DSE Chemistry | Acid-base Neutralisation and Titration: Tutor Lesson Plan

2026-11-03

Tutor lesson-prep guides · Secondary Chemistry

Students copy a formula and collapse the moment the acid is diprotic. This lesson builds one bridge: write the equation, read the mole ratio, then compare cV with cV, and finish by checking the pH and the mass balance.

Objectives — Students can: (1) write HCl + NaOH and state the 1:1 ratio; (2) find that 25.0 cm3 of 0.100 mol/dm3 NaOH needs 20.0 cm3 of 0.125 mol/dm3 HCl; (3) apply the 2:1 ratio for sulfuric acid and get 12.5 cm3; (4) average titres 20.0, 20.1 and 20.0 to 20.03 cm3 and explain why a 0.17 % difference from theory is acceptable.

01 Prerequisites (3-minute check)

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

Skill	Quick check	Pass mark
Molar concentration	How many moles are in 25.0 cm3 of 0.100 mol/dm3 NaOH?	0.00250 mol
Unit conversion	Convert 25.0 cm3 to dm3.	0.0250 dm3
Neutralisation	What are the products of HCl and NaOH?	A salt and water
pH	What is the pH of 0.050 mol/dm3 HCl?	1.30

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-8 min	Neutralisation	Ask what else is formed besides water	Write the full equation
8-18 min	Mole ratio	Show 1:1 and the 2:1 case for sulfuric acid	Mark the coefficients
18-30 min	Titration steps	Rinse, fill, read start and end volumes	Explain why the bottom of the meniscus
30-42 min	Concentration from titre	Back-calculate NaOH from 20.0 cm3 HCl	Repeat with units
42-50 min	Error and end-point	Compare indicator change with equivalence	State why results run high or low
50-60 min	Independent practice	Three questions in eight minutes	Check and label each error

03 Teaching script (read it out)

Scene 1: The mole ratio lives in the equation, not in a formula sheet

HCl + NaOH gives a 1:1 ratio, so n(HCl) equals n(NaOH). Sulfuric acid is different: one H2SO4 needs two NaOH. 25.0 cm3 of 0.100 mol/dm3 NaOH contains 0.00250 mol, so only 0.00125 mol of acid is needed, which is 0.0125 dm3 or 12.5 cm3. Insist that students write the equation first and copy the coefficients, never guess.

Have the student write the sulfuric acid case and then predict phosphoric acid.

Scene 2: A burette gives two readings, not one

Record the start and the end volume and subtract. Read the bottom of the meniscus at eye level. Show a start of 0.50 cm3 and an end of 21.10 cm3 and ask for the delivered volume: 20.60 cm3. Students routinely quote the end reading as the volume, and every later step inherits the error.

Ask the student to say “end minus start” and write both readings on the practical sheet.

Scene 3: Average only concordant titres, to three significant figures

20.0, 20.1 and 20.0 differ by at most 0.1 cm3, so they are concordant and average to 20.03 cm3. A 24.5 cm3 run means a rinsed jet or a fast tap and must be rejected, never averaged in. Theory for 0.125 mol/dm3 HCl against 25.0 cm3 of 0.100 mol/dm3 NaOH is 20.0 cm3, and 20.03 cm3 is 0.17 % away, which is acceptable; tell students that 1 % is normally tolerated and 5 % means a technique problem.

Ask the student to pick which of three runs to reject and justify it.

04 Question types and worked steps

Type	Example	Steps	Answer
Strong acid against strong base	25.0 cm3 of 0.100 M NaOH with 0.125 M HCl	Write 1:1, find n from cV, divide by the acid concentration	20.0 cm3
Diprotic acid	Same base with 0.100 M H2SO4	Write 2:1, halve the moles, then divide	12.5 cm3
Concentration from titres	25.0 cm3 NaOH needs 20.03 cm3 of 0.125 M HCl	n(HCl) = cV, divide by 0.0250, keep three significant figures	0.100 mol/dm3
pH and dilution	0.050 M HCl diluted ten times	pH = -log[H+]; tenfold dilution adds 1 to pH	1.30 to 2.30

Sanity check (teach this): Order-of-magnitude check: 0.00250 mol of NaCl weighs 0.146 g, small but weighable. A titre near 20 cm3 fits a 50 cm3 burette, so a result of 200 cm3 signals a unit slip. Diluting tenfold raises pH from 1.30 to 2.30: still acidic, as expected.

05 Independent practice (3 questions)

Question	Answer	What to watch
How much 0.150 M HCl neutralises 30.0 cm3 of 0.100 M NaOH?	20.0 cm3	Do they find n(NaOH) before dividing
How many moles of H+ are in 20.0 cm3 of 0.050 M H2SO4?	0.0020 mol	Do they multiply by two
What is the pH of 0.0050 M HCl?	2.30	Do they misplace the exponent

06 Common mistakes and what to say

Mistake	Student writes	Say this
Quoting the end reading as the volume	“The end reading is 21.10, so I used 21.10 cm3”	“That is where the liquid sits, not how much ran out; subtract the start and write both numbers down.”
Treating sulfuric acid as monoprotic	“H2SO4 and NaOH are 1:1”	“Look at the equation: one acid takes two hydroxide ions, so you are out by a factor of two.”
Using cm3 as dm3	“25.0 cm3 is 25.0 dm3”	“There are 1000 cm3 in a dm3; writing 25 makes the volume a thousand times too big.”
Averaging an obviously bad run	“20.0, 20.1 and 24.5 average to 21.5”	“A run 4 cm3 away is a technique fault; reject it and say why, do not average it in.”

07 Homework (paper, section and question numbers)

Chemistry chapter index — Revise mole calculations first, then questions 1 and 2
Mole and stoichiometry — Question 3 needs n = cV, units before numbers
DSE Chemistry paper — Do the titration section in 12 minutes and label each error

08 Exit check (last 5 minutes)

Question	Answer	Decision
Volume of 0.125 M HCl for 25.0 cm3 of 0.100 M NaOH	20.0 cm3	Error: ratio ignored
Volume of 0.100 M H2SO4 for the same base	12.5 cm3	Error: acid treated as monoprotic
pH of 0.0050 M HCl	2.30	Error: significant figures or log direction
Average of 20.0, 20.1 and 20.0	20.03 cm3	Error: bad run averaged in

09 Tutor FAQ

Which indicator should we choose?

Phenolphthalein or methyl orange both work for strong acid against strong base; a weak acid against strong base needs phenolphthalein because the equivalence point is alkaline.

What is the difference between end-point and equivalence point?

Equivalence is the exact mole ratio, a theoretical point; the end-point is the indicator colour change, an observation. The gap is the titration error.

Why use the same burette three times?

The same bore and the same reading bias make the runs comparable; changing burettes changes the ruler.

Find a tutor — free Tutor Registration →

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.