Answer:
Sodium carbonate reacts with hydrochloric acid:
N
a
2
C
O
3
(
s
)
+
2
H
C
l
(
a
q
)
→
2
N
a
C
l
(
a
q
)
+
C
O
2
(
g
)
+
H
2
O
(
l
)
We can find the number of moles of
HCl
before the reaction.
We then use the titration result to find the number of moles remaining after the reaction.
By subtracting the two we can get the number of moles of
HCl
which have reacted.
From the equation we can find the number of moles of
Na
2
CO
3
.
From this we get the mass of
Na
2
CO
3
.
We can then work out the percentage purity.
Concentration = moles of solute / volume of solution.
c
=
n
v
∴
n
=
c
×
v
∴
Initial moles of
H
C
l
=
0.1280
×
50.00
1000
=
6.400
×
10
−
3
The acid remaining is titrated with
NaOH
:
H
C
l
(
a
q
)
+
N
a
O
H
(
a
q
)
→
N
a
C
l
(
a
q
)
+
H
2
O
(
l
)
n
O
H
−
=
0.1220
×
30.10
1000
=
3.6722
×
10
−
3
Since they react in a 1:1 ratio the no. of moles of
HCl
must be the same:
n
H
C
l
=
3.6722
×
10
−
3
∴
The no. moles used up:
=
(
6.400
−
3.6722
)
×
10
−
3
=
2.7228
×
10
−
3
From the original equation you can see that the no. moles of
Na
2
CO
3
must be half of this.
∴
n
N
a
2
C
O
3
=
2.7228
×
10
−
3
2
=
1.3614
×
10
−
3
M
r
[
N
a
2
C
O
3
]
=
105.99
∴
m
a
s
s
N
a
2
C
O
3
=
1.3614
×
10
−
3
×
105.99
=
0.14429
g
∴
percentage purity
=
0.14429
0.3240
×
100
=
44.53
%
This technique is known as a "back titration".
