Class 12 Chemistry Practicals Volumetric Analysis Estimation Oxalic Acid

Ex.No: 03.   ESTIMATION OXALIC ACID

This Class 12 Chemistry practical record covers Experiment 3 on volumetric analysis, the estimation of oxalic acid dissolved in 500 ml of a given solution using standard ferrous ammonium sulphate (FAS) and potassium permanganate (KMnO₄) as the link solution. It also includes the standardisation of hydrochloric acid with sodium carbonate, with the normality calculations, weight calculation and result for each titration.

EXPERIMENT NO. 03: ESTIMATION OF OXALIC ACID

AIM

To estimate the amount of oxalic acid dissolved in 500 ml of the given solution volumetrically. For this you are given a standard solution of ferrous ammonium sulphate (FAS) of normality 0.1 N and potassium permanganate as the link solution.

PRINCIPLE

During these titrations, oxalic acid is oxidised to CO₂ and MnO₄^- ions (from KMnO₄) are reduced to Mn²⁺ ions.

Oxidation: 5(COOH)₂ + 2 KMnO₄ + 3H₂SO₄ → 10 CO₂ + K₂SO₄ + 2 MnSO₄ + 8H₂O
Reduction: MnO₄^- + 8H⁺ + 5e^- → Mn²⁺ + 4H₂O
Overall reaction: 5(COOH)₂ + 2 MnO₄^- + 6H⁺ → 10 CO₂ + 2 Mn²⁺ + 8H₂O

Since one mole of oxalic acid releases 2 moles of electrons, the equivalent weight of oxalic acid is 126 / 2 = 63 (oxalic acid is dihydrated, so its molar mass is 126).

PROCEDURE

TITRATION I: (Standard FAS) Vs (Link KMnO₄)

Burette is washed with water, rinsed with KMnO₄ solution and filled with the same KMnO₄ solution up to the zero mark. Exactly 20 ml of standard FAS solution is pipetted out into a clean, washed conical flask. To this FAS solution, approximately 20 ml of 2 N H₂SO₄ is added. This mixture is titrated against the KMnO₄ link solution from the burette. KMnO₄ is added drop wise till the appearance of a permanent pale pink colour. The burette reading is noted and the same procedure is repeated to get concordant values.

TITRATION II: (Link KMnO₄) Vs (Unknown Oxalic Acid)

Burette is washed with water, rinsed with KMnO₄ solution and filled with the same KMnO₄ solution up to the zero mark. Exactly 20 ml of unknown oxalic acid solution is pipetted out into a clean, washed conical flask. To this oxalic acid solution, approximately 20 ml of 2 N H₂SO₄ is added. This mixture is heated to 60°C to 70°C using a Bunsen burner, and that hot solution is titrated against the KMnO₄ link solution from the burette. KMnO₄ is added drop wise till the appearance of a permanent pale pink colour. Burette readings are noted and the same procedure is repeated to get concordant values.

TITRATION II: LINK KMnO₄ Vs UNKNOWN OXALIC ACID

Indicator: Self (KMnO₄)

CALCULATIONS

Volume of KMnO₄ link solution (V₂) = 20.8 ml
Normality of KMnO₄ link solution (N₂) = 0.0980 N
Volume of unknown oxalic acid solution (V₃) = 20 ml
Normality of unknown oxalic acid solution (N₃) = ? N

According to the law of normality equation: V₂ N₂ = V₃ N₃

1. N₃ = \( \frac{V_2 N_2}{V_3} \)
2. N₃ = \( \frac{20.8 \times 0.0980}{20} \) = 0.1019 N
3. Normality of unknown oxalic acid solution (N₃) = 0.1019 N.

ESTIMATION: WEIGHT CALCULATION

The amount of oxalic acid dissolved in one litre of the solution = Normality × Equivalent weight

The amount of oxalic acid dissolved in 500 ml of the solution = \( \frac{\text{Normality} \times \text{Equivalent weight} \times 500}{1000} \)

1. Equivalent weight of oxalic acid = 63
2. Amount in 500 ml = 0.1019 × 63 / 2 = 3.21 g

The amount of oxalic acid dissolved in 500 ml of the given solution = 3.21 g.

RESULT

1. Normality of KMnO₄ link solution = 0.0980 N.
2. Normality of unknown oxalic acid solution = 0.1019 N.
3. The amount of oxalic acid dissolved in 500 ml of the given solution = 3.21 g.

SHORT PROCEDURE: STANDARDISATION OF HYDROCHLORIC ACID

TITRATION I: Std Na₂CO₃ Vs Link HCl
Indicator: Methyl orange

Normality of sodium carbonate (Na₂CO₃) solution = 0.0948 N

CALCULATIONS

Volume of std Na₂CO₃ solution (V₁) = 20 ml
Normality of std Na₂CO₃ solution (N₁) = 0.0948 N
Volume of HCl link solution (V₂) = 20.3 ml
Normality of HCl link solution (N₂) = ? N

According to the law of normality equation: V₁ N₁ = V₂ N₂

1. N₂ = \( \frac{V_1 N_1}{V_2} \)
2. N₂ = \( \frac{20 \times 0.0948}{20.3} \) = 0.0934 N
3. Normality of hydrochloric acid (HCl), N₂ = 0.0934 N.