Vitamin C lab report sheet - Ciara Bell
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Lab 10. Determination of Vitamin C by Redox Titration
Name: Ciara Bell
Due Date: 11.08.2023
1.
Data and observations (46 points):
Mass of KIO
3:
(2 points)
-
1.1075g
Table 1: Data of the standardization of sodium thiosulfate. You can solve the average
and standard deviation using Excel (
22 points
):
Trial 1
Trial 2
Trial 3
Trial 4
Initial volume (mL)
0.20
0.05
0.12
0.09
Final Volume (mL)
32.10
31.10
30.67
21.70
Delivered
Volume
(mL)
31.90
31.05
30.55
21.61
Average (mL) and
standard
deviation
(mL)
n=3
x=31.17 mL
s
x
=0.6825
-
G
test
:
G
calc
=1.490
G
table
=1.463
1.490 > 1.463 = Reject Data
Table 2: Data of the back titration of vitamin C You can solve the average and
standard deviation using Excel (
22 points
):
Trial 1
Trial 2
Trial 3
Trial 4
Initial volume (mL)
0.02
0.45
0.30
0.23
Final Volume (mL)
16.05
13.87
13.25
14.94
Delivered
Volume
(mL)
16.03
13.42
12.95
14.71
Average (mL) and
standard deviation(mL)
n=4
x=14.28mL
1
s
x
=1.39
2.
Data treatment and calculations
Pre-lab questions (22 points):
(1)
In the lab you use potassium iodate (primary standard, KIO
3
) for vitamin C
determination. What are the molar masses of these two substances (2×2=4 points)
?
MM KI O
3
=
(
39.098
)
+
(
126.90
)
+
(
15.999
x
3
)
=
214.001
g
/
mol
MM C
6
H
8
O
6
=
(
12.011
x
6
)
+
(
1.008
x
8
)
+
(
15.999
x
6
)
=
176.124
g
/
mol
(2)
What is a back titration? How is the vitamin C concentration determined in this
experiment by a back titration (3×2 = 6 points)?
A back titration is a titration in which the molarity of an unknown compound is concluded by
reacting it with a known amount and concentration of the excess reagent. Vitamin C’s
concentration is determined by titrating it initially with KIO
3
and the doing a back titration
with KIO
3
to find the molarity and amount of Vitamin C.
(3)
Equation 2-4 are all redox reactions, identify the reducing and oxidizing agents in each of
the three equations (4×3=12 points)?
−
¿
+
3
H
2
O
+
¿
⇌
3
I
3
¿
−
¿
+
6
H
¿
−
¿
+
8
I
¿
Eq.
2:
I O
3
¿
The element I in IO
3
-
is reduced, so IO
3
-
is the oxidizing agent.
The element I in I
-
oxidized, so I
-
is the reducing agent.
2
+
¿
−
¿
+
2
H
¿
−
¿
+
H
2
O
⇌
C
6
H
6
O
6
+
3
I
¿
Eq.
3:
C
6
H
8
O
6
+
I
3
¿
The element I in I
3
-
is reduced, so I
3
-
is the oxidizing agent.
The element C in
C
6
H
8
O
6
oxidized, so
C
6
H
8
O
6
is the reducing agent.
−
¿
2
−
¿
+
3
I
¿
2
−
¿
⇌
S
4
O
6
¿
−
¿
+
2
S
2
O
3
¿
Eq .
4 :
I
3
¿
The element I in I
3
-
is reduced, so I
3
-
is the oxidizing agent.
The element S in
2
−
¿
S
2
O
3
¿
oxidized, so
2
−
¿
S
2
O
3
¿
is the reducing agent
3
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