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Some samples refuse to be titrated directly. Chalk does not dissolve in water. Ammonia escapes from an open flask. A solid oxide reacts with acid so slowly that the indicator changes colour before the reaction is done. The fix is a back titration: add a known excess of reagent, let it finish reacting, then titrate what is left. This post gives you the one equation behind it, two worked purity problems, the slips that cost marks and a short practice set.
Where does back titration appear in the syllabus?
You will not find the phrase "back titration" in the JEE (Main) 2026 syllabus, but the method sits inside three of its lines: "Chemical equations and stoichiometry" in Unit 1, "Quantitative analysis (basic principles only)" including the estimation of nitrogen, and "the titrimetric exercises" in practical chemistry. The NEET (UG) 2026 syllabus published by the National Medical Commission has the same quantitative-analysis and titrimetric lines.
NCERT uses the method without naming it. In the Class 11 unit Organic Chemistry: Some Basic Principles and Techniques (2026-27 reprint), Kjeldahl's method traps ammonia "in an excess of standard solution of sulphuric acid", and the ammonia is found by "estimating unreacted sulphuric acid left after the absorption of ammonia by titrating it with standard alkali solution". Johan Kjeldahl published the method in 1883. Our post on Kjeldahl and Dumas nitrogen estimation covers that case; here we use the same logic for purity.
Why not titrate the sample directly?
Use a back titration when:
the sample is an insoluble solid, such as calcium carbonate or magnesium oxide, that reacts slowly;
the substance is volatile, like ammonia, and must be trapped first;
the reaction needs time or heat, so a direct end point would come too early.
The whole calculation rests on one line: moles of reagent that reacted with the sample = moles added − moles left over.
Worked example 1: how pure is this chalk?
Sample numbers. A 0.500 g chalk sample is dissolved in 50.0 mL of 0.200 M HCl. After the fizzing stops, the excess acid needs 20.0 mL of 0.100 M NaOH. Assume the impurities do not react with acid.
HCl added = 0.0500 L × 0.200 mol/L = 0.0100 mol.
HCl left over = NaOH used = 0.0200 L × 0.100 mol/L = 0.00200 mol (HCl and NaOH react 1 : 1).
HCl reacted = 0.0100 − 0.00200 = 0.00800 mol.
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂, so CaCO₃ = 0.00800 ÷ 2 = 0.00400 mol.
Mass of CaCO₃ = 0.00400 × 100.09 g/mol = 0.400 g. Purity = 0.400 ÷ 0.500 × 100 = 80.1%.
In a real lab, dissolved CO₂ is itself an acid and would use up some NaOH, so it is driven off before titrating. Exam questions assume this has been done.
Worked example 2: magnesium oxide and a diprotic acid
Sample numbers. 1.00 g of impure MgO is warmed with 100.0 mL of 0.250 M H₂SO₄. The excess acid needs 30.0 mL of 0.500 M NaOH.
H₂SO₄ added = 0.1000 × 0.250 = 0.0250 mol, which supplies 2 × 0.0250 = 0.0500 mol of H⁺.
OH⁻ used = 0.0300 × 0.500 = 0.0150 mol, so H⁺ left over = 0.0150 mol.
H⁺ reacted = 0.0500 − 0.0150 = 0.0350 mol.
MgO + 2H⁺ → Mg²⁺ + H₂O, so MgO = 0.0175 mol.
Mass = 0.0175 × 40.31 g/mol = 0.705 g. Purity = 70.5%.
Counting H⁺ rather than H₂SO₄ is the equivalent-weight idea in disguise. If you prefer n-factors, our guide to n-factor and the equivalent concept gives the same answer in one line.
Which mistakes cost marks?
Skipping the subtraction. Using the NaOH moles as if they measured the sample. The NaOH measures what the sample did not use.
Forgetting the acid is diprotic. One H₂SO₄ gives two H⁺. Missing this in example 2 gives 0.0250 − 0.0150 = 0.0100 mol of H⁺ reacted and a purity of about 20%, a wrong answer that looks plausible.
Dropping the 1 : 2 ratio. CaCO₃ and MgO each use two H⁺. Halving at the right step is where most wrong answers come from.
Mixing mL and L. Convert every volume to litres before multiplying by molarity.
Assuming inert impurities when the question says otherwise. If the impurity is, say, MgCO₃, it also uses acid, and you need a second equation.
Rounding early. Keep three significant figures until the last line.
Can you run the calculation backwards?
A common variant runs the other way: given the purity, predict the titre. For a 0.500 g chalk sample that is 90.0% CaCO₃, with the same 50.0 mL of 0.200 M HCl, CaCO₃ = 0.450 ÷ 100.09 = 0.004496 mol, so HCl reacted = 0.008992 mol, HCl left = 0.001008 mol, and the NaOH needed is 10.1 mL of 0.100 M. Purer chalk leaves less acid, so the titre falls. Check that direction every time.
When you are stuck on which ratio applies at which step, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer, so you can compare your ledger with a full solution.
A practice routine for back titration
Redo both examples from a blank page, writing the added, left-over and reacted lines before any arithmetic.
Change example 2 to 100.0 mL of 0.500 M HCl, which supplies the same 0.0500 mol of H⁺, and confirm the purity stays at 70.5%.
Write two reverse problems of your own: pick a purity, then work out the titre.
Mix five back titrations with five direct titrations and decide which is which before solving.
Tonight, take one problem from your module, draw the three-line ledger first, and only then pick up the calculator.
