Page 297 - Physical chemistry understanding our chemical world

264 ACIDS AND BASES

Worked Example 6.10 The methanoic acid from a nettle sting is extracted into 50 cm 3
of water and neutralized in the laboratory by titrating with sodium hydroxide solution.
−3
The concentration of NaOH is 0.010 mol dm . The volume of NaOH solution needed to
3
neutralize the acid is 34.2cm . What is the concentration c of the acid?
Unlike the Worked Example 6.9, we do not know the number of moles n of either reactant,
we only know the volumes of each. But we do know one of the concentrations.

Answer strategy. (1) First, we calculate the amount n of hydroxide required to neutralize
the acid. (2) We equate this amount n with the amount of acid neutralized by the alkali.
(3) Knowing the amount of acid, we finally calculate its concentration.
(1) To determine the amount of alkali, we first remember the defi-
nition of concentration c as
amount,n
concentration,c = (6.42)
volume,V

and rearrange Equation (6.41) to make amount n the subject, i.e.
We could have achiev-
ed this conversion
with quantity calcu- n = c × V
lus: knowing there are
3
3
3
3
1000 cm per dm (so The volume V of alkali is 34.2cm . As there are 1000 cm in a litre,
3
3
3
10 −3 dm cm −3 ). In SI this volume equates to (34.2 ÷ 1000) dm = 0.0342 dm . Accordingly
units, we write the
3
volume as 34.2cm × n = 0.0342 dm × 0.010 mol dm −3 = 3.42 × 10 −4 mol
3
3
10 −3 dm cm −3 .The
3
units of cm and cm −3
(2) The reaction between the acid and alkali is a simple 1:1 reac-
cancel to yield V = −4 −4
3
0.0342 dm . tion, so 3.42 × 10 mol of alkali reacts with exactly 3.42 × 10 mol
of acid.
(3) The concentration of the acid is given by Equation (6.42) again.
Notice now the units of Inserting values:
3
dm and dm −3 cancel
out here.
3.42 × 10 −4 mol
concentration c = 3
0.05 dm
Notice how we con- −3 −3
verted the volume of c = 6.8 × 10 mol dm
3
acid solution (50 cm )
3
to 0.05 dm . After extraction, the concentration of the methanoic acid is 0.068 mol
−3
dm .
An altogether simpler and quicker way of calculating the concentration of an acid
during a titration is to employ the equation
(6.43)
c (acid) × V (acid) = c (alkali) × V (alkali)
where the V terms are volumes of solution and the c terms are concentrations.