Commercial aqueous hydrogen peroxide was diluted 150 times. 25cm3 of the sample solution was titrated with 0.02molar KMnO4 and 26.3cm3 was required to reach the end point. Calculate the molar concentration of hydrogen peroxide
2MnO4^- + 5H2O2 + 6H^+ ==>2Mn^2+ + 5O2 + 8H2O
Check that equation.
mols MnO4^- = M x L = approx 0.0005 but you must redo this to obtain a better answer.
Using the coefficients in the balanced equation, convert mols MnO4^- to mols H2O2. That will be approx 0.0005 x (5 mols H2O2/2 mols MnO4^-) = ??
Then M H2O2 = mols H2O2/L H2O2 = ??mols H2O2/0.025L = ?
Then multiply the final number you have by 150 to find the concn of the original commercial H2O2.
Post your work if you get stuck.
To calculate the molar concentration of hydrogen peroxide, we can use the equation:
M1V1 = M2V2
Where:
M1 = initial molar concentration of hydrogen peroxide
V1 = initial volume of hydrogen peroxide
M2 = final molar concentration of KMnO4 (potassium permanganate)
V2 = final volume of KMnO4 required to reach the end point
In this case, we are given:
V1 = 25 cm^3
V2 = 26.3 cm^3
M2 = 0.02 M
First, let's calculate the initial volume of hydrogen peroxide (V1) that was present before its dilution.
Since the commercial hydrogen peroxide was diluted 150 times, it means its concentration was reduced to 1/150 times. Therefore, we can calculate V1 as follows:
V1 = V2 * (1 + 150)
V1 = 26.3 cm^3 * (1 + 150)
V1 = 26.3 cm^3 * 151
V1 = 3961.3 cm^3
Now, let's calculate the molar concentration of hydrogen peroxide (M1) using the equation:
M1 = (M2 * V2) / V1
M1 = (0.02 M * 26.3 cm^3) / 3961.3 cm^3
M1 = 0.000132 M
Therefore, the molar concentration of hydrogen peroxide is approximately 0.000132 M.