Chemistry Reference
In-Depth Information
a tertiary standard, and so on. This process cannot continue indefinitely,
however, as errors creep in with every assay, and the results become less
reliable the farther the solution gets from the initial primary standard.
Worked example
A primary standard solution of Na
2
CO
3
was prepared and used to stand-
ardise a solution of H
2
SO
4
of unknown concentration. 25.0 mL of 1
M
(
f
1.000) Na
2
CO
3
was added by pipette to a conical flask and 24.60 mL of
H
2
SO
4
was required for neutralisation. Calculate the factor of the H
2
SO
4
solution.
From the reaction
H
2
O
it can be seen that 1 mole of sodium carbonate reacts with 1 mole of
sulfuric acid. Then
Na
2
CO
3
H
2
SO
4
1
Na
2
SO
4
CO
2
1 mole Na
2
CO
3
1 mole H
2
SO
4
1000 mL 1
M
Na
2
CO
3
1000 mL 1
M
H
2
SO
4
1 mL 1
M
H
2
SO
4
Since both solutions are 1
M
, the concentrations effectively cancel out
to leave the relationship
1 mL 1
M
Na
2
CO
3
(volume
factor) of Na
2
CO
3
(volume
factor) of H
2
SO
4
or, to put it another way,
(25 mL
f
(Na
2
CO
3
))
(24.60 mL
f
(H
2
SO
4
))
(25 mL
1.000)
(24.60 mL
f
(H
2
SO
4
))
and
f
(H
2
SO
4
) is given by 25
(1.000/24.6), so that
f
(H
2
SO
4
)
1.016
A moment's thought will confirm that the correct answer has been
achieved. The only calculation error that could be made in this simple
example is to get the factor upside-down (a so called 'inverted factor'). But,
in the reaction, 25 mL of a
f
1
.
000 solution of Na
2
CO
3
was neutralised
by
less than
25 mL of the acid. The acid must clearly be stronger than
f
1
.
000 if it required only 24.60 mL to neutralise the 25 mL of sodium
carbonate. A check of this type should be carried out after every volumetric
calculation. It is quick and easy to do and, to paraphrase the great Robert
Burns, 'It wad frae monie a blunder free us, An' foolish notion'.
