A 0.13 g sample of a purified carbonate was dissolved in 50.0 mL of 0.1 M HCl and boiled to eliminate CO2. Back titration of the excess HCl required 24.0 mL of 0.1 M NaOH. Identify the carbonate. Options:-
(a) CaCO3
(b) Na2C03
(c) MgCO3
(d) Insufficient Data
Answers
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
Answer:
A 0.13 g sample of purified carbonate was dissolved in 50.0 mL of 0.1 M HCl and boiled to eliminate CO2. Back titration of the excess HCl required 24.0 mL of 0.1 M NaOH. Identify the carbonate.
Explanation:
The carbonate is Na2CO3
Given Data-
Mass of purified Carbonate=0.13g
Molarity of Hcl=0.1M
The volume of HCl=50.00ml
Molarity of NaOH= 0.1M
The volume of NaOH= 24.0ml
In, Order to identify the carbonate in the given sample, we have to calculate the molar mass of carbonate salt and the carbonate salt cation.
For this, the amount of CO3, HCl and NaOH has to be obtained first.
HCl = Molarity of Hcl× Volume of HCl
=0.1 mmol HCl/ml HCl× 50.0ml HCl
=5.7mmol HCl
Amount of NaoH = 0.1 mmol NaOH/ml NaOH ×24.0ml NaOH
= 2.3726 mmol NaOH.