidity and Error Identification
AnalyseBand 4–5

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.

1

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

2

(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

3

(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

4

(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%

5

(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.

Worked Example 3, Extended Response: Oxalic Acid Standardisation, Error Analysis & Burette Technique (8 marks)
EvaluateBand 6(8 marks)

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.

1

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

2

(a) c(H₂C₂O₄):

n = 0.6320 / 126.1 = 5.012 × 10⁻³ mol

c = 5.012 × 10⁻³ / 0.1000 = 0.05012 mol/L

3

(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

4

(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.

5

(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.

Real-World Anchor
Winemaking and titration: The winemaker's 16.45 mL titre with 0.5000 mol/L NaOH corresponds to n(NaOH) = 0.5000 × 0.01645 = 8.225 × 10⁻³ mol. With a 1:1 ratio for acetic acid: n(CH₃COOH) = 8.225 × 10⁻³ mol; c = 8.225 × 10⁻³/0.01000 = 0.8225 mol/L; mass in 10 mL = 0.8225 × 0.01000 × 60.06 = 0.494 g; % = (0.494/10.05) × 100% = 4.92% ≈ 5.0% acidity. The same calculation you just performed on vinegar is exactly what professional oenologists perform on every blend.
Copy Into Your Books

Primary Standard Criteria

  • High purity, high molar mass, stable in air, non-hygroscopic, readily soluble
  • Na₂CO₃ standardises HCl (M = 106.0 g/mol); H₂C₂O₄·2H₂O standardises NaOH (M = 126.1 g/mol)
  • NaOH and HCl are NOT primary standards

Four-Step Titration

  • (1) Balanced equation + mole ratio
  • (2) n(known) = c × V [V in litres]
  • (3) n(unknown) = n(known) × ratio
  • (4) c(unknown) = n/V [V in litres]

Technique Rules

  • Burette: rinse with titrant; read bottom of meniscus; eye level; ±0.05 mL
  • Pipette: rinse with analyte; do not blow out last drop
  • Flask: rinse with distilled water only
  • Endpoint ≠ equivalence point

Concordant Titres

  • Within ±0.10 mL of each other
  • Average only concordant values
  • Exclude non-concordant titres
  • % acidity of vinegar: n × 60.06 / mass(sample) × 100%
True or false?
In a titration, the equivalence point is where moles of acid exactly equal moles of base reacted.
🔀 Sort the Steps +7 XP
Arrange these steps for a back-calculation titration (finding unknown concentration of acid from titration data) in the correct order:
Write the balanced neutralisation equation
Use mole ratio from balanced equation to find moles of acid
Calculate concentration of acid: c = n/V
Record the average titre from at least 3 concordant results
Calculate moles of base from n = cV

Complete the Learn phase to unlock Practice.

Practice questions

A
Activity A

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.

B
Activity B

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."

01
Multiple Choice
+5 XP

A fresh set drawn from this lesson's question bank, with feedback shown immediately. +5 XP per correct · +25 XP all correct

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.

Evaluatehigh-mark(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.

C
Calculation practice

Written calculation questions in the style and mark range of the HSC paper. Set your working out in full, because method marks are awarded for the steps as well as the final value.

AnalyseBand 6(7 marks) A student standardises a sodium hydroxide solution against 0.0500 mol/L oxalic acid, then uses it to analyse a vinegar sample. Titrating 25.00 mL aliquots of the oxalic acid gives titres of 22.15, 21.40, 21.45 and 21.50 mL. A 25.00 mL aliquot of vinegar, diluted from 10.00 mL to 100.0 mL, then requires 18.60 mL of the standardised NaOH. (a) Select the concordant titres and justify your selection. (b) Calculate the concentration of the NaOH solution. (c) Calculate the concentration of ethanoic acid in the undiluted vinegar. (d) Calculate the percentage by mass of ethanoic acid in the vinegar, assuming a density of 1.01 g/mL. (M(CH₃COOH) = 60.05 g/mol)

ApplyBand 4(4 marks) A 25.00 mL aliquot of a hydrochloric acid solution of unknown concentration is titrated against 0.1050 mol/L sodium carbonate solution. The average concordant titre is 17.85 mL. Write the balanced equation for the reaction and calculate the concentration of the hydrochloric acid.

Show All Answers

MC Explanations

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.

Q6 Sample Answer (4 marks)

(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).

Q7 Sample Answer (5 marks)

(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.

Q8 Sample Answer (6 marks)

(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.

Activity B, Spot & Fix Answers

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.

Retrieve and reflect

Check what actually stuck
Extended Response

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)

Key Fact Drills
What are the five criteria for a primary standard?
High purity, high molar mass, chemically stable in air, non-hygroscopic, readily soluble in water.
Why can NaOH NOT be a primary standard?
It absorbs CO₂ from air (NaOH + CO₂ → Na₂CO₃ + H₂O) and is hygroscopic, both change its effective concentration, making accurate weighing impossible.
What is the concordant titre criterion?
Titres must be within ±0.10 mL of each other. Only concordant titres are averaged; non-concordant titres are excluded.
For H₂SO₄ + 2NaOH, if n(NaOH) = 3.00 × 10⁻³ mol, what is n(H₂SO₄)?
n(H₂SO₄) = n(NaOH)/2 = 1.50 × 10⁻³ mol. The mole ratio is 1:2 (acid:base), so divide n(NaOH) by 2.
What indicator is correct for the vinegar + NaOH titration, and why?
Phenolphthalein (range 8.3–10.0). CH₃COOH + NaOH is weak acid + strong base; equivalence point pH ≈ 8.7 (CH₃COO⁻ hydrolyses), within phenolphthalein's range.
Now You Know, The Winemaker's Technique

Return to your Think First response about the winemaker's titration. Recall the two causes named at the start: a poorly standardised NaOH solution, and averaged non-concordant titres. Can you now explain every decision the winemaker made in precise technical terms?

  • Technique decisions: (1) Rinsing the burette with NaOH before filling, prevents dilution. (2) Recording to ±0.05 mL, ensures titre precision. (3) Reading at eye level with the meniscus, eliminates parallax. (4) Swirling (not stirring), prevents spills and mixes efficiently. (5) Phenolphthalein, correct for weak acid + strong base (tartaric acid + NaOH), EP pH > 7.
  • Why faint pink, not deep pink: Phenolphthalein is pink at all pH values above its transition range. A faint pink lasting 30 seconds is the true endpoint, the first permanent colour change. Deeper pink indicates excess NaOH beyond equivalence, overestimating acidity.
  • Why NaOH must be known precisely: n(tartaric acid) = n(NaOH) × (1/2). A 1% error in c(NaOH) produces a 1% error in the final acidity, significant for commercial winemaking, labelling, and regulatory compliance.