Write a SOP (standard
operating procedure)
for doing a titration
Do not use your books
Hints – diagram, use known/unknown/
tips
Assessment at AS
Unit F331
Chemistry for Life
Unit F332
Chemistry of Natural
Resources
Unit F333
Chemistry in Practice
(internal assessment)
Elements of Life
Developing Fuels
Elements from the Sea
The Atmosphere
Polymer Revolution
Five practical skills
15% of A level
25% of A level
10% of A level
Written paper
1 hr 15 min
Written paper (one
question on an ‘Advance
Notice’ passage)
1 hr 45 min
Internal assessment of
skills using materials and
mark schemes provided
by OCR
CI p371
Content
• Skill of carrying out a titration
• Calculate unknowns and percentage
uncertainty
Process
• Carry out a titration
Benefit
• Help in 10% of AS grade!
ES4.1
• Carry out titration
• Put your results on the board
• Use the booklet to help write your
results
The question:
10.0 cm3 of a solution of potassium hydroxide was
titrated with a 0.10 M solution of hydrochloric acid. 13.5
cm3 of the acid was required for neutralization.
Calculate the concentration of the potassium hydroxide
solution.
Step 1
Write down everything you know from the question. I put
the one we know most about on the left
Hydrochloric Acid
Volume used = 13.5 cm3
Concentration = 0.10M
Potassium hydroxide
Volume used = 10.0 cm3
Concentration = TBD
Step 2
Write the BALANCED equation (or as much as you know
about it).
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
This is necessary to check out the ratios of reactants – in
this case 1:1 so we don’t have to worry about any ratios.
Step 3
Calculate the Number of Moles used of the reactant you
know most about. In this case the HCl(aq)
The rule here is ;
Number of Moles = concentration
Step 3
Calculate the Number of Moles used of the reactant
you know most about. In this case the HCl(aq)
The rule here is ;
Number of Moles = volume × concentration
Never do cm3!
Step 4
Work out how many moles of the unknown you
have used, this is where you may need to multiply
up or down the number of moles, so if 1 mole of
HCl needed 2 mol of potassium hydroxide, then at
this point you would multiply the number of moles
of HCl by 2. In this case this is unnecessary.
Number of moles of HCl = 0.0014 mol
Number of moles KOH = 0.0014 mol – they react in a
one to one ratio
Step 5
Work out the concentration of the potassium
hydroxide.
Moles of KOH (same as HCl as 1:1) – 0.0014
Volume (q) 10cm3
Number of Moles = volume × concentration
Concentration = 0.14 M
CH3COOH + NaOH  H2O + NaOOCCH3
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0.1M NaOH
? CH3COOH
Volume 6cm3
10cm3
Balance equation
No. moles = Vol dm3 x conc.
?
= (6/1000) x 0.1Moles dm-3
?
= 6 x 104 Moles
1:1 ratio thus 6 x 104 Moles of acid
So Conc = Moles/vol dm3 = 6 x 10 4 / (10/1000)
= 0.06 Moles dm-3
Deduce the concentration of hydrochloric
acid if 20.0cm3 is neutralised by 25.0cm3
of sodium hydroxide at 0.2oo moldm-3?
Carry out a titration?
Confident with procedure?
Confident with calculations?
What to do if you’re not?
Best Choice – Salters A level / ES
Formula, equations, amount / Titrations
and titration calculations
Next lesson
is halogen
production
conference –
have poster
with you and
all 15.1-6 qs
+ ready to
speak