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

Given : Concentration of KMnO₄= 0.2 M = 0.2 moles/L Solution

Oxidizing Hydrogen Peroxide using Potassium Permanganate solution:

2MnO₄^- + 6H⁺ + 5H₂O₂ 8H₂O + 5O₂ + 2Mn²⁺

We can observe from the reaction that 2 moles of permanganate ion is required to neutralize 5 moles of Hudrogen peroxide.

To calculate the Molarity of the New Hydrogen Peroxide Solution:

Average Volume of 0.2M KMnO₄ = (18.02 + 18.01)/2 = 18.015 ml = 0.018015 L

Number of moles of KMnO₄= 0.018015 L * 0.2 mol/L = 3.603 * 10^-3 mol

Number of moles of H₂O₂ = (5/2) * 3.603 * 10^-3 = 9.0075 * 10^-3 mol ( From the balanced equation given above)

Given: Volume of Hydrogen Peroxide = 10 ml = 0.01 L

Molarity of new bottle = Concentration = moles of Solute/ litre of solution= number of moles of Hydrogen peroxide/ volume of hydrogen peroxide

= (9.0075 * 10^-3)/0.01 = 0.90075 M

Mass Percent of Hydrogen peroxide:

Given Molar mass : 34.0147 g/mol, density : 1.000g/ml

number of moles = mass of a given substance/ Molar mass

9.0075*10^-3 = mass of the substance/34.0147

mass of the substance = 0.306 g

mass of solution = 10/1.000 = 10 g

mass percent = (mass of solute/ total mass of solution)*100 = (0.306/10) *100 = 3.06%

Experiment 2:

To calculate the Molarity of the New Hydrogen Peroxide Solution:

Average Volume of 0.2M KMnO₄ = (12.63 + 12.65)/2 = 12.64 ml = 0.01264 L

Number of moles of KMnO₄= 0.01264 L * 0.2 mol/L = 2.528 * 10^-3 mol

Number of moles of H₂O₂ = (5/2) *2.528 * 10^-3 = 6.32 * 10^-3 mol ( From the balanced equation given above)

Given: Volume of Hydrogen Peroxide = 10 ml = 0.01 L

Molarity of new bottle = Concentration = number of moles of Hydrogen peroxide/ volume of hydrogen peroxide

= (6.32 * 10^-3)/0.01 = 0.632 M

Mass Percent of Hydrogen peroxide:

Given Molar mass : 34.0147 g/mol, density : 1.000g/ml

number of moles = mass of a given substance/ Molar mass

6.32*10^-3 = mass of the substance/34.0147

mass of the substance = 0.214 g

mass of solution = 10/1.000 = 10 g

mass percent = (mass of solute/ total mass of solution)*100 = (0.214/10) *100 = 2.14%