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TITRATION ANALYSES 265

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Worked Example 6.11 A titration is performed with 25 cm of NaOH neutralizing
3
−3
29.4cm of nitric acid. The concentration of NaOH is 0.02 mol dm . Calculate the
concentration of the acid.
We rearrange Equation (6.43), to make c (acid) the subject:
We obtain here a ratio
c (alkali) V (alkali)
c (acid) = (6.44) of volumes (V (alkali) ÷
V (acid)
V (acid) ), enabling us to
We then insert values into Equation (6.44): cancel the units of the
two volumes. Units
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0.02 mol dm × 25 cm 3 are irrelevant if both
c (acid) =
29.4cm 3 volumes have the same
c (acid) = 0.017 mol dm −3 units.

Justification Box 6.4
A definition of the point of ‘neutralization’ in words says, ‘at the neutralization point,
the number of moles of acid equals the number of moles of hydroxide’. We re-express
the definition as
n (acid) = n (alkali) (6.45)

Next, from Equation (6.42), we recall how the concentration of a solution c when
multiplied by its respective volume V equals the number of moles of solute: n = c × V .
Clearly, n (acid) = c (acid) × V (acid) ,and n (alkali) = c (alkali) × V (alkali) .
Accordingly, substituting for n (acid) and n (alkali) into Equation (6.45) yields Equa-
tion (6.43).

SAQ 6.11 What volume of NaOH (of concentration 0.07 mol dm −3 )is
required to neutralize 12 cm 3 of nitric acid of concentration 0.05 mol
dm −3 ?

Aside

Equation (6.43) is a simplified version of a more general equation:
c (alkali) V (alkali)
c (acid) = s (6.46)
V (acid)
where s is the so-called stoichiometric ratio.
For the calculation of a mono-protic acid with a mono-basic base, the stoichiometry
is simply 1:1 because 1 mol of acid reacts with 1 mol of base. We say the stoichiometric
ratio s = 1. The value of s will be two if sulphuric acid reacts with NaOH since 2 mol
of base are required to react fully with 1 mol of acid. For the reaction of NaOH with
citric acid, s = 3; and s = 4if theacidisH 4 EDTA.