Question

A certain amount of hydrogen peroxide was dissolved in 100 mL of water and then titrated with 1.68 M KMnO[sub]4[/sub]. How much H[sub]2[/sub]O[sub]2[/sub] was dissolved if the titration required 18.3 mL of the KMnO[sub]4[/sub] solution?


2KMnO[sub]4[/sub](aq) + H[sub]2[/sub]O[sub]2[/sub](aq) + 3H[sub]2[/sub]SO[sub]4[/sub](aq) → 3O[sub]2[/sub](g) + 2MnSO[sub]4[/sub](aq) + K[sub]2[/sub]SO[sub]4[/sub](aq) + 4H[sub]2[/sub]O(l)

Answer 1

Answer on Question #36760 - Chemistry - Inorganic Chemistry

Question:

A certain amount of hydrogen peroxide was dissolved in $100~\mathrm{mL}$ of water and then titrated with $1.68\mathrm{M}$ $\mathrm{KMnO_4}$ . How much $\mathrm{H}_2\mathrm{O}_2$ was dissolved if the titration required $18.3~\mathrm{mL}$ of the $\mathrm{KMnO_4}$ solution?

$$2 \mathrm {K M n O} _ {4} (\mathrm {a q}) + \mathrm {H} _ {2} \mathrm {O} _ {2} (\mathrm {a q}) + 3 \mathrm {H} _ {2} \mathrm {S O} _ {4} (\mathrm {a q}) \rightarrow 3 \mathrm {O} _ {2} (\mathrm {g}) + 2 \mathrm {M n S O} _ {4} (\mathrm {a q}) + \mathrm {K} _ {2} \mathrm {S O} _ {4} (\mathrm {a q}) + 4 \mathrm {H} _ {2} \mathrm {O} (\mathrm {l})$$ $$V \left(H _ {2} O\right) = 1 0 0 m l;$$ $$C (K M n O _ {4}) = 1. 6 8 M;$$ $$V (K M n O _ {4}) = 1 8. 3 m l.$$ $$\mathrm {m} \left(\mathrm {H} _ {2} \mathrm {O} _ {2}\right) -?$$

Solution:

In the direct titration the equivalent number of moles of the substance (hydrogen peroxide) equals to the equivalent number of moles of the titrant (potassium permanganate). In permanganometric titration such a reaction takes place

$$5 \mathrm {H} _ {2} \mathrm {O} _ {2} + 2 \mathrm {M n O} _ {4} ^ {-} + 6 \mathrm {H} ^ {+} \rightarrow 5 \mathrm {O} _ {2} + 8 \mathrm {H} _ {2} \mathrm {O} + 2 \mathrm {M n} ^ {2 +}$$

The half-reactions corresponding to this reaction are:

$$\mathrm {M n O} _ {4} ^ {-} + 8 \mathrm {H} ^ {+} + 5 \mathrm {e} \rightarrow \mathrm {M n} ^ {2 +} + 4 \mathrm {H} _ {2}$$ $$\mathrm {H} _ {2} \mathrm {O} _ {2} - 2 \mathrm {e} \rightarrow \mathrm {O} _ {2} + 2 \mathrm {H} ^ {+}$$

$f_{x}(\mathrm{KMnO}_{4}) = 1 / 5$ , and $f_{x}(\mathrm{H}_{2}\mathrm{O}_{2}) = 1 / 2$ , since one electron is chemically equivalent to conventional particles $1 / 5\mathrm{KMnO}_{4}$ and $\frac{1}{2}\mathrm{H}_{2}\mathrm{O}_{2}$ .

Consequently: $n\left( {1/{5KMnO}_{4}}\right) = 5 \cdot c\left( {{KMnO}_{4}}\right) \cdot V\left( {{KMnO}_{4}}\right)$

$$n \left(^ {1} / _ {2} H _ {2} O _ {2}\right) = m \left(H _ {2} O _ {2}\right) / M \left(^ {1} / _ {2} H _ {2} O _ {2}\right)$$

As far as $n\left( {1/{5KMnO}_{4}}\right) = n\left( {1/{2H}_{2}{O}_{2}}\right)$ ,after the transformation we can find out that:

$$m \left(\mathrm {H} _ {2} \mathrm {O} _ {2}\right) = 5 \cdot c \left(\mathrm {K M n O} _ {4}\right) \cdot V \left(\mathrm {K M n O} _ {4}\right) \cdot \mathrm {M} \left(^ {1} / _ {2} \mathrm {H} _ {2} \mathrm {O} _ {2}\right) = 5 \cdot 1. 6 8 \cdot 0. 0 1 8 3 \cdot 1 8. 0 1 5 = 2. 6 1 4 \mathrm {g}$$

Answer: $m = 2.77$ g