Senior Chemistry | Salt Preparation and Neutralisation Prep Guide (for tutors) | Mole ratio, crystallisation and yield in three steps

2026-11-06

Tutor lesson-prep guides · Secondary Chemistry

Students can write acid plus base gives salt plus water, but guess the volume of acid and the mass of salt. This lesson strings mole ratio, solubility and yield into one calculation route.

Objectives — Students can: (1) use the neutralisation mole ratio to find the acid volume, including halving it for a dibasic acid; (2) convert moles to theoretical mass and then to percentage yield; (3) use solubility to choose between evaporation and cooling crystallisation and explain why the solution must not be evaporated to dryness.

01 Prerequisites (3-minute check)

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

Skill	Quick check	Pass mark
Moles	What unit must V be in for n = cV	dm3, which is litres
Neutralisation	Mole ratio of NaOH to H2SO4	2 to 1
Relative mass	Mr of NaCl	58.5

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-8 min	Opening diagnostic	How many moles in 25 cm3 of 0.100 M NaOH	Hands up
8-22 min	Mole ratio	0.0025 mol NaOH needs 25.0 cm3 of HCl but only 12.5 cm3 of H2SO4	Work it once themselves
22-36 min	Mass and yield	Theoretical NaCl 0.146 g; obtained 0.112 g, yield 76.6 per cent	Compute the percentage
36-48 min	Crystallisation	At 25 C, 36 g dissolves in 100 g water, so 18 g in 50 g	Decide if solid appears
48-56 min	Never evaporate to dryness	CuSO4.5H2O has Mr 249.6 and water is 36.1 per cent	State what heating does
56-60 min	Exit check	Collect and mark three items	Work alone

03 Teaching script (read it out)

Scene 1: The mole ratio comes from the equation, not the volume

n = cV, and V must be in dm3: 0.100 times 0.0250 gives 0.0025 mol of NaOH.
HCl reacts one to one, so it needs the same 0.0025 mol, a volume of 0.0025 divided by 0.100 = 0.0250 dm3, which is 25.0 cm3.
H2SO4 is dibasic and supplies two moles of hydrogen ions per mole, so only half the moles are needed: 12.5 cm3. Method: write the equation, count coefficients, then substitute. Skipping the equation guarantees an error.

Students say: equal volumes of HCl and H2SO4 at the same concentration, which neutralises more? They must answer twice as much for H2SO4.

Scene 2: From moles to grams, then to percentage yield

0.0025 mol of NaCl with Mr = 23.0 + 35.5 = 58.5 gives a theoretical mass of 0.0025 times 58.5 = 0.146 g.
After evaporation 0.112 g was weighed, so the yield is 0.112 divided by 0.146 times 100, which is 76.6 per cent.
Losses come from three places: residue left in the transfer, spitting during evaporation, and incomplete reaction. Anything above 100 per cent means the sample is wet or impure.

Students do: change the obtained mass to 0.130 g, recompute the yield and say whether 90 per cent is suspicious.

Scene 3: Crystallisation uses a solubility difference, not dryness

Sodium chloride dissolves to about 36 g per 100 g of water at 25 C, so 50 g holds 18 g. The 0.146 g produced is far below that, so the solution is unsaturated and most of the water must be removed before crystals appear.
For salts whose solubility rises with temperature, such as copper sulphate, heat to concentrate and then cool for larger, purer crystals. For salts like sodium chloride, evaporate instead.
Never evaporate to dryness: the Mr of CuSO4.5H2O is 249.6, of which water is 90.0, or 36.1 per cent. Heating dry removes the water of crystallisation, so the mass is no longer that of the hydrate.

Students say: why not evaporate to dryness for a pure salt? They must mention losing water of crystallisation and concentrating impurities.

04 Question types and worked steps

Type	Example	Steps	Answer
Acid volume	25 cm3 of 0.100 M NaOH against 0.100 M H2SO4	Write the equation, count coefficients, halve	12.5 cm3
Theoretical mass	0.0025 mol of NaCl	n times Mr	0.146 g
Yield	Theory 0.146 g, obtained 0.112 g	Obtained over theory times 100	76.6 per cent
Solubility	50 g water, 36 g per 100 g water	Scale by proportion	18 g

Sanity check (teach this): Sanity check: 0.100 M times 25 cm3, which is 0.0250 dm3, gives 0.0025 mol, matching the hand calculation; the H2SO4 volume 12.5 cm3 is exactly half of the HCl volume 25.0 cm3, as a dibasic acid requires; 0.0025 mol times Mr 58.5 gives 0.146 g, a sensible millimole-scale mass; the yield 76.6 per cent is below 100, the right direction; 18 g is half of 36 g, matching the water being halved from 100 g; water is 36.1 per cent of the hydrate, close to the usual pentahydrate figure.

05 Independent practice (3 questions)

Question	Answer	What to watch
Acid volume, three items	One monobasic and one dibasic acid	Is the coefficient division done
Mass and yield, three items	Different moles and obtained masses	Is the correct Mr used
Solubility, two items	Different water masses and temperatures	Is the proportion right
Water of crystallisation, one item	Percentage from Mr	Are all five water molecules counted

06 Common mistakes and what to say

Mistake	Student writes	Say this
Using cm3 directly	0.100 times 25 gives 2.5 mol	V must be in dm3; 25 cm3 is 0.0250 dm3, so the answer is out by a factor of 1000.
One to one for a dibasic acid	Using 25.0 cm3 of H2SO4 as well	H2SO4 supplies two hydrogen ions, so half the moles are needed: 12.5 cm3. Writing the equation prevents this.
Theory over obtained	Writing 0.146 over 0.112 to get 130 per cent	Yield is obtained over theory; reversing it produces an impossible figure above 100 per cent.
Evaporating the hydrate dry	Evaporating copper sulphate solution to complete dryness	That removes the water of crystallisation and concentrates impurities, raising the mass; evaporate to the point of crystallisation and then cool.

07 Homework (paper, section and question numbers)

Secondary chemistry chapter index — All chapters at a glance
Prep guide master index — Compare similar chapters
Redo this chapter — Two acid-volume items, two yield items and one hydrate item

08 Exit check (last 5 minutes)

Question	Answer	Decision
Moles in 25.0 cm3 of 0.100 M NaOH	0.0025 mol	Answering 2.5 means the 1000 division was missed
Volume of 0.100 M H2SO4 needed for the same base	12.5 cm3	Answering 25.0 sends you back to act one
Why not evaporate the salt solution to dryness	It loses water of crystallisation and concentrates impurities	Answering it would burn sends you back to act three

09 Tutor FAQ

Is an indicator always needed?

Phenolphthalein or methyl orange both work for acid-base titration. A pH meter or conductivity probe also finds the end point, but exam questions usually name the indicator and its colour change.

Can a yield exceed 100 per cent?

Not in theory. A figure above 100 means the product is wet, impure, or a titre was misread, so students should recheck the data rather than write it down.

Can every salt be crystallised by evaporation?

No. Salts whose solubility changes strongly with temperature, such as potassium nitrate and copper sulphate, give purer crystals on cooling; only salts like sodium chloride use evaporation.

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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.