Question

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 titration. If you had to perform three fine titrations, disregard the one that was different.

Answer 1

Answer on Question #82647 – Chemistry – General Chemistry

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 titration. If you had to perform three fine titrations, disregard the one that was different.

Solution:

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

In this titration you start off with a fixed volume of hydrogen peroxide (2 different samples - old hydrogen peroxide should be less reactive, require less KMnO₄) but unknown concentration and a known concentration of potassium permanganate. You add the permanganate until all of the H₂O₂ is neutralized. 2 moles of permanganate ion are neutralized by 5 moles of hydrogen peroxide. The reaction requires sulfuric acid H₂SO₄.

So we start by determining the number of moles of KMnO₄ (permanganate). You added 18.08 mL (= 0.01808 L) of a 0.2 M or 0.2 moles/L solution. The number of moles is given by the equation n = C × V.

For the first titration with the new H₂O₂ you had n = 0.01808 L × 0.2 mol/L = 3.6 × 10⁻³ mol.

Based on the ratio that 2 molecules of permanganate neutralise 5 molecules of H₂O₂, the number of moles of H₂O₂ (that were neutralized) is 5/2 × the number of moles of permanganate n (KMnO₄) = 2 mol; n (H₂O₂) = 5 mol

n (KMnO₄) = 3.6 × 10⁻³ mol, n (H₂O₂) = 9 × 10⁻³ mol

The molarity (which is the same as concentration) is therefore

If n = C × V, then C = n/V, the volume used was 10 mL or 0.01 L

c (H₂O₂) = 9 × 10⁻³ mol / 0.010 L = 0.9 M

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