Multi-Step Mole Calculations
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Lesson overview
Acid–base titration technique and calculations: concordant titres, the mole-ratio formula, and chaining moles, mass and concentration to find purity, molar mass and water of crystallisation.
Go to lesson quiz →Exam guidance · JAMB UTME Speed Lens
Core idea (Mole Concept, Molar Mass & Stoichiometry): Convert every mass, gas volume, or concentration to moles first, apply the balanced mole ratio, then convert to the target unit.
Pacing
~40 seconds per MCQ (180 questions in 120 minutes, no negative marking).
How JAMB marks this
Fast dimensional/sign elimination and standard formula shortcuts — skip multi-line algebra when substitution tests options faster.
Watch out for
UTME distractors routinely include the intermediate value (e.g., forgetting to convert cm³ to dm³ or °C to Kelvin).
Lesson content
Overview
Volumetric analysis, or titration, measures exactly how much of one solution reacts with a known volume of another. Food and drug laboratories use it to check the acid content of vinegar or the purity of an antacid, and the practical chemistry examination is built around it. UTME titration questions are multi-step mole calculations: from a titre to moles, through a mole ratio, then to a concentration, a mass, a percentage purity or a formula. In this lesson you will learn the technique, how to choose concordant titres, the formula that links two solutions, and how to chain several mole steps without losing track.
Key concept: titration
A titration is a procedure in which a solution of known concentration (the standard solution) is added from a burette to a measured volume of another solution until the reaction is just complete. The end point is the point at which the indicator changes colour, showing that the reaction is complete. The equivalence point is when the reactants have been mixed in exactly the ratio of the equation; a good indicator makes the two coincide. The titre is the volume of solution added from the burette to reach the end point. Concordant titres are titres that agree within 0.10 cm³ (some schemes allow 0.20 cm³); the average of concordant titres is used in calculations. The first rough titre is usually excluded. For the reaction aA + bB → products, with A in the burette, the mole ratio formula is CₐVₐ ÷ CᵦVᵦ = a ÷ b, where C is concentration and V is volume.
Technique and the calculation chain
Rinse the burette with the solution it will hold and the pipette with the solution it will deliver, so that any water left inside does not dilute them. Rinse the conical flask with distilled water only; extra water there does not change the number of moles in the pipetted portion. Remove the funnel from the burette before reading, read at eye level from the bottom of the meniscus, and record readings to two decimal places (0.05 cm³). Indicators: methyl orange suits titrations involving a strong acid with either a strong or a weak base (for example HCl with Na₂CO₃ or NH₃); phenolphthalein suits titrations of a weak acid with a strong base (ethanoic acid with NaOH); either works for strong acid with strong base. No indicator gives a sharp end point for a weak acid with a weak base. The calculation chain is the same every time. First, find the moles of the known solution: n = C × V (V in dm³). Second, use the mole ratio from the balanced equation to find the moles of the unknown in the portion titrated. Third, scale up if the portion was taken from a larger volume (for example 25.0 cm³ out of 250 cm³ is one-tenth). Finally, convert to whatever is asked: concentration (n ÷ V), mass concentration (× M), molar mass (mass ÷ n), percentage purity, or the number of water molecules of crystallisation. Useful equations: HCl + NaOH → NaCl + H₂O (1 : 1); H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O (1 : 2); Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂ (1 : 2, methyl orange).
Formulae to know
n = C × V (V in dm³ = cm³ ÷ 1000) CₐVₐ ÷ CᵦVᵦ = nₐ ÷ nᵦ (coefficients of acid and base in the equation) Scaling: moles in whole solution = moles in portion × (total volume ÷ portion volume) Molar mass = mass ÷ moles; % purity = (mass of pure substance ÷ mass of sample) × 100 Water of crystallisation: M(hydrate) = M(anhydrous) + 18x
Worked example 1: choosing titres and finding a concentration
25.0 cm³ of sodium hydroxide solution is titrated with 0.100 mol/dm³ HCl. The titres are: rough 20.60, then 20.10, 20.00 and 20.05 cm³ (see the bar chart). Find the concentration of the NaOH.
Active Worked Example · Predict Step 1
4 stepsWrite down the formula, rule, or first move you would make before revealing Step 1.
Worked example 2: a dibasic acid
20.0 cm³ of tetraoxosulphate(VI) acid neutralises 25.0 cm³ of 0.200 mol/dm³ sodium hydroxide. Find the concentration of the acid in mol/dm³ and g/dm³. (H = 1, S = 32, O = 16.)
Active Worked Example · Predict Step 1
4 stepsWrite down the formula, rule, or first move you would make before revealing Step 1.
Worked example 3 (UTME standard): water of crystallisation by titration
2.86 g of washing soda, Na₂CO₃·xH₂O, is dissolved and made up to 250 cm³. 25.0 cm³ of this solution requires 20.0 cm³ of 0.100 mol/dm³ HCl, using methyl orange. Find x. (Na = 23, C = 12, O = 16, H = 1.)
Active Worked Example · Predict Step 1
4 stepsWrite down the formula, rule, or first move you would make before revealing Step 1.
Common traps
• Exclude the rough titre and average only concordant results. • Use the mole ratio: H₂SO₄ and Na₂CO₃ each react with two moles of NaOH or HCl. • Do not forget to scale up when a portion is taken from a larger volume. • Convert cm³ to dm³ in n = C × V. • Record burette readings to two decimal places, ending in 0 or 5.
Summary
• Titration finds an unknown concentration from a standard solution; use concordant titres. • Rinse burette and pipette with their own solutions; the flask with distilled water. • Chain: n = CV → mole ratio → scale up → concentration, mass, molar mass, purity or formula. • CₐVₐ ÷ CᵦVᵦ = nₐ ÷ nᵦ. • Methyl orange for strong acids (including with Na₂CO₃ and NH₃); phenolphthalein for weak acids with strong bases.
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