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

Back Titration & Conductometric Titration

A pharmaceutical manufacturer needs to verify that each antacid tablet contains exactly the stated amount of calcium carbonate — but CaCO₃ is insoluble and reacts too slowly with weak acid for a direct titration. Back titration solves both problems simultaneously, and it is used in quality control laboratories processing thousands of tablets per day.

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Think First

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?

📐

Key Relationships — This Lesson

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 with analyte]

Step 3: n(HCl)_reacted = n(HCl)_total − n(HCl)_excess

The defining step of back titration

Step 4: n(analyte) = n(HCl)_reacted / (mole ratio of analyte/HCl)

e.g. CaCO₃ + 2HCl → mole ratio is 1:2

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

κ = G = 1/R

Conductance; units: siemens (S)

📖 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

📚 Core Content

Key Terms — scan these before reading

excessGreater than the total, an error has occurred in the calculation — this is physically impossible (you cannot have more a.

Brønsted-Lowry acidA proton (H⁺) donor in an acid-base reaction.

Brønsted-Lowry baseA proton (H⁺) acceptor in an acid-base reaction.

Conjugate acid-base pairTwo species differing by one H⁺ that interconvert.

pHThe negative logarithm of hydronium ion concentration.

BufferA solution resisting pH change upon addition of small amounts of acid or base.

A direct titration assumes the analyte dissolves readily, reacts cleanly and completely with the titrant, and has a sharp, detectable endpoint — back titration is the method of choice when any of these three assumptions 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 and make direct titration unreliable or impossible:

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 (dissolve fully, react completely); (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.

Slow reaction

Why direct titration fails: Endpoint occurs before reaction is complete

How back titration solves it: Add excess; wait for complete reaction; titrate excess

Example: CaCO₃ in antacid or eggshell

Insoluble analyte

Why direct titration fails: Cannot form solution for direct titration

How back titration solves it: Dissolve in excess standard acid first

Example: Limestone, chalk, bone ash

No indicator

Why direct titration fails: Analyte is coloured; weak/weak system

How back titration solves it: Titrate excess of well-defined standard reagent

Example: Coloured solutions, turbid suspensions

Gaseous or volatile analyte

Why direct titration fails: Cannot hold in solution for titration

How back titration solves it: Absorb in excess base; titrate excess

Example: Ammonia gas, SO₂ emissions

Must DoThe 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 or unstandardised, the subtraction n(reacted) = n(total) − n(excess) contains an error in the first term that propagates through the entire calculation.

Common ErrorStudents perform the back titration subtraction in the wrong order, writing n(reacted) = n(excess) − n(total) — which gives a negative moles value and immediately reveals the error. The correct subtraction is always n(reacted) = n(total) − n(excess), because the total moles added was more than the excess moles remaining. Physically: you started with more than what is left.

InsightBack titration can also be used in the reverse direction — where a base is added in excess to a weakly basic analyte and the excess base is then titrated with standard acid. For example, determining the nitrogen content of a fertiliser by the Kjeldahl method: organic nitrogen is converted to NH₃, which is absorbed in excess standard H₂SO₄, and the excess H₂SO₄ is then back-titrated with standard NaOH. This is the same four-step logic applied to a base-in-excess system rather than an acid-in-excess system.

The Back Titration Calculation — Method in Full Detail

Four steps to find the analyte mass and percentage

The four-step calculation method for back titration is identical in logic to a direct titration calculation — except that 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 — titre V(NaOH) mL at concentration c(NaOH) mol/L.

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 DoIn 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. Inverting this ratio (n(CaCO₃) = 2 × n(HCl)) is the most common mole ratio error in back titration calculations and doubles the final answer. Always write the balanced equation, identify the mole ratio, and apply it explicitly.

Common ErrorStudents mix up n(HCl)_total and n(HCl)_excess in the subtraction, writing n(reacted) = n(NaOH titre) − n(HCl initial) — which is n(excess) − n(total), a negative number. The correct order is always n(total) − n(excess) = n(reacted). If the excess is greater than the total, an error has occurred in the calculation — this is physically impossible (you cannot have more acid remaining than you started with).

Exam TipIn acid-base calculations, always write the balanced equation first, identify the conjugate pair, and state any assumptions before substituting into the Ka expression.

Common Back Titration Applications — Antacids, Eggshell, and Limestone

Applying the four-step method to real-world samples

The three most frequently tested back titration applications in HSC chemistry share the same four-step calculation structure — what changes is only the identity of the analyte, the balanced equation, and the mole ratio in Step 4.

Application 1 — Antacid tablet (CaCO₃ content): A calcium carbonate antacid tablet (labelled 500 mg CaCO₃) is crushed and treated with excess standard HCl. Balanced equation: CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g). Mole ratio: 1 mol CaCO₃ : 2 mol HCl. The excess HCl is back-titrated with standard NaOH. The result gives % CaCO₃ in the tablet, which is compared to the labelled content.
Application 2 — Eggshell composition: Natural eggshell is approximately 95% CaCO₃. The same back titration procedure (excess HCl, then back-titrate with NaOH) determines the percentage CaCO₃ in a crushed eggshell sample. The result varies between species, age of egg, and diet of the hen — making this a quantitative investigation into real biological variation.
Application 3 — Limestone purity: Limestone (primarily CaCO₃ with impurities such as MgCO₃, SiO₂) is analysed for CaCO₃ content by back titration with HCl. Industrial limestone used in flue gas desulfurisation (L04) must meet a minimum purity specification — back titration is the quality control method.
Application 4 — NaHCO₃ in baking soda: NaHCO₃ + HCl → NaCl + H₂O + CO₂. Mole ratio 1:1. This is a simpler back titration because the ratio is 1:1 — but the same four steps apply.

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

Must DoIn all three 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 in student responses.

Common ErrorStudents calculate n(CaCO₃) = n(HCl)_reacted (using a 1:1 ratio instead of 1:2), effectively halving the amount of HCl reacted required to neutralise the carbonate. This gives a CaCO₃ mass that is half the correct value, and a percentage CaCO₃ that is also half the true value. The balanced equation check — always write CaCO₃ + 2HCl — prevents this error every time.

InsightThe CO₂ produced in CaCO₃ + 2HCl reactions during a back titration must be accounted for — if the flask is sealed or the CO₂ is absorbed into solution, it can react with water to form H₂CO₃, which then reacts with NaOH during the back-titration: H₂CO₃ + 2NaOH → Na₂CO₃ + 2H₂O. This would consume extra NaOH, leading to an overestimate of n(HCl)_excess and an underestimate of n(HCl)_reacted, giving a falsely low % CaCO₃. To prevent this, the flask is boiled (or gently heated) after adding HCl to drive off all CO₂ as gas before the back-titration begins. This step is required in rigorous laboratory practice.

Conductometric Titration — Principle and Conductance Curve Shape

Replacing colour changes with electrical conductance

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. High conductance means many fast-moving ions; low conductance means few or slow-moving ions. The molar conductivities of common ions at 25°C (in approximate units of S·cm²·mol⁻¹):

H⁺ (hydronium) = 350 (highest — the Grotthuss mechanism allows proton "hopping" through the hydrogen bond network of water, far faster than classical ion migration)
OH⁻ = 198 (second highest — 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⁺ (very high molar conductivity) and Cl⁻ → high initial conductance.
As NaOH is added: H⁺ + OH⁻ → H₂O. H⁺ ions (κ = 350) are replaced by Na⁺ ions (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⁺ has been converted to H₂O; only Na⁺ and Cl⁻ remain in solution → 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 (or check-mark 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 point of the two linear segments of the V).

Start

What happens: Only H⁺ and Cl⁻

Ion change: H⁺ (κ=350) + Cl⁻ (κ=76)

Conductance trend: High conductance

Adding NaOH

What happens: H⁺ + OH⁻ → H₂O

Ion change: H⁺ replaced by Na⁺ (κ=50

Conductance trend: Decreasing conductance

Equivalence point

What happens: All H⁺ consumed

Ion change: Only Na⁺ and Cl⁻

Conductance trend: Minimum conductance

Excess NaOH

What happens: Na⁺ and OH⁻ added

Ion change: OH⁻ (κ=198) increases conductance

Conductance trend: Increasing conductance

Must DoThe equivalence point in conductometric titration is the minimum of the conductance curve — NOT the minimum pH (which is the EP in a potentiometric titration). 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 from a conductometric titration.

Common ErrorStudents say conductance increases throughout the titration "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 in conductance per ion replaced. Only after all H⁺ has been replaced do the added Na⁺ and OH⁻ ions produce a net increase. "More ions added = higher conductance" is a naive application of the principle that ignores the simultaneous removal of H⁺.

Conductometric Titration — Weak Acid and Advantages

Why the shape changes for weak acids

The conductance curve for a weak acid + strong base titration has a more complex shape than the strong acid case — and understanding why the shape changes reveals the deeper connection between conductance, equilibrium, and ion concentration in weak acid systems.

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 CH₃COOH (mostly intact, contributing few ions) to CH₃COO⁻ (a new anion) and Na⁺ (another cation) → the number of ions in solution increases. Conductance rises during the titration because ions are being added faster than any highly-conducting species is being removed (no H⁺ to replace — the solution was barely ionised to start with).
At the equivalence point: all CH₃COOH converted to CH₃COO⁻ + Na⁺ — maximum ion concentration for this reaction → maximum conductance (relative maximum or inflection).
After equivalence: excess NaOH adds Na⁺ and OH⁻ — OH⁻ has high conductance (κ=198) → conductance continues to increase but with a steeper slope.

The EP is identified as the kink or change in slope at the point where the gradient of the conductance curve increases sharply (intersection of two linear segments).

For weak acid + weak base: both reactants have low conductance initially; the curve shape is more complex; the EP is identified by the minimum conductance gradient.

Strong acid + strong base

Conductance curve shape: V-shape (decreasing then increasing)

EP identification: Minimum conductance

Advantage over indicator: Objective measurement; works for coloured solutions

Weak acid + strong base

Conductance curve shape: Gradually rising then steeper rise

EP identification: Kink/slope change (intersection of two linear segments)

Advantage over indicator: Works when indicator endpoint is in buffer region

Weak acid + weak base

Conductance curve shape: Gentle rise throughout

EP identification: Minimum gradient or intersection

Advantage over indicator: Only method giving usable EP for this combination

Turbid/coloured solutions

Conductance curve shape: Any of the above shapes

EP identification: Mathematical intersection

Advantage over indicator: Colour of indicator obscured by solution colour

Worked Example 2 — Intermediate: Mole ratio ≠ 1:2 & Conductance

(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 & 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:
Ratio Al(OH)₃ : HCl = 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). 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). 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. (b) H⁺ (κ=350) is replaced by Na⁺ (κ=50), reducing total conductance until the EP minimum.

Worked Example 3 — Hard: Extended Response Design

(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: 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. The graph initially shows moderate conductance (Na⁺, HCO₃⁻). As HCl is added, HCO₃⁻ is converted to H₂O and CO₂, while Cl⁻ is added. Conductance may decrease slightly or remain flat. At the EP, only NaCl remains (minimum conductance). After the EP, excess H⁺ causes a steep rise. The EP is the minimum/intersection.

(d) Advantages & Limitations: Advantages: (1) Works for coloured/turbid solutions where indicators fail; (2) Works for weak base + strong acid without a sharp pH jump. Limitation: Requires specialised equipment (conductance meter) and temperature control (ion mobility is temperature-dependent).

Answer: Full 8-mark response covering procedure, justification for back titration, conductometric curve shape (EP at minimum/intersection), and advantages/limitations.

Test Your Knowledge

ApplyBand 4

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?

n(total) = 0.01600 mol; n(excess) = 4.920 × 10⁻³ mol; n(reacted) = 1.108 × 10⁻² mol; n(CaCO₃) = 5.540 × 10⁻³ mol; mass = 0.5546 g

n(total) = 16.00 mol; n(CaCO₃) = 8.00 mol; mass = 800.8 g

n(total) = 0.01600 mol; n(reacted) = 0.01600 mol; n(CaCO₃) = 8.00 × 10⁻³ mol; mass = 0.8008 g

n(reacted) = 1.108 × 10⁻² mol; n(CaCO₃) = 2.216 × 10⁻² mol; mass = 2.218 g

n(total) = 16.00 mol; n(CaCO₃) = 8.00 mol; mass = 800.8 g

n(total) = 0.01600 mol; n(reacted) = 0.01600 mol; n(CaCO₃) = 8.00 × 10⁻³ mol; mass = 0.8008 g

n(reacted) = 1.108 × 10⁻² mol; n(CaCO₃) = 2.216 × 10⁻² mol; mass = 2.218 g

UnderstandBand 4

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?

Conductance increases throughout as more ions (Na⁺ and OH⁻) are added

Conductance decreases to a minimum at the EP (H⁺ replaced by Na⁺), then increases (excess OH⁻ added)

Conductance remains constant until the EP, then drops sharply

Conductance increases to a maximum at the EP then decreases

Conductance decreases to a minimum at the EP (H⁺ replaced by Na⁺), then increases (excess OH⁻ added)

Conductance remains constant until the EP, then drops sharply

Conductance increases to a maximum at the EP then decreases

✅ Comprehensive Answers

▼

🔬 Activity 1 — Calculation Drill

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 data, but the mathematical steps are correct).

📈 Activity 2 — Graph Analysis

The curve shape will be a gradual rise, followed by a steeper rise after the equivalence point. Conductance increases before the EP because the weak acid (CH₃COOH) is mostly un-ionised initially (low conductance). As NaOH is added, it converts the intact molecules into highly conductive ions (CH₃COO⁻ and Na⁺), increasing the total ion concentration and thus the conductance.

❓ Multiple Choice

1. A — Correct 4-step calculation including the division by 2 for the mole ratio.

2. B — H⁺ (high conductance) is replaced by Na⁺ (low conductance) until the EP (minimum), then excess OH⁻ increases conductance.

3. B — Direct titration of insoluble analytes like CaCO₃ is unreliable due to slow reaction and CO₂ interference.

4. C — Weak acids are mostly intact molecules. Neutralisation creates ionic salts, increasing total ion concentration.

5. A — Dissolved CO₂ forms H₂CO₃, which reacts with NaOH during back-titration, causing an overestimate of excess acid and an underestimate of the analyte.

📝 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). This sevenfold reduction in conductivity per ion replaced causes the 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, and then back-titrates the excess HCl cleanly with NaOH. [1] Because the CO₂ can be boiled off before the back-titration, there is no interference at the endpoint. [1] Therefore, back titration is more appropriate and reliable for this specific analysis. [1]

Back Titration & Conductometric Titration

Download this lesson's worksheet

Key Relationships — This Lesson

📖 Know

💡 Understand

✅ Can Do

Slow reaction

Insoluble analyte

No indicator

Gaseous or volatile analyte

The Back Titration Calculation — Method in Full Detail

Common Back Titration Applications — Antacids, Eggshell, and Limestone

Conductometric Titration — Principle and Conductance Curve Shape

Start

Adding NaOH

Equivalence point

Excess NaOH

Conductometric Titration — Weak Acid and Advantages

Strong acid + strong base

Weak acid + strong base

Weak acid + weak base

Turbid/coloured solutions

⚠️ Common Misconceptions — Module 6 Lesson 18

📓 Copy Into Your Books

Why Back Titrate?

The 4-Step Calculation

Conductometric Titration

Key Mole Ratios

Back Titration Steps

Conductometric Curves

Test Your Knowledge

Extended Questions

Revisit Your Thinking

✅ Comprehensive Answers

🔬 Activity 1 — Calculation Drill

📈 Activity 2 — Graph Analysis

❓ Multiple Choice

📝 Short Answer Model Answers

Mark lesson as complete

Questions