A 20.0 mL solution containing 0.2 g impure H2O2 reacts completely with 0.316 g of KMnO4 in acid solu

Q: Why do we use equivalents instead of moles?

Using equivalents allows us to bypass the tedious process of writing and balancing the full complex redox equation. By definition, substances always react in a 1:1 ratio of their equivalents.

Q: How did we get the n-factor of 5 for $KMnO_4$?

In an acidic medium, the Manganese (Mn) in $KMnO_4$ has an oxidation state of +7. It gets reduced to $Mn^{2+}$. The change in oxidation state is $|7 - 2| = 5$, which represents the 5 electrons gained.

Q: How did we get the n-factor of 2 for $H_2O_2$?

In $H_2O_2$, oxygen is in a -1 oxidation state (peroxide). It gets oxidized to $O_2$ gas, where oxygen is in a 0 oxidation state. Since there are two oxygen atoms per molecule, the total change is $2 \times |(-1) - 0| = 2$.

Q: What does 'impure sample' mean in this context?

It means the 0.2g sample contains $H_2O_2$ mixed with some unreactive substances (like water or stabilizers). The $KMnO_4$ only reacts with the pure $H_2O_2$, allowing us to calculate exactly how much pure $H_2O_2$ was in that 0.2g.

Q: Why is the volume of 20.0 mL not used in the calculation?

The volume is extraneous information in this specific setup. Since we are given the absolute mass of the impure sample and the absolute mass of the titrant, we can directly equate their equivalents. We would only need the volume if we were asked to find the molarity or normality of the solution.

Question

A $20.0	ext{ mL}$ solution containing $0.2	ext{ g}$ impure $	ext{H}_2	ext{O}_2$ reacts completely with $0.316	ext{ g}$ of $	ext{KMnO}_4$ in acid solution. The purity of $	ext{H}_2	ext{O}_2$ (in $\%$) is ............. (molecular weight of $	ext{H}_2	ext{O}_2 = 34$; molecular weight of $	ext{KMnO}_4 = 158$).

Sigma Percentile · Accepted Answer

Welcome to a classic problem in stoichiometry and redox titrations! This question tests your ability to navigate the Law of Chemical Equivalence without getting bogged down by balancing complex chemical equations. Let's dive into the fascinating world of electron exchange.

### Analyzing the Setup

We are given a $20.0	ext{ mL}$ solution containing $0.2	ext{ g}$ of an **impure** sample of hydrogen peroxide ($H_2O_2$). This sample reacts completely with $0.316	ext{ g}$ of potassium permanganate ($KMnO_4$) in an acidic medium. Our ultimate goal is to find the percentage purity of the $H_2O_2$ sample.

What does "impure" mean here? It means that out of the $0.2	ext{ g}$ of powder or liquid we weighed out, only a fraction of it is actual, reactive $H_2O_2$. The rest is just unreactive filler or water. The $KMnO_4$ will only react with the pure $H_2O_2$.

### The Magic of n-factors

To solve this efficiently, we use the **Law of Chemical Equivalence**, which states that substances react in equal numbers of equivalents. To find equivalents, we need the $n$-factor (valency factor) for both reactants.

In an acidic medium, $KMnO_4$ is a powerful oxidizing agent. The manganese atom goes from an oxidation state of $+7$ to $+2$:

$$Mn^{+7} + 5e^- ightarrow Mn^{+2}$$

Because it gains 5 electrons, the **$n$-factor of $KMnO_4$ is 5**.

On the other side, $H_2O_2$ acts as a reducing agent. The oxygen atoms in the peroxide linkage have an oxidation state of $-1$ and are oxidized to $0$ in oxygen gas ($O_2$):

$$O_2^{-1} ightarrow O_2^{0} + 2e^-$$

Since two electrons are lost per molecule of $H_2O_2$, the **$n$-factor of $H_2O_2$ is 2**.

### The Master Equation

Now, we equate the equivalents of both substances:

$$	ext{Equivalents of } H_2O_2 = 	ext{Equivalents of } KMnO_4$$

We know that $	ext{Equivalents} = \frac{	ext{Given Mass}}{	ext{Equivalent Weight}}$, and $	ext{Equivalent Weight} = \frac{	ext{Molar Mass}}{n	ext{-factor}}$. Let's plug in our variables:

$$\frac{W_{H_2O_2}}{34 / 2} = \frac{0.316}{158 / 5}$$

Here, $W_{H_2O_2}$ is the mass of the **pure** hydrogen peroxide that actually participated in the reaction.

### Final Calculation

Let's solve for $W_{H_2O_2}$:

$$W_{H_2O_2} = \frac{0.316}{158} 	imes 5 	imes \frac{34}{2}$$

Notice how beautifully the numbers are designed to cancel out. $\frac{0.316}{158}$ is exactly $0.002$.

$$W_{H_2O_2} = 0.002 	imes 5 	imes 17$$

$$W_{H_2O_2} = 0.01 	imes 17 = 0.17	ext{ g}$$

So, out of the $0.2	ext{ g}$ impure sample, exactly $0.17	ext{ g}$ was pure $H_2O_2$.

Finally, we calculate the percentage purity:

$$\%	ext{ Purity} = \frac{	ext{Mass of pure } H_2O_2}{	ext{Mass of impure sample}} 	imes 100$$

$$\%	ext{ Purity} = \frac{0.17}{0.2} 	imes 100 = 85\%$$

The purity of the hydrogen peroxide sample is **$85\%$**.

Enter Numerical Value:

Visualized Solution

\text{Analyzing the Given Data}

\text{Redox Reaction and n-factors}

\text{Law of Chemical Equivalence}

\text{Substituting the Values}

\text{Calculating Pure Mass of } \text{H}_2\text{O}_2

\text{Percentage Purity}

\text{Reflections and Variations}

The Sigma Insight: Stoichiometric and Volumetric Calculations

Analyzing the Setup

The Magic of n-factors

The Master Equation

Final Calculation

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