Estimation of oxalic acid by titrating with KMnO4.

1. AIM: Estimation of oxalic acid by titrating with
KMnO4.
THEORY:
The titration of potassium permanganate (KMnO4) against oxalic acid (C2H2O4) is an
example of a redox reaction. In this titration, Mn is reduced and C is oxidized from
KMnO4 and (COOH)2, respectively. Initially, KMnO4 reacts with oxalate ions to form
Mn2+, K+, CO2, and water, which result in a colourless solution. After complete
consumption of oxalate ions at the endpoint, an extra drop of KMnO4 turns the
solution pink which indicates complete oxidation of oxalate ions. As the reaction is
sluggish at room temperature oxalic acid along with sulphuric acid is heated to about
60 °C before the titration. The aqueous solution of KMnO4 needs to be standardized
using sodium oxalate.
1

2. Observations: Standardization of KMnO4
• Burette solution: ~0.05N KMnO4 solution
• Conical flask solution: 10 mL of 0.05N Na2C2O4
+ 20 mL 2N H2SO4+ heat (~60 °C)
• Indicator: KMnO4 acts as self indicator
• Colour change: Colourless to light pink
• Reaction:
2(MnO4)- + 5(C2O4)2- + 16H+ → 2Mn2+ + 10CO2 + 8H2O
Burette
reading
Piolet
reading
I (mL) II (mL) III (mL) Constant
(mL)
Final
9.0-10.0 mL
20.0 30.2 40.3
10.1
Initial 10.0 20.1 30.2
Difference 10.0 10.1 10.1
Observations table

3. Observations: Estimation of oxalic acid
• Burette solution: ~0.05N KMnO4 solution
• Conical flask solution: 10 mL H2C2O4 + 20 mL 2N H2SO4+ heat (~60 °C)
• Indicator: KMnO4 acts as self indicator
• Colour change: Colourless to light pink
• Reaction:
2KMnO4 + 5H2C2O4 + 3H2SO4 → 2MnSO4 + K2SO4 + 10CO2 + 8H2O
Burette
reading
Piolet
reading
I (mL) II (mL) III (mL) Constant
(mL)
Final
9.0-10.0 mL
20.0 30.0 40.0
10.0
Initial 10.0 20.0 30.0
Difference 10.0 10.0 10.0
Observations table

4. Calculations:
1. Equivalent weight (E.W.) of oxalic acid
(H2C2O4.2H2O)
Molecular weight (M. W.) = (2 ×1) + (2×12) + (4×16) + (4×1) + (2 ×16)
= 126
Equivalent weight = (126/2) = 126 g/2 = 63 g

5. 2. Preparation of 100 mL of 0.05N N sodium oxalate
solution.
Molecular weight (M.W.( = 134 g
Equivalent weight = (M. W./2) = 134 g/2 = 67 g
1000 mL = 1N Na2C2O4 = 67 g
100 mL = 0.05N Na2C2O4 = ‘x’ g
Therefore, x = (0.05N X 100 mL X 67 g)/ (1N X 1000 mL) = 0.335 g
0.335 g of sodium oxalate dissolved and diluted up to the mark to prepare
0.05N standard solution.

6. 3. Exact normality of ~0.05N KMnO4.
10 mL of 0.05N Na2C2O4.2H2O solution requires 10.1 mL
of 0.05N KMnO4 solution
(sodium oxalate) N1V1 = N2V2(B.R of standardization) (KMnO4)
0.05 N X 10.0 = N2 X 10.1
N2 = 0.0495 N
Exact normativity (strength) of KMnO4 is 0.0495 N

7. 4. Strength of unknown concentration of oxalic acid.
10 mL of unknown concentration of oxalic acid solution requires 10
mL of 0.0495N KMnO4 solution
(oxalic acid) N1V1 = N2V2(B.R of estimation) (KMnO4 solution)
N1× 10 = 0.0495× 10
N1 = 0.0495 N
Strength of unknown concentration of oxalic acid is 0.0495 N

8. 4. Amount of oxalic acid in 100 mL.
Amount of oxalic acid in 1000 mL = N X E. W.
= 0.0495 X 63 g
= 3.119 g/1000 mL
So,
3.119 g → 1000 mL
0.3119 g → 100 mL
0.312 g → 100 mL
100 mL of given solution contains 0.312 g of oxalic acid.

9. Results:
• The exact normality of given KMnO4 solution
is 0.0495N.
• Amount of oxalic acid in 100 mL is 0.312 g.