The active agent in many hair bleaches is hydrogen peroxide. The amount of H2O2 in 14.8 g of hair bleach was determined by titration with a standard potassium permanganate solution: 2MnO4-aq+5H2O2aq+6H+aq→5O2g+2Mn2+aq+8H2Ol (a) How many moles of MnO4- were required for the titration if 43.2mL of 0.105M KMnO4 was needed to reach the end point? (b) How many moles of H2O2 were present in the 14.8-g sample of bleach? (c) How many grams of H2O2 were in the samp...

Q4.80P - Chemistry: Molecular Nature Of Matter And Change

Q4.80P

Question

The active agent in many hair bleaches is hydrogen peroxide. The amount of H₂O₂ in 14.8 g of hair bleach was determined by titration with a standard potassium permanganate solution:

$2 M n O_{4}^{-} (a q) + 5 H_{2} O_{2} (a q) + 6 H^{+} (a q) \to 5 O_{2} (g) + 2 M n^{2 +} (a q) + 8 H_{2} O (l)$

(a) How many moles of MnO₄^- were required for the titration if 43.2mL of 0.105M KMnO₄ was needed to reach the end point?

(b) How many moles of H₂O₂ were present in the 14.8-g sample of bleach?

(c) How many grams of H₂O₂ were in the sample?

(d) What is the mass percentage of H₂O₂ in the sample?

(e) What is the reducing agent in the redox reaction?

Step-by-Step Solution

Verified

Answer

You need to find how many moles of MnO₄^- were required for the titration if 43.2mL of 0.105M KMnO₄ was needed to reach the end point.

1Step 1: Calculation of the required moles of MnO 4 -

As you know,

$n_{M n O_{4}^{-}} = V_{M n O_{4}^{-}} \times M_{M n O_{4}^{-}}$

Where,

$n_{M n O_{4}^{-}}$ = moles of MnO₄^-

$V_{M n O_{4}^{-}}$ = volume of MnO₄^-

$M_{M n O_{4}^{-}}$ = Molarity of MnO₄^-

According to the question;

$V_{M n O_{4}^{-}}$ = $V_{k M n O_{4}^{-}}$ = 43.2mL = 0.0432L

$M_{M n O_{4}^{-}}$ = $M_{K M n O_{4}^{-}}$ = 0.105 mol/L

Now, you can write

$n_{M n O_{4}^{-}} = V_{K M n O_{4}^{-}} \times M_{K M n O_{4}^{-}} n_{M n O_{4}^{-}} = 0.0432 \times 0.105 (m o l) n_{M n O_{4}^{-}} = 0.004536 (m o l)$

2Step 2: Conclusion

0.004536 mol of MnO₄^- were required for the titration if 43.2mL of 0.105M KMnO₄ was needed to reach the end point.

Answer of subpart (b)

Answer: You need to find how many moles of H₂O₂ were present in the 14.8-g sample of bleach

3Step 3: Calculation of required moles of H 2 O 2

According to the balanced equation

2 M n O_{4}^{-} (a q) + 5 H_{2} O_{2} (a q) + 6 H^{+} (a q) \to 5 O_{2} (g) + 2 M n^{2 +} (a q) + 8 H_{2} O (l)

2 moles of MnO₄^- reacts with 5 moles of H₂O₂

Hence, moles of H₂O₂ ( $n_{H_{2} O_{2}^{}}$ ) can be calculated as

$n_{H_{2} O_{2}^{}} = n_{M n O_{4}^{-}} \times \frac{5}{2} n_{H_{2} O_{2}^{}} = 0.004536 \times \frac{5}{2} (m o l) n_{H_{2} O_{2}^{}} = 0.01134 (m o l)$