Every acid-base calculation follows the same four lines: equation, moles of the known substance, mole ratio, then the required quantity. The ratio line is where most errors enter.
This lesson is part of acids, bases and salts in SPM Chemistry. If moles still feel unfamiliar, go through moles, formulas and equations first.
What is the four-line routine?
Use these lines in order.
- Write the balanced equation.
- Find the moles of the substance you know: moles = concentration × volume in dm³, or mass ÷ molar mass.
- Use the coefficients to find the moles of the substance you want.
- Convert moles to the quantity asked for.
Keep the lines separate in your working. The marker can then award method marks even if a later line has an arithmetic slip.
Worked example 1: a mass of alkali
Question. What mass of sodium hydroxide neutralises 25.0 cm³ of 0.200 mol dm⁻³ hydrochloric acid? (Relative atomic masses: H = 1, O = 16, Na = 23.)
Equation: HCl + NaOH → NaCl + H₂O.
Moles of HCl = 0.0250 × 0.200 = 0.00500 mol. The ratio is 1 : 1, so moles of NaOH = 0.00500 mol. Molar mass of NaOH = 40 g mol⁻¹, so mass = 0.00500 × 40 = 0.200 g.
Worked example 2: a concentration from a titration
Question. 25.0 cm³ of 0.100 mol dm⁻³ sodium hydroxide needs 20.0 cm³ of sulfuric acid for complete neutralisation. Find the concentration of the acid.
Equation: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Moles of NaOH = 0.0250 × 0.100 = 0.00250 mol.
The ratio of acid to alkali is 1 : 2, so moles of H₂SO₄ = 0.00250 ÷ 2 = 0.00125 mol. Concentration = 0.00125 ÷ 0.0200 = 0.0625 mol dm⁻³.
Worked example 3: the mass of a salt
Question. Excess magnesium oxide reacts with 50.0 cm³ of 1.00 mol dm⁻³ hydrochloric acid. What is the maximum mass of magnesium chloride formed? (Mg = 24, Cl = 35.5.)
Equation: MgO + 2HCl → MgCl₂ + H₂O. Moles of HCl = 0.0500 × 1.00 = 0.0500 mol.
The ratio of HCl to MgCl₂ is 2 : 1, so moles of MgCl₂ = 0.0250 mol. Molar mass = 24 + 2(35.5) = 95 g mol⁻¹, so mass = 0.0250 × 95 = 2.375 g, which is 2.38 g to three significant figures.
The acid is the limiting reactant because the magnesium oxide is in excess, so the acid decides the yield.
The mistake to avoid
The common mistake is to use M₁V₁ = M₂V₂ for every titration. Apply it to example 2 and you get 25.0 × 0.100 ÷ 20.0 = 0.125 mol dm⁻³, which is exactly double the correct answer. The 1 : 2 ratio was ignored.
| Method | Answer for example 2 | Correct? |
|---|---|---|
| M₁V₁ = M₂V₂ | 0.125 mol dm⁻³ | No, ratio ignored |
| Moles and ratio | 0.0625 mol dm⁻³ | Yes |
The mole and stoichiometry steps tool lets you enter your own values and see each line.
Check yourself
20.0 cm³ of 0.500 mol dm⁻³ sulfuric acid neutralises sodium hydroxide of concentration 0.400 mol dm⁻³. Find the volume of sodium hydroxide needed.
Answer
Equation: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.
Moles of H₂SO₄ = 0.0200 × 0.500 = 0.0100 mol. Moles of NaOH = 2 × 0.0100 = 0.0200 mol. Volume = 0.0200 ÷ 0.400 = 0.0500 dm³, which is 50.0 cm³.
What to study next
Practise mixed questions in the acids, bases and salts practice set. For data where the acid’s basicity is not stated, try interpreting a neutralisation dataset without assuming every acid is monoprotic.
To have a teacher watch your working with you, see online one-to-one Chemistry tuition.