Ch 2 · Acid–Base Titrations · IGCSE Chemistry 0620
Topic 2 of 5 in Experimental Techniques and Chemical Analysis — IGCSE Chemistry 0620 — 48 lessons.
Introduction
A titration finds the exact volume of one solution that reacts with a measured volume of another, and from it a concentration. It is the most examined practical in chemistry: Paper 6 asks you to read burettes and average titres, and Papers 2 and 4 ask for the calculation. This lesson covers the method, the end-point, the recording rules and both levels of calculation.
Key ideas — The method
The first titration is usually a rough run to find the approximate end-point.
🧑🏫 Teacher's explanation
Imagine you are trying to balance a seesaw perfectly. If you add too much weight to one side, it tips instantly. In a titration, we are trying to find the exact moment when the number of acid particles perfectly matches the number of alkali particles. We use a burette to add the acid very slowly, drop by drop, so we don't 'overshoot' the balance point.
📖 Definitions
🔍 Worked example — Easy
A student performs a titration using 25.0 cm^3 of 0.10 mol/dm^3 NaOH and HCl. The initial burette reading is 0.00 cm^3 and the final reading is 22.45 cm^3. Calculate the volume of acid used.
🔍 Worked example — Standard
In a titration, three titres are obtained: 24.50 cm^3, 24.65 cm^3, and 24.55 cm^3. Calculate the mean titre using only concordant results.
🔍 Worked example — Exam-style
A 25.0 cm^3 sample of 0.20 mol/dm^3 KOH is titrated with 0.10 mol/dm^3 H_2SO_4. The volume of acid used is 50.00 cm^3. Calculate the concentration of the acid in mol/dm^3.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — Identifying the end-point
The end-point is the point at which the indicator changes colour, showing that the acid has exactly neutralised the alkali. Only indicators with a sharp colour change are suitable:
🧑🏫 Teacher's explanation
Imagine you are filling a glass with water from a tap that has a very slight drip. You want to stop exactly when the water reaches the rim without spilling. In a titration, the 'indicator' is like a sensor that flashes a light the moment the water reaches that perfect level. If the sensor is slow or changes color gradually, you will likely overfill the glass (overshoot the end-point).
📖 Definitions
🔍 Worked example — Easy
A student performs a titration using methyl orange. The initial color is yellow. The student stops the titration when the solution turns from yellow to a distinct orange. If the initial volume in the burette was 0.00 cm^3 and the final volume is 24.45 cm^3, what is the volume of acid added?
🔍 Worked example — Standard
A student obtains three titration results: 24.50 cm^3, 24.65 cm^3, and 24.55 cm^3. Calculate the mean titre using only concordant results.
🔍 Worked example — Exam-style
A 25.0 cm^3 sample of 0.10 mol/dm^3 HCl is titrated against 0.20 mol/dm^3 NaOH. Calculate the volume of NaOH required to reach the end-point.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — Recording results and the mean titre
Burette readings are recorded to the nearest 0.05 cm³, so every reading ends in .00 or .05 (for example 23.45 cm³). The titre is the final reading minus the initial reading, and results go in a table with rows for final volume, initial volume and titre.
🧑🏫 Teacher's explanation
Imagine you are trying to measure exactly how much water is in a leaking bucket. If you just take one quick measurement, you might be off because the water is still moving. In chemistry, we don't trust a single measurement. Instead, we perform the titration multiple times to make sure our results are consistent.
📖 Definitions
🔍 Worked example — Easy
A student records the following initial and final burette readings: Initial = 0.00 cm^3, Final = 24.35 cm^3. Calculate the titre.
🔍 Worked example — Standard
A student performs four titrations and obtains the following titres: 23.50 cm^3, 22.10 cm^3, 22.15 cm^3, and 22.20 cm^3. Calculate the mean titre using only concordant results.
🔍 Worked example — Exam-style
A student records the following titration results: 25.40 cm^3, 25.55 cm^3, 25.50 cm^3, and 25.90 cm^3. Calculate the mean titre to 2 decimal places.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — Titration calculations
For two solutions that react in a 1 : 1 ratio, the concentration and volume are linked by simple proportion. If 25.0 cm³ of sodium hydroxide of concentration 0.100 mol/dm³ is neutralised by 20.0 cm³ of hydrochloric acid, then because HCl + NaOH arrow NaCl + H_2O is 1 : 1, the acid is more concentrated in the ratio 25.0 : 20.0, giving 0.100 × 25.0/20.0 = 0.125 mol/dm³.
🧑🏫 Teacher's explanation
Think of a titration like a recipe for making a perfectly flavored drink. You have a glass of juice (the analyte) and you are adding syrup (the titrant) drop by drop. You want to find out exactly how strong the juice is. To do this, you need to know exactly how much syrup you added to reach the 'perfect' taste (the end-point) and exactly how strong that syrup is.
📖 Definitions
🔍 Worked example — Easy
A student titrates 25.0 cm³ of HCl with 0.100 mol/dm³ NaOH. The volume of NaOH used is 20.0 cm³. Calculate the concentration of HCl. The reaction is HCl + NaOH arrow NaCl + H_2O.
🔍 Worked example — Standard
25.0 cm³ of NaOH is neutralized by 22.50 cm³ of 0.0500 mol/dm³ H_2SO_4. Calculate the concentration of NaOH. The equation is H_2SO_4 + 2NaOH arrow Na_2SO_4 + 2H_2O.
🔍 Worked example — Exam-style
A student performs three titrations. The volumes of NaOH used are 22.50 cm³, 22.60 cm³, 22.55 cm³, and 23.10 cm³. Calculate the mean titre to be used for further calculations.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Worked example
Question. A student titrates 25.0 cm³ of aqueous sodium hydroxide with hydrochloric acid of concentration 0.120 mol/dm³ using thymolphthalein. The titres were 24.90, 24.35, 24.30 and 24.40 cm³. (a) State the colour change at the end-point. [1] (b) Identify the concordant titres and calculate the mean titre. [2] (c) (Supplement) Calculate the concentration of the sodium hydroxide in mol/dm³. [3]
🔍 Worked example — Easy
A student performs a titration. The volumes of acid added are 22.50 cm^3, 22.55 cm^3, 22.50 cm^3, and 23.10 cm^3. Calculate the mean titre using only concordant results, recording to 2 decimal places.
🔍 Worked example — Standard
Calculate the moles of H_2SO_4 in 25.0 cm^3 of 0.500 mol/dm^3 solution.
🔍 Worked example — Exam-style
A student titrates 25.0 cm^3 of NaOH with 0.120 mol/dm^3 HCl. The concordant titre is 24.35 cm^3. Calculate the concentration of NaOH. The reaction is NaOH + HCl arrow NaCl + H_2O.
⚠️ Common errors
Common mistakes
Practice
Answers Pipette 25.0 cm³ of potassium hydroxide into a conical flask [B1]; add a few drops of methyl orange or thymolphthalein [B1]; fill a burette with the acid and record the initial reading [B1]; add acid with swirling, dropwise near the end-point, until the indicator just changes colour, and record the final reading [B1]; repeat until concordant titres are obtained [B1]. It changes gradually through several colours so there is no sharp end-point. [B1] 17.75, 17.70 and 17.80 cm³ [B1]; mean = 17.75 cm³ [B1]. Moles of HCl = 0.100 × 0.0200 = 2.00 × 10^-3 mol [M1]; moles of Na_2CO_3 = 1.00 × 10^-3 mol [M1]; concentration = 1.00 × 10^-3/0.0250 = 0.0400 mol/dm³ [A1].