DSE Chemistry | Neutralisation and Titration: Mole Ratios, Indicators and Purity — Tutor Prep Guide

DSE Chemistry | Neutralisation and Titration: Mole Ratios, Indicators and Purity — Tutor Prep Guide | One equation covers three question types

2026-11-05

Tutor lesson-prep guides · Secondary Chemistry

Students can recite acid plus base gives salt and water, then treat every mole ratio as one to one. This lesson puts the coefficients back into the equation.

Objectives — Students can: (1) write a neutralisation equation and read the acid to base mole ratio, including one to two cases such as sodium carbonate; (2) use n = cV to find an unknown concentration and a sample purity from titration volumes; (3) explain why phenolphthalein suits strong acid with strong base while methyl orange suits a weak base titration.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill	Quick check	Pass mark
Mole calculations	What one mole represents and how mass links to it	n = m / M and n = cV
Acid and base definitions	Where the hydrogen and hydroxide ions come from	Acids release H+, bases release OH-
Volume units	How cubic centimetres convert to cubic decimetres	Divide by 1000

02 Lesson flow (45 minutes)

Time	Stage	Tutor does	Student does
0-5 min	Open with a ratio	Ask how much hydrochloric acid 25 cm3 of 0.100 M sodium carbonate needs, no hint	They answer 25 cm3, assuming one to one
5-15 min	Write the equation, circle the coefficient	CO3 2- plus 2H+ gives CO2 and H2O, circle the 2	The 2 is circled and justified
15-27 min	The four-step titration	Equation, moles, ratio, divide by volume	Four steps give 25 cm3
27-38 min	Purity back-calculation	1.20 g sample in 250 cm3, 25 cm3 titrated	Purity 80.0 percent
38-45 min	Indicators and exit questions	Phenolphthalein 8.3 to 10.0, methyl orange 3.1 to 4.4	All three right

03 Teaching script (read it out)

Scene 1: Circle the coefficient before saying neutralisation

Write CO3 2- plus 2H+ gives CO2 and H2O across the board, circle the 2 in red, and only then say the word neutralisation. The common failure is not acid-base theory but treating every reaction as one to one. Once the coefficient is circled, n(acid) equals two times n(base) is the only line to remember. Anchor it in numbers: 25 cm3 of 0.100 M sodium carbonate holds 0.0025 mol, which needs 0.0050 mol of acid, so 0.200 M hydrochloric acid delivers 25 cm3.
Change the concentration or the volume and only the numbers move; the coefficient never does.

Ask the student: if the carbonate were replaced by sodium hydroxide, how would the volume change, and why?

Scene 2: The titration is a four-step assembly line

Step one, write the equation and circle the coefficient. Step two, find the known moles with n = cV. Step three, apply the ratio to get the unknown moles. Step four, divide by the unknown volume for the concentration. Fixed order on the page means marking shows exactly which step broke.
Purity questions add one step: 24 cm3 of 0.100 M acid is 0.0024 mol, the aliquot is 25 of 250 cm3, so the flask holds 0.0240 mol, which times 40 gives 0.96 g, and against 1.20 g of sample that is 80.0 percent. Make the student say why the factor of ten appears; failing that, proportion is not yet secure.

Ask the student: why does purity use the moles in the whole flask rather than in the pipette?

Scene 3: An indicator is a chosen endpoint colour, not decoration

Phenolphthalein turns from colourless to pink between pH 8.3 and 10.0, which suits base in the burette. Methyl orange turns red to yellow between 3.1 and 4.4, which suits acid in the burette for a weak base. The wrong indicator puts the endpoint far from the equivalence point, and the error direction is predictable.
Close with a dilution: 10 cm3 of 1.0 M acid made up to 250 cm3 gives 0.040 M and pH 1.40, while 0.010 M sodium hydroxide sits at pH 12.00. Side by side, students see that pH is logarithmic, not linear.

Ask the student: why is 0.010 M acid at pH 2 but 0.010 M alkali at pH 12?

04 Question types and worked steps

Type	Example	Steps	Answer
Mole ratio question	Decide the multiplier from the equation	Write the equation, circle the coefficient, then multiply	Carbonate to acid is one to two
Concentration from titration	Find an unknown concentration	Known moles, ratio, divide by volume	Convert to dm3 first
Purity calculation	Find sample purity from titration data	Flask moles, times molar mass, divide by sample mass	Scale the aliquot back to the flask
Indicator choice	Pick an indicator for a titration	Locate the equivalence pH, then match the colour range	Phenolphthalein 8.3 to 10.0

Sanity check (teach this): Three numbers must agree: 0.0025 mol of carbonate needs 0.0050 mol of acid, and that divided by 0.200 M is 25 cm3, matching the given volume. The purity line checks the same way: 0.0240 mol times 40 is 0.96 g, and over 1.20 g that is 80.0 percent. A purity above 100 percent means the scaling was applied the wrong way round.

05 Independent practice (3 questions)

Question	Answer	What to watch
How much 0.200 M acid does 25.0 cm3 of 0.100 M carbonate need?	25.0 cm3	Whether the coefficient was applied
A 1.20 g impure sample is made up to 250 cm3; 25 cm3 needs 24 cm3 of 0.100 M acid. Find the purity.	80.0 percent	Whether the aliquot was scaled back
10.0 cm3 of 1.0 M acid is diluted to 250 cm3. Find the pH.	1.40	The concentration is 0.040 M, not 1.0 M

06 Common mistakes and what to say

Mistake	Student writes	Say this
Treating every neutralisation as one to one	Answering the carbonate question without the factor two	The coefficient comes from the equation, not from habit.
Forgetting to scale the aliquot	Reporting a purity of 8 percent	You titrated one tenth, so multiply the flask by ten.
Putting cm3 straight into cV	Moles a thousand times too large	Divide by 1000 before multiplying by concentration.
Using phenolphthalein everywhere	Writing phenolphthalein for an acid-side equivalence	An acid-side endpoint needs methyl orange.

07 Homework (paper, section and question numbers)

DSE Chemistry acid-base practice paper — Questions 1 to 3: write the equations and circle the coefficients; question 3 needs one decimal place in cm3.
Same paper, purity and pH — Question 5: purity from 24 cm3 of titrant and explain the factor of ten; question 6: pH after dilution.

08 Exit check (last 5 minutes)

Question	Answer	Decision
Carbonate to acid mole ratio	One to two	One to one means the equation was never written
First step of a purity calculation	Moles in the titrated aliquot, then scale to the flask	Dividing by sample mass first means the flow is missing
Phenolphthalein range	pH 8.3 to 10.0	Missing this means the indicator section needs a rerun

09 Tutor FAQ

Why titrate three times and average?

Each endpoint reading carries a random error of about 0.05 cm3; averaging reduces it, and a clear outlier is discarded with a stated reason.

Can a pH meter replace an indicator?

Yes, and it is more precise; DSE still examines indicators because it tests judgement about the equivalence point.

Why must purity be below 100 percent?

The sample contains impurities, so the measured sodium hydroxide mass is always below the sample mass.

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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.