Question

2 MnO4 - (aq) + 5 H2O2(aq) + 6 H+ (aq) → 5 O2(g) + 2 Mn2+(aq) + 8 H2O(l)

Some potassium permanganate solution has been standardized at 0.01989 mol/L. A 25.00 mLsample of a very dilute hydrogen peroxide solution is pipetted with a volumetric pipette into a flask. 4 mL of 4.0 M H2SO4 is added. The potassium permanganate is put into a buret with a beginning volume of 1.36 mL. When the endpoint of the titration is reached (the last drop no longer becomes colorless, but instead turns the whole flask pale pink and it stays pink for 30 seconds), the level of potassium permanganate solution is now 27.47 mL.

a. What volume of potassium permanganate solution was used in the experiment?

b. How many moles of potassium permanganate were used in the experiment?

c. How many moles of hydrogen peroxide were dissolved in the 25.00 mL of solution?

d. What was the mass of that many moles of hydrogen peroxide?

e. If the density of the hydrogen peroxide solution is 1.00 g/ml, what was the mass of the original sample of the hydrogen peroxide?

f. What was the % mass of H2O2 in the dilute solution?

This question is verbatim question from a lab report. This problem is very confusing and any help would be very much appreciated.

Answer 1

(a.) initial buret volume = 1.36 mL

final buret volume = 27.47 mL

volume of KMnO₄ used = (final buret volume) - (initial buret volume)

volume of KMnO₄ used = (27.47 mL) - (1.36 mL)

volume of KMnO₄ used = 26.11 mL

volume of KMnO₄ used = 26.11 mL * (1.00 L / 1000 mL)

volume of KMnO₄ used = 0.02611 L

(b) moles of KMnO₄ used = (molarity KMnO₄) * (volume of KMnO₄ used in Liter)

moles of KMnO₄ used = (0.01989 mol/L) * (0.02611 L)

moles of KMnO₄ used = 5.193 x 10^-4 mol

(c) moles H₂O₂ = (5/2) * (moles of KMnO₄ used)

moles H₂O₂ = (2.5) * (5.193 x 10^-4 mol)

moles H₂O₂ = 1.298 x 10^-3 mol

(d) mass H₂O₂ = (moles H₂O₂) * (molar mass H₂O₂)

mass H₂O₂ = (1.298 x 10^-3 mol) * (34.01 g/mol)

mass H₂O₂ = 0.04416 g

(e) volume of H₂O₂ dilute solution = 25.00 mL

mass of H₂O₂ dilute solution = (volume of H₂O₂ dilute solution) * (density of H₂O₂ dilute solution)

mass of H₂O₂ dilute solution = (25.00 mL) * (1.00 g/mL)

mass of H₂O₂ dilute solution = 25.00 g

(f) % mass H₂O₂ = (mass H₂O₂ / mass of H₂O₂ dilute solution) * 100

% mass H₂O₂ = (0.04416 g / 25.00 g) * 100

% mass H₂O₂ = 0.1766 %