In order to determine the concentration of a commercial

Question

sample of hydrogen peroxide (H2O2
), 15 mL of it was
diluted to 300 mL. Then 20 mL of diluted solution was
acidified and titrated with a 0.021 M potassium
permanganate solution (KMnO4
). An average volume of
18.72 mL was required. Determine the concentration of
the original hydrogen peroxide
the answer is 0.982 M

Answer 1

To solve this problem, we can use the equation:

M1V1 = M2V2

Where:
M1 = concentration of the original hydrogen peroxide solution
V1 = volume of the original hydrogen peroxide solution (15 mL)
M2 = concentration of the potassium permanganate solution (0.021 M)
V2 = volume of the potassium permanganate solution used in the titration (18.72 mL)

First, we need to calculate the moles of KMnO4 used in the titration:
M2 = 0.021 M
V2 = 18.72 mL = 0.01872 L
Moles of KMnO4 = M2 x V2
Moles of KMnO4 = 0.021 M x 0.01872 L = 0.00039312 moles

Since the balanced chemical equation for the reaction between hydrogen peroxide and potassium permanganate is:

5 H2O2 + 2 KMnO4 + 3 H2SO4 → 5 O2 + 2 MnSO4 + K2SO4 + 8 H2O

We know that 2 moles of KMnO4 react with 5 moles of H2O2. Therefore, the moles of H2O2 in the original solution can be calculated as:

Moles of H2O2 = (5/2) x Moles of KMnO4
Moles of H2O2 = (5/2) x 0.00039312 = 0.0009828 mol

Now we can calculate the concentration of the original hydrogen peroxide solution:

M1 = Moles of H2O2 / V1
M1 = 0.0009828 mol / 0.015 L = 0.982 M

Therefore, the concentration of the original hydrogen peroxide solution is 0.982 M.