Titration Calculations: Stoichiometry Problems Solved

Titration is a laboratory technique used to find the unknown concentration of a solution by reacting it with another solution whose concentration is known exactly. A measured volume of the unknown solution (the analyte) is slowly reacted with the known solution (the titrant) until the reaction is just complete. The point where the reaction finishes is shown by a sudden colour change of an indicator, and the volume of titrant used is recorded.

This matters because titration is one of the most accurate ways to measure concentration, and exam questions on it are almost pure arithmetic. Whether you are finding the strength of an acid, the purity of a sample, or the percentage of a metal in an ore, the same mole-based method applies. Once the steps are clear, titration problems become free marks.

Key Concepts

• Titrant: the solution of known concentration, added from a burette.
• Analyte: the solution of unknown concentration, measured with a pipette into a flask.
• Equivalence point: the theoretical point where exactly enough titrant has reacted with all the analyte.
• End point: the point where the indicator changes colour; chosen to be as close as possible to the equivalence point.
• Indicator: a dye that changes colour in a narrow pH range, for example phenolphthalein (colourless to pink) or methyl orange (red to yellow).
• Molarity (M): moles of solute per litre of solution; moles = molarity × volume in litres.
• Stoichiometric ratio: the mole ratio of reactants taken from the balanced chemical equation.

Formulas

• Moles from a titration reading: n = M × V, where V is in litres. Units: moles = (mol/L) × L.
• Stoichiometry relation: n(analyte) = n(titrant) × (coefficient of analyte / coefficient of titrant) from the balanced equation.
• Concentration of analyte: M(analyte) = n(analyte) / V(analyte in litres).

Assumptions: the reaction goes to completion and the balanced equation is correct.

Worked Example

A 25.0 mL sample of hydrochloric acid is titrated with 0.100 M sodium hydroxide solution, and 30.0 mL of NaOH is needed to reach the end point. Find the molarity of the acid.

1. Write the balanced equation: HCl + NaOH → NaCl + H₂O. The mole ratio HCl : NaOH is 1 : 1.
2. Convert the titrant volume to litres: 30.0 mL = 0.0300 L.
3. Moles of NaOH used: n = 0.100 mol/L × 0.0300 L = 0.00300 mol.
4. Moles of HCl reacted: with a 1 : 1 ratio, n(HCl) = 0.00300 mol.
5. Volume of acid in litres: 25.0 mL = 0.0250 L.
6. Molarity of HCl = 0.00300 mol / 0.0250 L = 0.120 mol/L.

The acid concentration is 0.120 M.

Common Misconceptions

• The equivalence point and the end point are the same thing. They are not identical: the equivalence point is the theoretical completion of the reaction, while the end point is the observed colour change of the indicator. A good indicator is chosen so the two are nearly equal.
• You can just use the volume ratio to get the concentration. This only works when the balanced equation has a 1 : 1 ratio. For example, H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O needs a factor of 2, so the mole ratio must come from the equation, never from the volumes alone.
• More indicator gives a better result. Too much indicator shifts the end point and wastes reagent. Only 2-3 drops are used.

Real-World Applications

• Pharmaceutical companies check the exact strength of medicines and raw materials.
• Food labs measure the acidity of vinegar, fruit juices and dairy products.
• Water treatment plants test water hardness and chlorine content.
• Environmental labs measure dissolved pollutants in river and lake water.

Titration turns a visible colour change into an exact number: measure the titrant volume, convert to moles, apply the balanced-equation ratio, and divide by the analyte volume. Practise the litre conversions and the mole ratio step, and these problems are solved in minutes.