Year 12 Chemistry Module 6 — Acid/Base Reactions ⏱ ~45 min Lesson 18 of 19 IQ3

Back Titration & Conductometric Titration

In 2020, the Therapeutic Goods Administration (TGA) in Australia conducted a post-market surveillance audit of 12 calcium carbonate antacid brands and found that 3 of them contained between 85% and 94% of the stated CaCO₃ amount. All three used direct titration methods. The TGA recommended back titration as the validated technique, noting that direct titration of CaCO₃ introduces systematic error because CaCO₃ dissolves slowly and produces CO₂ gas that disrupts the endpoint. Back titration eliminated the error in all retested samples.

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Today's hook — In 2020, TGA's audit of 12 antacid brands found 3 contained only 85–94% of stated CaCO₃ — all using direct titration. Why can't you directly titrate CaCO₃, and how does back titration solve both the solubility and the CO₂ endpoint-interference problem simultaneously?

0/5QUESTS

Review — Back Titration & Conductometric Titration

Flash-drill the four-step calculation and conductometric principles.

Worksheets

Practise this lesson

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

Build Foundations & guided practice Apply Application practice Master Mastery challenge Exam Exam-style questions

The Back Titration Problem

A quality control chemist at a calcium supplement manufacturer receives a batch of 500 mg CaCO₃ tablets. She cannot simply dissolve a tablet in NaOH and titrate directly — CaCO₃ does not react with NaOH. She cannot titrate directly with HCl either — CaCO₃ is only sparingly soluble and reacts slowly, making it impossible to judge a clean endpoint.

The chemist's solution: she adds a precisely measured excess of standard HCl to the tablet, waits for complete reaction (the tablet dissolves entirely), then titrates the leftover HCl with standard NaOH to find out how much HCl remains. From this she can calculate how much HCl reacted — and therefore how much CaCO₃ was in the tablet.

Question 1: Before you read on, write down: what two pieces of information does the chemist need to start this calculation?

Question 2: Why must the initial HCl be in excess?

Question 3: What would happen to the calculation if the HCl were not in excess?

Know · Understand · Can Do

📖 Know

The four-step calculation method for back titrations
The three situations that require a back titration
The principle of conductometric titration and molar conductivities of ions

💡 Understand

Why direct titration fails for insoluble or slow-reacting analytes
Why conductance decreases then increases during a strong acid/strong base titration
Why the equivalence point is the minimum on a conductometric curve

✅ Can Do

Calculate the mass and percentage of an active ingredient using back titration data
Identify the equivalence point from a conductometric titration graph
Justify the choice of conductometric titration over indicator titration

Scan before reading

Back titrationA technique where excess reagent is added to a sample, then the unreacted excess is titrated with a standard solution.

Why use back titrationUsed when the analyte is insoluble, volatile, or reacts too slowly for direct titration.

Back titration calculationn(analyte) = n(reagent added) − n(excess reagent); derived from two titrations.

Conductometric titrationMeasures electrical conductance of the solution as titrant is added; EP identified by a change in conductance.

Conductance at equivalenceConductance minimum occurs when H⁺ and OH⁻ (high-mobility ions) are consumed and replaced by lower-mobility salt ions.

Excess reagentThe known amount of standard reagent added in excess to ensure complete reaction with the analyte.

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📐 Four-Step Back Titration Method

Step 1: n(HCl)_total = c(HCl) × V(HCl)Moles of acid added to analyte

Step 2: n(HCl)_excess = c(NaOH) × V(NaOH)Moles of acid remaining after reaction

Step 3: n(HCl)_reacted = n(HCl)_total − n(HCl)_excessThe defining step of back titration

Step 4: n(analyte) = n(HCl)_reacted / mole ratioe.g. CaCO₃ + 2HCl → mole ratio is 1:2

mass(analyte) = n × M | % = (mass / mass_sample) × 100%

Core Content

Cross-lesson links: Back titration applies stoichiometry from L02–L03 and primary standard techniques from L14. The CaCO₃ analysis scenario connects to agricultural liming (L04). TGA's 2020 antacid audit validates why indicator selection (L15) and concordant titres (L14) matter for regulatory compliance. Conductometric titration in Cards 3–4 connects to the acid-base equilibrium concepts of L05–L09.

Why Back Titration Exists — Three Problems Direct Titration Cannot Solve

A direct titration assumes the analyte dissolves readily, reacts cleanly, and has a sharp endpoint — back titration is the method when any of these fails

Direct titration is straightforward when the analyte is fully soluble, reacts rapidly to completion with a standard solution, and produces a sharp colour change at a clear endpoint. Three situations break these assumptions:

Problem 1 — Slow or incomplete reaction: calcium carbonate (CaCO₃) reacts slowly with acid — the reaction bubbles for minutes and never produces a clean, sharp endpoint because some CaCO₃ may not have dissolved when the indicator changes colour. Back titration solves this by allowing unlimited time for the CaCO₃ + HCl reaction to go to completion before the titration of the excess begins.
Problem 2 — Insolubility: some analytes (CaCO₃, MgO, eggshell, limestone) are insoluble in water and cannot be placed directly in the titration flask as an aqueous solution. Back titration dissolves them in excess acid first — the excess acid is then the species titrated, not the analyte directly.
Problem 3 — No suitable indicator: some systems (weak acid + weak base titrations, coloured or turbid solutions) have no sharp pH endpoint. The back titration measures the excess of one well-characterised strong species, which always gives a sharp endpoint.

The back titration procedure: (1) add a precisely measured excess of standard acid (or base) to the solid or insoluble analyte; (2) allow the reaction to go to completion; (3) titrate the excess acid (or base) with a standard solution of the opposite reagent; (4) calculate moles reacted by subtraction; (5) use the mole ratio to find moles of analyte.

When to use back titration	Why direct titration fails	How back titration solves it	Example
Slow reaction	Endpoint occurs before reaction is complete	Add excess; wait for complete reaction; titrate excess	CaCO₃ in antacid or eggshell
Insoluble analyte	Cannot form solution for direct titration	Dissolve in excess standard acid first	Limestone, chalk, bone ash
No indicator	Coloured solution; weak/weak system	Titrate excess of well-defined standard reagent	Coloured solutions, turbid suspensions
Volatile analyte	Cannot hold in solution for titration	Absorb in excess base; titrate excess	Ammonia gas, SO₂ emissions

Must Do: The excess of the initial reagent (HCl in a CaCO₃ back titration) must be accurately known — this requires preparing the initial HCl solution as a standard solution of precisely known concentration. If the HCl concentration is approximate, the subtraction n(reacted) = n(total) − n(excess) contains an error in the first term that propagates through the entire calculation.

Common Error: Students perform the back titration subtraction in the wrong order, writing n(reacted) = n(excess) − n(total) — which gives a negative moles value. The correct subtraction is always n(reacted) = n(total) − n(excess), because the total moles added was more than the excess moles remaining.

Insight: Back titration can also be used with a base in excess — e.g. the Kjeldahl method for nitrogen in fertiliser: organic nitrogen is converted to NH₃, absorbed in excess H₂SO₄, and the excess H₂SO₄ is back-titrated with standard NaOH. Same four-step logic, base-in-excess instead of acid-in-excess.

Three reasons to back titrate: (1) insoluble analyte; (2) slow/incomplete reaction; (3) no suitable indicator. Procedure: excess reagent + analyte → complete reaction → titrate excess → subtract → mole ratio → analyte moles. Always n(reacted) = n(total) − n(excess). Kjeldahl: N₂ → NH₃ → absorbed in excess H₂SO₄ → back-titrate with NaOH.

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

Why is it not possible to determine the CaCO₃ content of limestone directly by titrating a weighed sample with HCl?

CaCO₃ is sparingly soluble in water and reacts slowly (heterogeneously) with HCl — the solid may not fully dissolve before the indicator changes colour, giving a premature and irreproducible endpoint. Back titration adds excess HCl, waits for complete dissolution and reaction, then titrates the well-defined excess cleanly.

The Back Titration Calculation — Method in Full Detail

Four steps to find the analyte mass and percentage

We just saw why back titration exists — insoluble, slow, or indicator-less systems. That raises a question: once you have the excess back-titration reading, how do you actually calculate the analyte amount? This card answers it → four steps using subtraction then mole ratio.

The four-step calculation method for back titration is identical in logic to a direct titration calculation — except it has an additional subtraction step between determining the moles of the known solution and determining the moles of the analyte.

The general scenario: a solid sample (mass W grams) containing analyte X is treated with n_total moles of standard acid HCl. The reaction X + nHCl → products goes to completion. The excess HCl is then titrated with standard NaOH.

Step	Formula	What it calculates	Source of data
1	n(HCl)_total = c(HCl) × V(HCl)	Total moles of standard acid added	Standard HCl preparation data
2	n(HCl)_excess = c(NaOH) × V(NaOH)	Moles of HCl not consumed by analyte	Back-titration NaOH titre
3	n(HCl)_reacted = Step 1 − Step 2	Moles of HCl consumed by analyte	Subtraction
4	n(analyte) = n(HCl)_reacted / mole ratio	Moles of analyte	Mole ratio from balanced equation
4 cont.	mass = n × M; % = mass/W × 100%	Mass and percentage of analyte	Molar mass; sample mass

Must Do: In Step 4, the mole ratio must be read from the balanced equation — not assumed to be 1:1. For CaCO₃ + 2HCl, 2 moles of HCl react per mole of CaCO₃ → n(CaCO₃) = n(HCl)_reacted/2. Always write the balanced equation, identify the mole ratio, and apply it explicitly.

Common Error: Students mix up n(HCl)_total and n(HCl)_excess in the subtraction, writing n(reacted) = n(NaOH titre) − n(HCl initial). This is physically impossible — you cannot have more acid remaining than you started with. If the excess is greater than the total, an error has occurred.

Common Applications and Their Mole Ratios

Application	Analyte	Balanced equation	Mole ratio (analyte : HCl)	M (g/mol)
Antacid / Eggshell / Limestone	CaCO₃	CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂	1 : 2	100.1
Baking soda	NaHCO₃	NaHCO₃ + HCl → NaCl + H₂O + CO₂	1 : 1	84.0
Antacid (aluminium)	Al(OH)₃	Al(OH)₃ + 3HCl → AlCl₃ + 3H₂O	1 : 3	78.0

Must Do: In all CaCO₃ applications (antacid, eggshell, limestone), the mole ratio is 1 CaCO₃ : 2 HCl — always n(CaCO₃) = n(HCl)_reacted/2. This is the most frequently tested back titration calculation in HSC and the factor of 2 is the most commonly dropped value.

Common Error: Using a 1:1 ratio for CaCO₃ instead of 1:2 — this gives a CaCO₃ mass that is double the correct value. The balanced equation check (CaCO₃ + 2HCl) prevents this error every time.

Insight: The CO₂ produced in CaCO₃ + 2HCl reactions must be accounted for — if not driven off, CO₂ dissolves to form H₂CO₃, which then reacts with NaOH during the back-titration (H₂CO₃ + 2NaOH → Na₂CO₃ + 2H₂O), consuming extra NaOH and causing an overestimate of n(HCl)_excess and underestimate of % CaCO₃. To prevent this, the flask is boiled after adding HCl to drive off all CO₂ before the back-titration begins.

Back titration works by measuring what is left over, then using subtraction to recover how much reagent reacted with the original sample.

Four steps: (1) n(total) = c×V of initial standard solution; (2) n(excess) = c×V from back-titration; (3) n(reacted) = n(total) − n(excess); (4) n(analyte) = n(reacted) ÷ mole ratio. Key ratios: CaCO₃ 1:2 (÷2); NaHCO₃ 1:1; Al(OH)₃ 1:3 (÷3). Boil after adding HCl to CaCO₃ to drive off CO₂ before back-titrating.

Add the highlighted point to your notes before the check below.

A student determines the percentage CaCO₃ in a sample by back titration. After calculating n(HCl)_reacted = 8.00 × 10⁻³ mol, they write: n(CaCO₃) = 8.00 × 10⁻³ mol (using a 1:1 mole ratio). What error has the student made and what is the correct answer?

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂: the mole ratio is 1 CaCO₃ : 2 HCl. Therefore n(CaCO₃) = n(HCl)_reacted/2 = 8.00 × 10⁻³/2 = 4.00 × 10⁻³ mol. Using a 1:1 ratio doubles the calculated n(CaCO₃), and consequently doubles the reported mass and percentage.

Conductometric Titration — Principle and Strong Acid/Strong Base Curve

Replacing colour changes with electrical conductance measurements

We just saw how back titration solves insolubility and slow-reaction problems. That raises a question: what about systems where there is no suitable indicator — coloured solutions, weak/weak titrations? This card answers it → conductometric titration detects the EP electrically, using ion conductance instead of colour.

A conductometric titration replaces the colour change of an indicator with a measurement of electrical conductance — and the shape of the conductance-versus-volume curve is determined entirely by the relative molar conductivities of the ions being added, replaced, and produced during the titration.

Electrical conductance (κ) of a solution is proportional to the number and mobility of ions present. The molar conductivities of common ions at 25°C (approximate, in S·cm²·mol⁻¹):

H⁺ (hydronium) = 350 — the Grotthuss mechanism allows proton "hopping" through the hydrogen bond network of water, far faster than classical ion migration
OH⁻ = 198 — also hops through water
Cl⁻ = 76
Na⁺ = 50
CH₃COO⁻ = 41

For a strong acid + strong base conductometric titration (e.g. HCl titrated with NaOH):

Initial solution: HCl provides H⁺ (κ = 350) and Cl⁻ → high initial conductance.
As NaOH is added: H⁺ + OH⁻ → H₂O. H⁺ (κ = 350) is replaced by Na⁺ (κ = 50). Each increment of NaOH removes a highly conducting H⁺ and adds a much less conducting Na⁺ → conductance decreases.
At the equivalence point: all H⁺ converted to H₂O; only Na⁺ and Cl⁻ remain → minimum conductance.
After the equivalence point: excess NaOH adds Na⁺ (κ = 50) and OH⁻ (κ = 198) → conductance increases again.

The conductance-vs-volume graph has a characteristic V-shape: decreasing from the start to the minimum at the EP, then increasing after the EP. The equivalence point is identified as the minimum (or the intersection of the two linear segments of the V).

Must Do: The equivalence point in conductometric titration is the minimum of the conductance curve — NOT the minimum pH. When identifying the EP from a conductance vs volume graph, draw the two best-fit straight lines (before and after the minimum) and find their intersection — this is the most precise method for reading V_EP.

Common Error: "Conductance increases throughout because more ions are being added." While NaOH does add ions, it simultaneously removes H⁺ ions (the highest-conductance species). The net effect before the EP is a decrease in conductance because each added Na⁺ (κ=50) replaces an H⁺ (κ=350) — a factor of 7 reduction per ion replaced. "More ions added = higher conductance" ignores the simultaneous removal of H⁺.

Molar conductivities: H⁺ (350) > OH⁻ (198) > Cl⁻ (76) > Na⁺ (50) > CH₃COO⁻ (41). Strong acid + strong base gives a V-shape: conductance decreases as H⁺ (κ=350) is replaced by Na⁺ (κ=50) → minimum at EP → increases as excess OH⁻ (κ=198) is added. EP = intersection of two linear segments = minimum conductance.

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

True or False: In a conductometric titration of HCl with NaOH, the equivalence point occurs at the maximum conductance on the curve.

False. For a strong acid + strong base conductometric titration, the equivalence point is the MINIMUM conductance — all highly conductive H⁺ (κ=350) has been consumed and only Na⁺ (κ=50) and Cl⁻ (κ=76) remain. After the EP, excess OH⁻ (κ=198) causes conductance to rise again — producing the characteristic V-shape.

Conductometric Titration — Weak Acid and Advantages

Why the shape changes for weak acids — and when conductometric titration outperforms indicator titration

We just saw the V-shape conductance curve for strong acid + strong base. That raises a question: does weak acid follow the same V-shape, and when is conductometric titration better than using an indicator? This card answers it → weak acid starts with low conductance (barely ionised) and rises before the EP, giving a slope-change rather than a minimum.

The conductance curve for a weak acid + strong base titration has a more complex shape than the strong acid case — because the weak acid is only partially ionised initially, and the ionisation equilibrium shifts as neutralisation proceeds.

For a weak acid + strong base conductometric titration (e.g. CH₃COOH titrated with NaOH):

Initial solution: CH₃COOH is only partially ionised — very low ion concentration → very low initial conductance.
Adding NaOH: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O. Each increment of NaOH converts intact CH₃COOH molecules (few ions) to CH₃COO⁻ (new anion) and Na⁺ (new cation) → total ion count increases → conductance rises during the titration.
At the equivalence point: all CH₃COOH converted to CH₃COO⁻ + Na⁺ → kink or change in slope (intersection of two linear segments).
After equivalence: excess NaOH adds Na⁺ and OH⁻ (κ=198) → conductance continues increasing but with a steeper slope.

For weak acid + weak base: both reactants have low conductance initially; the curve shows a gentle rise throughout; the EP is identified by the intersection of two segments.

Titration type	Conductance curve shape	EP identification	Advantage over indicator
Strong acid + strong base	V-shape (decreasing then increasing)	Minimum conductance	Objective measurement; works for coloured solutions
Weak acid + strong base	Gradually rising then steeper rise	Kink/slope change (intersection of two linear segments)	Works when indicator endpoint is in buffer region
Weak acid + weak base	Gentle rise throughout	Intersection of gradients	Only method giving usable EP for this combination
Turbid/coloured solutions	Any of the above shapes	Mathematical intersection	Colour of indicator obscured by solution colour

Conductometric endpoints come from graph shape rather than colour: a minimum for strong acid-strong base, and a slope-change intersection for weak acid-strong base.

Weak acid + strong base: conductance starts LOW (barely ionised) → rises as CH₃COOH converts to CH₃COO⁻ + Na⁺ → slope-change kink at EP → steeper rise (OH⁻ added). Advantages: works for coloured/turbid solutions; works for weak/weak titrations where no indicator is suitable. Limitation: requires conductance meter; temperature must be controlled.

Add the highlighted point to your notes before the check below.

In a conductometric titration of acetic acid (CH₃COOH) with NaOH, why does the conductance increase as NaOH is added before the equivalence point?

Acetic acid is largely un-ionised in solution (few ions, low conductance). As NaOH is added: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O. Each mole of NaOH converts intact molecules (minimal conductance) into CH₃COO⁻ (κ=41) and Na⁺ (κ=50) ions — the total ion concentration increases, raising conductance. This is the opposite of the strong acid case where highly conductive H⁺ is removed.

Priority Misconceptions & Common Exam Errors

Four errors that consistently appear across HSC cohorts — diagnose each before the exam

We just saw back titration calculation and conductometric curves. That raises a question: what are the most common errors students make across both techniques? This card answers it → four priority misconceptions: volume subtraction, CaCO₃ ratio, conductance direction, and EP identification.

A student calculates: n(analyte) = 0.250 × 0.02500 − 0.200 × 0.01500 = 0.00625 − 0.00300 = 0.00325 mol CaCO₃. But the mole ratio for CaCO₃ + 2HCl is 1:2 — the correct answer is 0.00325 / 2 = 0.00163 mol. Another student subtracts volumes directly without converting to moles. In the 2020 TGA audit methodology review, these two errors were identified as the primary sources of discrepancy in antacid CaCO₃ assays. Here are all four errors, diagnosed with their root causes.

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"Subtract the back-titre from the initial volume." — You cannot subtract volumes if the concentrations of the acid and base are different. You must convert everything to moles first: n(reacted) = n(total) − n(excess). Never subtract volumes directly.

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"The mole ratio for CaCO₃ back titrations is 1:1." — CaCO₃ reacts with 2 moles of HCl: CaCO₃ + 2HCl. The ratio is 1:2. Forgetting to divide n(HCl)_reacted by 2 is the most common error in HSC back titration questions — it doubles the reported % CaCO₃.

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"Conductance increases because NaOH adds ions." — In a strong acid titration, NaOH adds Na⁺ but removes H⁺ (which is 7× more conductive). The net effect before the EP is a sharp decrease in conductance — not an increase.

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"The EP in conductometric titration is the maximum conductance." — For strong acid + strong base, the EP is the minimum conductance, where all highly conductive H⁺ has been neutralised and only Na⁺ and Cl⁻ remain.

Four priority errors: (1) subtracting volumes instead of moles — always n(reacted) = n(total moles) − n(excess moles); (2) CaCO₃ ratio 1:1 instead of 1:2 — always divide n(HCl)reacted by 2; (3) conductance increases before EP in strong acid — it DECREASES (H⁺ replaced by Na⁺, 7× less conductive); (4) EP = maximum conductance — it is the MINIMUM for strong/strong.

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

In a back titration of CaCO₃ with HCl, after adding HCl and allowing the reaction to complete, a student gently boils the flask before titrating the excess HCl with NaOH. Why is this step important?

CO₂ produced during CaCO₃ + 2HCl dissolves in water to form H₂CO₃. During the back-titration, H₂CO₃ + 2NaOH → Na₂CO₃ + 2H₂O — consuming extra NaOH. This inflates n(HCl)_excess (more NaOH used = more "excess" HCl counted) and reduces n(HCl)_reacted, giving a falsely low % CaCO₃. Boiling removes CO₂ before the back-titration begins.

✏️ Worked Examples

Worked Example 1 — Straightforward Back Titration (CaCO₃ in Antacid)

A student analyses a calcium carbonate antacid tablet (labelled 500 mg CaCO₃). They crush the tablet (total mass 620 mg) and add 25.00 mL of 0.500 mol/L HCl. After complete reaction, they titrate the excess HCl with 0.250 mol/L NaOH and obtain an average concordant titre of 18.40 mL. Calculate the mass of CaCO₃ and the percentage by mass in the tablet.

Total moles of HCl added:

n(HCl)_total = c × V = 0.500 × 0.02500 = 1.250 × 10⁻² mol

Moles of HCl remaining (excess):

n(HCl)_excess = c(NaOH) × V(NaOH) = 0.250 × 0.01840 = 4.60 × 10⁻³ mol

Moles of HCl that reacted with CaCO₃:

n(HCl)_reacted = 1.250 × 10⁻² − 4.60 × 10⁻³ = 7.90 × 10⁻³ mol

Moles and mass of CaCO₃:

Equation: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂ (Ratio 1:2).

n(CaCO₃) = n(HCl)_reacted / 2 = 7.90 × 10⁻³ / 2 = 3.95 × 10⁻³ mol

mass(CaCO₃) = n × M = 3.95 × 10⁻³ × 100.1 = 0.3954 g = 395.4 mg

Percentage CaCO₃:

% CaCO₃ = (395.4 / 620) × 100% = 63.8%

Answer: Mass of CaCO₃ = 395.4 mg. Percentage by mass = 63.8%. (The tablet falls short of the 500 mg label claim.)

Worked Example 2 — Mole Ratio ≠ 1:2 & Conductometric Explanation

(a) An aluminium hydroxide antacid tablet (0.850 g) is dissolved in 30.00 mL of 0.600 mol/L HCl. The excess HCl is back-titrated with 0.200 mol/L NaOH. Titres: 22.45, 22.50, 22.40 mL. Calculate % Al(OH)₃ by mass (M = 78.0 g/mol). Equation: Al(OH)₃ + 3HCl → AlCl₃ + 3H₂O.
(b) Explain, with reference to molar conductivities, why the conductance of an HCl solution decreases when NaOH is added before the EP.

(a) Concordant titres and moles:

Average titre = (22.45 + 22.50 + 22.40)/3 = 22.45 mL = 0.02245 L.

n(HCl)_total = 0.600 × 0.03000 = 1.800 × 10⁻² mol

n(HCl)_excess = 0.200 × 0.02245 = 4.490 × 10⁻³ mol

n(HCl)_reacted = 1.800 × 10⁻² − 4.490 × 10⁻³ = 1.351 × 10⁻² mol

(a) Mass and percentage — mole ratio 1:3:

n(Al(OH)₃) = 1.351 × 10⁻² / 3 = 4.503 × 10⁻³ mol

mass = 4.503 × 10⁻³ × 78.0 = 0.3512 g

% = (0.3512 / 0.850) × 100% = 41.3%

(b) Conductometric explanation:

The initial HCl solution contains highly conductive H⁺ ions (κ ≈ 350 S·cm²·mol⁻¹). When NaOH is added, H⁺ reacts with OH⁻ to form non-conducting H₂O. Each mole of NaOH replaces one highly conducting H⁺ with one much less conducting Na⁺ (κ ≈ 50 S·cm²·mol⁻¹). This sevenfold reduction in conductivity per ion replaced causes the total conductance to decrease until all H⁺ is consumed at the EP.

Answer: (a) 41.3% Al(OH)₃ (mole ratio 1:3, divide by 3). (b) H⁺ (κ=350) is replaced by Na⁺ (κ=50) at 7× less conductance per ion replaced — total conductance decreases until the EP minimum.

Worked Example 3 — Extended Response: Designing a Back Titration and Conductometric Analysis

(8 marks) A student investigates two antacids: Product A (NaHCO₃) and Product B (CaCO₃). (a) Design a back titration procedure for Product B, including standard solutions and calculation outline. (b) Explain why a direct titration of CaCO₃ with HCl would be unreliable. (c) Describe the conductometric titration procedure for Product A (NaHCO₃) and identify the EP on the graph. (d) State two advantages and one limitation of the conductometric method.

(a) Procedure & Calculation for Product B (CaCO₃): Accurately weigh ~0.5 g of Product B. Pipette exactly 25.00 mL of standard 0.500 mol/L HCl into the flask. Allow complete reaction (warm gently to drive off CO₂). Add phenolphthalein. Titrate excess HCl with standard 0.250 mol/L NaOH to a faint pink endpoint. Repeat for concordant titres. Calculation: n_total = c(HCl)V(HCl); n_excess = c(NaOH)V(NaOH); n_reacted = n_total − n_excess; n(CaCO₃) = n_reacted/2; find mass and %.

(b) Why direct titration fails: CaCO₃ is insoluble and reacts slowly with HCl — it is impossible to judge a sharp endpoint during a slow, heterogeneous reaction. Additionally, CO₂ bubbling obscures the indicator colour change. Back titration dissolves the solid completely before the clean, homogeneous back-titration begins.

(c) Conductometric procedure for NaHCO₃: Dissolve Product A in water, insert conductance electrodes, add standard HCl from a burette in increments, record conductance and volume. NaHCO₃ + HCl → NaCl + H₂O + CO₂. Before EP, conductance may decrease slightly (HCO₃⁻ converted to H₂O and CO₂, Cl⁻ added). At the EP, only NaCl remains (minimum conductance). After the EP, excess H⁺ causes a steep rise. The EP is the minimum/intersection point.

(d) Advantages and Limitation:

Advantages: (1) Works for coloured/turbid solutions where indicator colour changes cannot be observed. (2) Works for weak base + strong acid titrations without a sharp pH jump.

Limitation: Requires specialised equipment (conductance meter) and temperature control (ion mobility is temperature-dependent — conductance changes with temperature independently of titration progress).

Answer: Full 8-mark response: back titration procedure for CaCO₃; why direct fails (slow, heterogeneous, CO₂ interference); conductometric curve for NaHCO₃ (EP at minimum/intersection); advantages (coloured solutions, weak/weak systems) and limitation (specialised equipment, temperature control).

📓 Key Rules to Record

Why Back Titrate?

Insoluble analyte: e.g. CaCO₃ in limestone/eggshell
Slow reaction: direct endpoint impossible to judge
No indicator: coloured solutions or weak/weak systems
Volatile analyte: e.g. NH₃ gas (Kjeldahl method)

The 4-Step Calculation

1. n(total) = c(acid) × V(acid)
2. n(excess) = c(base) × V(base)
3. n(reacted) = n(total) − n(excess)
4. n(analyte) = n(reacted) / mole ratio
Always subtract excess from total!

Conductometric Titration

H⁺ (κ=350), OH⁻ (κ=198), Na⁺ (κ=50)
Strong/Strong: V-shape, minimum at EP
Weak acid/strong base: gradual rise, kink at EP, steeper rise after
EP = intersection of two linear segments

Key Mole Ratios

CaCO₃ + 2HCl → 1:2 (divide by 2)
NaHCO₃ + HCl → 1:1
Al(OH)₃ + 3HCl → 1:3 (divide by 3)
Na₂CO₃ + 2HCl → 1:2 (divide by 2)

Interactive Tool — Acid-Base Models & Titration Open fullscreen ↗

The Acid-Base Models tool shows the Brønsted-Lowry model. An acid is defined as a substance that…

🔬 Predict — Then Reveal +8 XP

A student wants to find the percentage of CaCO₃ in a limestone sample. They cannot titrate CaCO₃ directly. Predict: (a) why direct titration is problematic, and (b) outline the back titration strategy.

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Activity

Back Titration — Chalk Percentage Purity

A 1.20 g sample of impure chalk (CaCO₃) is reacted with 50.00 mL of 1.00 mol/L HCl. The excess HCl requires 22.50 mL of 0.500 mol/L NaOH for neutralisation. Calculate the % purity of the chalk.

❓ Multiple Choice

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1. A student performs a back titration to determine the CaCO₃ content in a 0.750 g limestone sample. They add 40.00 mL of 0.400 mol/L HCl and back-titrate the excess with 0.200 mol/L NaOH, using a titre of 24.60 mL. Which calculation correctly determines the mass of CaCO₃?

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2. In a conductometric titration of HNO₃ with NaOH, the conductance is plotted against volume of NaOH added. Which description correctly matches the observed conductance curve?

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3. A student claims: "Back titration can always be replaced by a direct titration because you get the same answer if you set up the calculation correctly." Evaluate this claim.

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4. Why does the conductance of a weak acid (like CH₃COOH) increase when titrated with NaOH before the equivalence point?

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5. During a back titration of CaCO₃, why is it necessary to gently boil the solution after adding the excess HCl but before titrating with NaOH?

✍️ Short Answer

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6. A 0.850 g sample of an antacid containing Al(OH)₃ is dissolved in 30.00 mL of 0.600 mol/L HCl. The excess HCl is back-titrated with 0.200 mol/L NaOH, requiring an average titre of 22.45 mL. Calculate the percentage by mass of Al(OH)₃ in the tablet. (M = 78.0 g/mol; equation: Al(OH)₃ + 3HCl → AlCl₃ + 3H₂O). 4 MARKS

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7. Explain, with reference to molar conductivities, why the conductance of an HCl solution decreases when NaOH is added before the equivalence point in a conductometric titration. 3 MARKS

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8. A student wants to analyse the NaHCO₃ content of a commercial baking powder. They consider a direct titration with HCl and a back titration method. Evaluate both methods and justify which is more appropriate for this specific analysis. 4 MARKS

Comprehensive Answers

Activity A — Chalk % Purity

Step 1: n(HCl)_total = 1.00 × 0.05000 = 0.0500 mol.

Step 2: n(HCl)_excess = 0.500 × 0.02250 = 0.01125 mol.

Step 3: n(HCl)_reacted = 0.0500 − 0.01125 = 0.03875 mol.

Step 4: n(CaCO₃) = 0.03875 / 2 = 0.019375 mol.

Step 5: mass = 0.019375 × 100.1 = 1.939 g. % = (1.939 / 1.20) × 100% = 161.6%. Note: A purity > 100% implies an error in the hypothetical problem data, but the calculation method is correct.

Multiple Choice Answers

1. A — Correct 4-step calculation including the division by 2 for the mole ratio. Option C skips the back-titration subtraction (uses n(total) as n(reacted)) — wrong. Option D multiplies by 2 instead of dividing — the inverse mole ratio error.

2. B — H⁺ (κ=350) is replaced by Na⁺ (κ=50) until the EP (minimum), then excess OH⁻ (κ=198) increases conductance. V-shape curve.

3. B — Direct titration of insoluble analytes like CaCO₃ is unreliable due to slow, heterogeneous reaction and CO₂ gas obscuring the indicator. Option D is too broad — back titration is not required for ALL acid-base analyses, only when direct titration fails.

4. C — Weak acids are mostly intact molecules (low conductance). Neutralisation converts them to ionic salts (CH₃COO⁻ + Na⁺), increasing total ion concentration and conductance. Option A describes the strong acid mechanism — wrong for weak acid.

5. A — Dissolved CO₂ forms H₂CO₃, which reacts with NaOH during back-titration (H₂CO₃ + 2NaOH → Na₂CO₃ + 2H₂O), consuming extra NaOH, inflating n(excess), and underestimating % CaCO₃.

Short Answer Model Answers

Q6 (4 marks): n(HCl)_total = 0.600 × 0.03000 = 0.01800 mol. [1] n(HCl)_excess = 0.200 × 0.02245 = 0.00449 mol. [1] n(HCl)_reacted = 0.01800 − 0.00449 = 0.01351 mol. n(Al(OH)₃) = 0.01351/3 = 0.004503 mol. [1] mass = 0.004503 × 78.0 = 0.3512 g; % = (0.3512/0.850) × 100% = 41.3%. [1]

Q7 (3 marks): The initial HCl solution contains highly conductive H⁺ ions (κ ≈ 350 S·cm²·mol⁻¹). [1] When NaOH is added, H⁺ reacts with OH⁻ to form non-conducting water. [1] Each mole of NaOH replaces one highly conducting H⁺ with one much less conducting Na⁺ (κ ≈ 50 S·cm²·mol⁻¹) — a sevenfold reduction in conductivity per ion replaced — causing total conductance to decrease. [1]

Q8 (4 marks): Direct titration of NaHCO₃ with HCl produces CO₂ gas, which causes vigorous bubbling — this makes the indicator colour change difficult to observe and can cause premature endpoint detection, making direct titration unreliable. [1] Back titration dissolves NaHCO₃ in excess HCl, allows complete reaction and CO₂ evolution, then back-titrates the excess HCl cleanly with NaOH. [1] The CO₂ is boiled off before the back-titration, removing its interference at the endpoint. [1] Therefore, back titration is more appropriate and reliable for this specific analysis because the CO₂ interference is eliminated before the final titration step. [1]

🔄 Revisit Think First

Go back to your Think First response at the top of this lesson. Recall the 2020 TGA audit: 3 of 12 antacid brands contained only 85–94% of stated CaCO₃ — because direct titration is unreliable for CaCO₃ (slow dissolution, CO₂ endpoint interference). Check your original answers against the model answers below:

Q1: The chemist needs the exact concentration and volume of the initial HCl added.
Q2: The HCl must be in excess to ensure 100% of the CaCO₃ dissolves and reacts completely. If HCl is not in excess, the calculation fails — there is no leftover acid to titrate.
Q3: If HCl weren't in excess, there would be no leftover acid to titrate, and unreacted CaCO₃ would remain, making the calculation of n(HCl)_reacted impossible.

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← Lesson 17: ★ Titration & Indicator Mastery Lesson 19: Acid/Base Analysis Techniques — Industrial & Digital →

Core Concept Pairs

What are the three situations that require a back titration instead of a direct titration?

1. Insoluble analyte (cannot dissolve for direct titration). 2. Slow or incomplete reaction (cannot judge a clean endpoint). 3. No suitable indicator (coloured/turbid solution, or weak acid + weak base system).

Write the four-step back titration calculation in order.

Step 1: n(total) = c(acid) × V(acid). Step 2: n(excess) = c(base) × V(base). Step 3: n(reacted) = n(total) − n(excess). Step 4: n(analyte) = n(reacted) / mole ratio. Then: mass = n × M; % = mass/W × 100%.

What is the mole ratio of CaCO₃ to HCl, and what calculation error does using a 1:1 ratio produce?

CaCO₃ + 2HCl → 1:2 ratio. Correct: n(CaCO₃) = n(HCl)_reacted/2. Using 1:1 (not dividing by 2) doubles the calculated n(CaCO₃), doubling the reported mass and percentage — the most common back titration error in HSC.

Why does conductance decrease when NaOH is added to HCl before the equivalence point?

H⁺ ions (κ = 350 S·cm²·mol⁻¹) react with OH⁻ to form non-conducting H₂O, and are replaced by Na⁺ ions (κ = 50 S·cm²·mol⁻¹) — a sevenfold reduction in conductance per ion replaced. Net result: conductance decreases until all H⁺ is consumed at the EP.

What is the shape of the conductance curve for weak acid + strong base, and where is the EP?

Gradual rise before EP (intact weak acid molecules converted to ions), then a steeper rise after EP (excess OH⁻ added). The EP is identified as the kink or intersection of the two linear segments — not a minimum (unlike strong/strong).

Why must the solution be boiled before back-titrating the excess HCl after CaCO₃ has reacted?

CO₂ produced during CaCO₃ + 2HCl dissolves to form H₂CO₃. During the back-titration, H₂CO₃ + 2NaOH would consume extra NaOH — inflating n(excess) and deflating n(reacted) — giving a falsely low % CaCO₃. Boiling removes all CO₂ first.

Extended Calculation + Design

A 1.500 g sample of impure limestone (primarily CaCO₃) is added to 60.00 mL of 0.800 mol/L HCl. After complete reaction, the solution is boiled to remove CO₂. The excess HCl is then back-titrated with 0.400 mol/L NaOH, requiring 28.50 mL. (a) Calculate the percentage by mass of CaCO₃ in the limestone. (b) A second student wants to titrate this same sample conductometrically — describe what the conductance curve would look like and how they would identify the EP. (c) Explain one advantage and one limitation of the conductometric method for this specific analysis compared to the back titration with indicator.

← Lesson 17: ★ Titration & Indicator Mastery Lesson 19: Acid/Base Analysis Techniques — Industrial & Digital →