Titration 2

Question 1

Aim

Answer 1

To prepare a standard solution of oxalic acid (M/40) and determine volumetrically the molarity and strength of the given KMnO₄ solution.

Question 2

Theory

Answer 2

Ionic equation:
2MnO₄⁻ + 16H⁺ + 5C2 O4^2⁻ → 2Mn²⁺ + 8H₂O +10CO2

Question 3

Procedure

Answer 3

Burette – Potassium permanganate
Conical flask – 10 mL of oxalic acid (M/40) + 10 mL of 2M H₂SO₄
Indicator – KMnO₄ is a self-indicator
Endpoint – Colorless to pale pink color

Question 4

To prepare a standard solution of oxalic acid (M/40):

Answer 4

M = (WB × 1000) / (MB × V)

1/40 = (WB × 1000) / (126 × 250)

WB = (0.025 × 153) / 2
WB = 1.575 / 2
WB = 0.7875 g

Therefore, 0.7875 g of oxalic acid is needed to make (M/40) oxalic acid solution in 250 ml.

Question 5

Observation table

Answer 5

Refer to notebook

Question 6

Calculations

Answer 6

From the equation, it is clear that 2 moles of KMnO₄ react with 5 moles of oxalic acid.

M KMnO₄ × V_KMnO₄ / (M_oxalic acid × V_oxalic acid) = 2 / 5

M_KMnO₄ × V_KMnO₄ / (1/40 × 10) = 2 / 5

M KMnO₄ = (2 × 10) / (40 × V_KMnO₄)

Strength = Molarity × Molar Mass of KMnO₄
Strength = M_KMnO₄ × 158

Question 7

Results

Answer 7

The concentration is ________ M
The strength is ________g/l