Question

\(20 \mathrm{ml}\) of \(\mathrm{H}_{2} \mathrm{O}_{2}\) after acidification with dil \(\mathrm{H}_{2} \mathrm{SO}_{4}\) required \(30 \mathrm{ml}\) of \(\frac{\mathrm{N}}{12} \mathrm{KMnO}_{4}\) for complete oxidation. The strength of \(\mathrm{H}_{2} \mathrm{O}_{2}\) solution is [Molar mass of \(\left.\mathrm{H}_{2} \mathrm{O}_{2}=34\right]\) (1) \(2 \mathrm{~g} / \mathrm{L}\) (2) \(4 \mathrm{~g} / \mathrm{L}\) (3) \(8 \mathrm{~g} / \mathrm{L}\) (4) \(6 \mathrm{~g} / \mathrm{L}\)

Short answer

The strength of \( \text{H}_2\text{O}_2 \) solution is \( 2 \text{ g/L} \).

Step-by-step solution

- Write the balanced reaction

The balanced chemical reaction for the oxidation of \( \text{H}_2\text{O}_2 \) by \( \text{KMnO}_4 \) in acidic medium is: \[ 2 \text{MnO}_4^- + 5 \text{H}_2\text{O}_2 + 6 \text{H}^+ \rightarrow 2 \text{Mn}^{2+} + 5 \text{O}_2 + 8 \text{H}_2\text{O} \]

- Identify normality of \( \text{KMnO}_4 \)

\( \text{KMnO}_4 \) solution is given as \( \frac{\text{N}}{12} \). So, the normality \( = \frac{1}{12} \text{ N} \).

- Determine the milliequivalents of \( \text{KMnO}_4 \)

Milliequivalents \( \text{KMnO}_4 = \text{Volume (ml)} \times \text{Normality} \) \[ = 30 \text{ ml} \times \frac{1}{12} \text{ N} = 2.5 \text{ meq} \]

- Calculate milliequivalents of \( \text{H}_2\text{O}_2 \)

Milliequivalents of \( \text{H}_2\text{O}_2 \) used will be the same as \( \text{KMnO}_4 \), therefore it is \( 2.5 \text{ meq} \).

- Find normality of \( \text{H}_2\text{O}_2 \)

\( \text{Normality of } \text{H}_2\text{O}_2 \) \[ = \frac{\text{Milliequivalents}}{\text{Volume (L)}} = \frac{2.5}{0.02 \text{ L}} = 0.125 \text{ N} \]

- Convert normality to molarity

For \( \text{H}_2\text{O}_2, \text{Normality} = 2 \times \text{Molarity} \), thus: \[ \text{Molarity} = \frac{0.125}{2} = 0.0625 \text{ M} \]

- Calculate the strength in g/L

\( \text{Strength (g/L)} = \text{Molarity} \times \text{Molar Mass of } \text{H}_2\text{O}_2 \) \[ = 0.0625 \text{ M} \times 34 \text{ g/mol} = 2.125 \text{ g/L} \] The closest answer is \( 2 \text{ g/L} \).

Key concepts

chemical reactions

Chemical reactions involve the transformation of one or more substances into new substances. A balanced chemical equation ensures that the number of atoms for each element is the same on both sides of the reaction. This complies with the law of conservation of mass. In our exercise, \( \text{H}_2\text{O}_2 \) (hydrogen peroxide) reacts with \( \text{KMnO}_4 \) (potassium permanganate) in the presence of an acid (sulfuric acid, \( \text{H}_2\text{SO}_4 \)). The balanced equation is:
\[ 2 \text{MnO}_4^- + 5 \text{H}_2\text{O}_2 + 6 \text{H}^+ \rightarrow 2 \text{Mn}^{2+} + 5 \text{O}_2 + 8 \text{H}_2\text{O} \] This equation tells us that every 2 moles of permanganate ions react with 5 moles of hydrogen peroxide.
Balancing equations is crucial because it allows us to predict the quantities of reactants needed and products formed in a chemical reaction.

oxidation

Oxidation is a chemical process where a substance loses electrons. This is often accompanied by an increase in oxidation state. In the given exercise, \( \text{H}_2\text{O}_2 \) is oxidized to \( \text{O}_2 \), while \( \text{KMnO}_4 \) is reduced from \( \text{MnO}_4^- \) to \( \text{Mn}^{2+} \). Here's a breakdown:
- \( \text{H}_2\text{O}_2 \) loses electrons (oxidation):
\[ \text{H}_2\text{O}_2 \rightarrow \text{O}_2 + 2\text{H}^+ + 2e^- \]
- \( \text{KMnO}_4 \) gains electrons (reduction):
\[ \text{MnO}_4^- + 8 \text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} \]
Understanding oxidation and reduction in terms of electron transfer helps in balancing redox reactions and calculating correct stoichiometric coefficients.

molarity calculations

Molarity (M) is a measure of concentration defined as moles of solute per liter of solution. To find the molarity, we need to convert the given normality (N) of \( \text{KMnO}_4 \) in our problem:
Normality (N) is given by:
\[ N \times \text{Equivalent weight} = M \times \text{Molecular weight} \]
Given that the normality is \( \frac{1}{12} \text{N} \) and using the equivalent weight of \( \text{KMnO}_4 \), we start by calculating the milliequivalents (meq) used. The volume used is 30 ml:
\[ 30 \text{ ml} \times \frac{1}{12} \text{ N} = 2.5 \text{ meq} \] This is the same meq as \( \text{H}_2\text{O}_2 \), leading to further calculations to obtain the normality of \( \text{H}_2\text{O}_2 \):
\[ \frac{2.5}{0.02 \text{ L}} = 0.125 \text{ N} \]
Finally, we can convert normality to molarity:
\[ \text{Molarity} = \frac{0.125}{2} = 0.0625 \text{ M} \]
This demonstrates the steps involved in converting between concentration units.

titration analysis

Titration is a quantitative chemical analysis method used to determine the concentration of an identified analyte. During titration, a solution of known concentration (titrant) is gradually added to a known volume of the analyte until the reaction reaches completion. In our case:
- \( \text{KMnO}_4 \) is the titrant with known normality \( \frac{N}{12} \)
- \( \text{H}_2\text{O}_2 \) is the analyte in the acidic medium
The end point is determined when the \( \text{H}_2\text{O}_2 \) is completely oxidized by \( \text{KMnO}_4 \). The volume measurements and concentrations help us calculate the strength in g/L. From the titration data, we found:
\[ \text{Strength (g/L)} = \text{Molarity} \times \text{Molar Mass of } \text{H}_2\text{O}_2 \]
\[ 0.0625 \text{ M} \times 34 \text{ g/mol} = 2.125 \text{ g/L} \]
Titration analysis provides precise and accurate results, essential for quantitative chemistry.