Question

For your titrations of the hydrogen peroxide in a new bottle, record the following data in the table below. If you had to perform three fine titrations, disregard the one that was different.

initial volume of .2M KMnO4 in the burette (mL)	50.00
volume of hydrogen peroxide solution added to the flask (mL)	10.00
volume of .2M KMnO4 dispensed in the coarse titration (mL)	19.06
volume of .2M KMnO4 dispensed in the first fine titration (mL)	18.02
volume of .2M KMnO4 dispensed in the second fine titration (mL)	18.01

Data Analysis

For your titrations of the hydrogen peroxide in a new bottle, calculate the molarity of the new hydrogen peroxide solution using the average volume of permanganate solution dispensed in the fine titrations. If you had to perform three fine titrations, disregard the one that was different.

Calculate the mass percent of the hydrogen peroxide in the new bottle. The molar mass of H₂O₂ is 34.0147 g/mol and the density of the solution is 1.000 g/mL.

Experiment 2: Titrate the Hydrogen Peroxide in an Old Bottle

Lab Results

For your titrations of the hydrogen peroxide in a new bottle, record the following data in the table below. If you had to perform three fine titrations, disregard the one that was different.

initial volume of .2M KMnO4 in the burette (mL)	50.00
volume of hydrogen peroxide solution added to the flask (mL)	10.00
volume of .2M KMnO4 dispensed in the coarse titration (mL)	13.37
volume of .2M KMnO4 dispensed in the first fine titration (mL)	12.63
volume of .2M KMnO4 dispensed in the second fine titration (mL)	12.65

Data Analysis

For your titrations of the hydrogen peroxide in an old bottle, calculate the molarity of the old hydrogen peroxide solution using the average volume of permanganate solution dispensed in the fine titrations. If you had to perform three fine titrations, disregard the one that was different.

Calculate the mass percent of the hydrogen peroxide in the old bottle. The molar mass of H₂O₂ is 34.0147 g/mol and the density of the solution is 1.000 g/mL.

Answer 1

Average volume of KMnO₄ = (18.01+18.02)/2

= 18.015 ml

Molarity of KMnO₄ = 0.2 M

Milli-Moles of KMnO₄ = 0.2 * 18.015

Consider Molarity of H₂O₂ = x M

Following is the balanced reaction

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

Hence ( Moles of KMnO₄ )/2 = (Moles of H₂O₂ )/5

5* moles of KMnO4 = 2* Moles of H₂O₂

  5 * 0.2 * 18.015 = 2 * x *10

x= 0.9 M

Hence Molarity of H₂O₂ is 0.9M

Now mass%

1000 ml of H₂O₂ solution contain 0.9 moles of H₂O₂ . Hence wt of H₂O₂ =0.9 * 34 = 30.6 gm

wt of H₂O₂ sloution = 1000 gm

  

mass % of H₂O₂ = 3%