Chemistry Reference
In-Depth Information
Now you can work out the number of moles of
the acid using the formula given in Chapter 4, p. 63.
In the example:
M
1
= 0.10 mol dm
−3
V
1
= 21.0 cm
3
M
acid
= 1 mole
M
2
= unknown
V
2
= 25.0 cm
3
M
alkali
= 1 mole
Substituting in the equation:
0.10 × 21.0
1
moles =
volume
1000
× concentration
= 21.0 ×
0.10
1000
= 2.1 × 10
−3
number of moles
=
number of moles
of hydrochloric acid of sodium hydroxide
Therefore, the number of moles of sodium
hydroxide
= 2.1 × 10
−3
2.1 × 10
−3
moles of sodium hydroxide is present in
25.0 cm
3
of solution.
Therefore, in 1 cm
3
of sodium hydroxide solution we
have
2.1 × 10
−3
25.0
M
2
× 25.0
1
=
Rearranging:
M
2
=
0.10 × 21.0 × 1
1 × 25.0
M
2
= 0.084
The concentration of the sodium hydroxide solution
is 0.084 mol dm
−3
.
Another example of a titration calculation could
involve a neutralisation reaction in which the ratio
of the number of moles of acid to alkali is not 1 : 1.
The example below shows how such a calculation
could be carried out.
moles
Therefore, in 1 litre of sodium hydroxide solution
we have
2.1 × 10
−3
25.0
Example
In a titration to find the concentration of a solution
of sulfuric acid, 25 cm
3
of it were just neutralised
by 20.15 cm
3
of a 0.2 mol dm
−3
solution of sodium
hydroxide. What is the concentration of the sulfuric
acid used?
First, write out the balanced chemical equation for
the reaction taking place.
× 1000 = 0.084 mole
The concentration of sodium hydroxide solution is
0.084 mol dm
−3
.
You can simplify the calculation by substituting in
the following mathematical equation:
M
1
V
1
M
acid
M
2
V
2
M
alkali
=
sulfuric acid + sodium hydroxide → sodium sulfate + water
H
2
SO
4
+ 2NaOH → Na
2
SO
4
+ 2H
2
O
From this balanced equation it can be seen that
1 mole of sulfuric acid reacts with 2 moles of sodium
hydroxide.
Therefore, the number of moles of sodium
hydroxide used
where:
M
1
= concentration of the acid used
V
1
= volume of acid used (cm
3
)
M
acid
= number of moles of acid shown in the
chemical equation
M
2
= concentration of the alkali used
V
2
= volume of the alkali used (cm
3
)
M
alkali
= number of moles of alkali shown in the
chemical equation.
0.2
1000
= 4.03 × 10
−3
= 20.15 ×
