National 5 - Calculations from Volumetric titrations
Example
0.1 mol l-1 HCl
20cm3 NaOH
+ Universal indicator
In this reaction 10.0cm3 of acid was needed to neutralise 20cm3 of the alkali.
When doing a titration, the first titre is a rough value and is not used in calculating an
average. Only 2 titres that are within 0.2cm3 of each other are used to calculate the
average volume needed for neutralisation.
Step 1 – Balanced Equation to work out the ratio
HCl
+
NaOH
1
:
1
NaCl
+
H 2O
Step 2 – Calculate unknown value from reactant that you have 2 values for.
We have a volume and a Concentration for HCl in this reaction.
n=c x v
n = 0.1 x 0.01
n = 0.001 moles of HCl
Remember to convert
volume into Litres
Step 3 – Calculate the unknown value for the other reactant using n from step 2
c =
n
___
v
=
0.001
___
0.02
= 0.05 mol l-1
Because they react in
a 1:1 ratio, the value of
n for HCl is the same
as n for NaOH.
Examples for you to try
1 - What is the concentration of a Sodium hydroxide solution if 20cm3 of
the solution is neutralised by
A – 20cm3 of 1 mol l-1 Hydrochloric acid
B – 40cm3 of 2 mol l-1 Nitric acid
C – 40cm3 of 0.25 mol l-1 Sulfuric acid (Careful – not a 1:1 ratio)
2 - What Volume of 0.1 mol l-1 Hydrochloric acid is needed to neutralise
A – 15cm3 of 0.02 mol l-1 Potassium hydroxide
B – 40cm3 of 0.5 mol l-1 Lithium hydroxide
C – 20cm3 of 0.05 mol l-1 Barium hydroxide solution
(Careful – not a 1:1 ratio)
Answers
Question
A
B
C
1
1 mol l-1
4 mol l-1
1 mol l-1
2
0.003 Litres (3cm3)
0.2 Litres (200cm3)
0.02 Litres (20cm3)
Alternative method
CA
x
VA
Conc
Volume of
Of Acid x
Acid
x
x
nHA
Number of H+
ions in formula
=
CB
Conc of
= Base
x
x
VB
x
nOHB
Volume
Number of OHof Base x ions in formula