Back Titration Calculations: Solved Step-by-Step

By Aniket Bhardwaj · 11 November 2026 · Solutions

Back titration, or indirect titration, is an essential analytical technique in competitive chemistry examinations. In direct titrations, titrant is added straight to an analyte until reaching equivalence. However, direct titration fails when an analyte is water-insoluble, reacts slowly, is volatile, or lacks a sharp indicator colour transition. To resolve this, a known excess of standard reagent is added to consume the analyte completely. The unreacted excess is then back-titrated with a second standard solution. Mastering back titration calculations trains students to handle stoichiometry, excess reagents, and percentage purity accurately.

Why and When Back Titration Is Used

Back titration is preferred in four standard laboratory situations:

Insoluble solids: Substances like calcium carbonate (CaCO3) dissolve poorly in water but react completely with excess strong acid.
Slow reactions: When an analyte reacts slowly at room temperature, heating with excess reagent drives the reaction to completion.
Volatile analytes: Analytes like ammonia (NH3) released in nitrogen estimation must be trapped in excess acid to prevent loss.
Weak endpoints: Titrating unreacted strong acid with strong base yields a sharper endpoint than titrating a weak analyte directly.

Core Principles and Key Formulas

Back titration relies on mole conservation across three species: the analyte, Reagent 1 (added in excess), and Reagent 2 (the back titrant).

Total Moles of Reagent 1 Added = Moles Consumed by Analyte + Moles Unreacted (Excess)
Moles Consumed = Total Moles Added - Moles Unreacted
Moles Unreacted = Volume of Reagent 2 (L) × Molarity of Reagent 2 (mol/L) × Stoichiometric Ratio
Percentage Purity (%) = [Mass of Pure Analyte (g) / Mass of Impure Sample (g)] × 100

Variables and units: Volume (V) is in litres (L) or millilitres (mL); concentration (M) is in mol/L or mmol/mL; mass (m) is in grams (g); and amount (n) is in moles (mol) or millimoles (mmol, where 1 mmol = 10^-3 mol).

Exam details

Source: practice problem written for this article (not a past-paper question)
Relevant exams: CBSE Class 11, Class 12, JEE Main, JEE Advanced, NEET, CUET
Topic: Volumetric Analysis and Back Titration Calculations
Difficulty: Medium to Hard

Step-by-Step Solved Examples

Question: A 1.500 g sample of impure limestone (CaCO3) was treated with 50.00 mL of 0.6000 M HCl. After CO2 was expelled by boiling, unreacted excess HCl required 15.00 mL of 0.4000 M NaOH solution to reach the phenolphthalein end point. Calculate the mass of pure CaCO3 and its percentage purity. (Take IUPAC atomic masses: Ca = 40.08 g/mol, C = 12.01 g/mol, O = 16.00 g/mol, giving CaCO3 = 100.09 g/mol; or rounded school textbook value CaCO3 = 100.0 g/mol).

Solution:

Step 1: Balanced chemical equations:
CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

Step 2: Total initial HCl added:
Initial mmol of HCl = 50.00 mL × 0.6000 mmol/mL = 30.00 mmol.

Step 3: Unreacted excess HCl:
mmol of NaOH used = 15.00 mL × 0.4000 mmol/mL = 6.000 mmol.
From 1:1 neutralisation stoichiometry, unreacted HCl = 6.000 mmol.

Step 4: HCl consumed by CaCO3:
HCl consumed = 30.00 mmol - 6.000 mmol = 24.00 mmol = 0.02400 mol.

Step 5: Moles and mass of pure CaCO3:
From stoichiometry, 1 mol CaCO3 reacts with 2 mol HCl.
Moles of CaCO3 = 24.00 mmol / 2 = 12.00 mmol = 0.01200 mol.
Mass of pure CaCO3 (at 100.09 g/mol) = 0.01200 mol × 100.09 g/mol = 1.201 g.
(At rounded 100.0 g/mol: 0.01200 mol × 100.0 g/mol = 1.200 g).

Step 6: Percentage purity:
Percentage purity = (1.201 g / 1.500 g) × 100 = 80.07% (or 80.00% using 100.0 g/mol).

Second route verification (Milliequivalents):
mEq CaCO3 = (50.00 × 0.6000) - (15.00 × 0.4000) = 30.00 - 6.000 = 24.00 mEq.
Mass = (24.00 / 1000) × (100.09 / 2) = 1.201 g. Both routes match.

Answer: 1.201 g pure CaCO3, representing 80.07% purity (80.00% with standard rounded atomic masses).

Question: A 0.5000 g fertiliser sample containing (NH4)2SO4 was boiled with 50.00 mL of 0.2000 M NaOH until all NH3 was expelled. The cooled solution was titrated with 0.1500 M HCl, requiring 24.40 mL to reach the methyl orange end point. Calculate the mass and percentage of (NH4)2SO4 in the fertiliser. (Take IUPAC atomic masses: H = 1.008 g/mol, N = 14.01 g/mol, O = 16.00 g/mol, S = 32.06 g/mol, giving (NH4)2SO4 = 132.14 g/mol; or rounded 132.0 g/mol).

Solution:

Step 1: Balanced chemical equations:
(NH4)2SO4(s) + 2 NaOH(aq) → Na2SO4(aq) + 2 NH3(g) + 2 H2O(l)
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

Step 2: Total initial NaOH added:
Initial mmol of NaOH = 50.00 mL × 0.2000 mmol/mL = 10.00 mmol.

Step 3: Unreacted excess NaOH:
mmol of HCl used = 24.40 mL × 0.1500 mmol/mL = 3.660 mmol.
From 1:1 stoichiometry, unreacted NaOH = 3.660 mmol.

Step 4: NaOH consumed by (NH4)2SO4:
NaOH consumed = 10.00 mmol - 3.660 mmol = 6.340 mmol = 0.006340 mol.

Step 5: Moles and mass of (NH4)2SO4:
1 mol (NH4)2SO4 reacts with 2 mol NaOH.
Moles of (NH4)2SO4 = 6.340 mmol / 2 = 3.170 mmol = 0.003170 mol.
Mass of (NH4)2SO4 (at 132.14 g/mol) = 0.003170 mol × 132.14 g/mol = 0.4189 g.
(At rounded 132.0 g/mol: 0.003170 mol × 132.0 g/mol = 0.4184 g).

Step 6: Percentage by mass:
Percentage of (NH4)2SO4 = (0.4189 g / 0.5000 g) × 100 = 83.78% (or 83.68% using 132.0 g/mol).

Second route verification (Milliequivalents):
mEq salt = (50.00 × 0.2000) - (24.40 × 0.1500) = 10.00 - 3.660 = 6.340 mEq.
Mass = (6.340 / 1000) × (132.14 / 2) = 0.4189 g. Both methods verify the result.

Answer: 0.4189 g of (NH4)2SO4, representing 83.78% by mass (83.68% with rounded values).

Common Mistakes in Back Titration Calculations

Ignoring stoichiometric coefficients: Treating every reaction as 1:1 causes errors. CaCO3 consumes two moles of HCl; omitting this factor halves or doubles the calculated analyte amount.
Confusing excess and consumed reagent: Back titration measures unreacted excess reagent. It must always be subtracted from total initial reagent before calculating analyte quantity.
Mixing millilitres and litres: Millilitres times molarity yields millimoles (mmol), whereas litres times molarity yields moles (mol). Mixing units causes a 1000-fold discrepancy.
Neglecting aliquot scaling factors: If a sample is diluted to 250 mL and a 25 mL portion is titrated, scale the result by 10 (250/25) to obtain total sample content.
Failing to remove dissolved CO2: In carbonate titrations, dissolved CO2 forms carbonic acid, consuming extra base and distorting results. Gentle boiling prevents this error.

Exam Relevance and Question Patterns

Back titration appears regularly across school boards and competitive exams. The table below summarises how this topic is evaluated.

Examination Level	Question Format	Key Focus Areas Tested
CBSE Class 11 and Class 12	Structured numerical problems and practical viva voce	Acid-base back titration, percentage purity of limestone, antacid analysis, molarity calculations
JEE Main and CUET-UG	Multiple-choice and numerical value questions	Limiting and excess reagents, multi-step stoichiometry, Kjeldahl nitrogen estimation
JEE Advanced	Multi-concept passages and comprehensive numericals	Volhard precipitation titration, carbonate-bicarbonate mixtures, multi-stage industrial reactions
NEET	Direct objective numericals	Fast millimole subtraction, antacid neutralisation capacity, fertilizer purity determination

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