Question

A \(5.0\text{ cm}^3\) solution of \(\text{H}_2\text{O}_2\) liberates \(1.27\text{ g}\) of iodine from an acidified \(\text{KI}\) solution. The percentage strength of \(\text{H}_2\text{O}_2\) is close to

• A. $11.2$
• B. $5.8$
• C. $1.9$
• D. $3.4$

Hint:

Remember the stoichiometric equivalence for hydrogen peroxide reactions:
- $1\text{ mol of }\text{H}_2\text{O}_2 \equiv 1\text{ mol of }\text{I}_2$
Calculating moles first avoids any confusion with normality or equivalent weights, which can sometimes lead to minor errors.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to find the percentage strength (weight/volume, $\% \text{ w/v}$) of a $5.0\text{ cm}^3$ (which is $5.0\text{ mL}$) $\text{H}_2\text{O}_2$ solution that reacts completely with acidified potassium iodide ($\text{KI}$) to liberate $1.27\text{ g}$ of iodine ($\text{I}_2$).


Step 2: Key Formula or Approach:
1. Write the balanced redox equation between $\text{H}_2\text{O}_2$ and $\text{I}^-$ in acidic medium.
2. Calculate the moles of $\text{I}_2$ liberated to find the moles of $\text{H}_2\text{O}_2$ reacted.
3. Determine the mass of $\text{H}_2\text{O}_2$ and express it as a percentage of the solution volume ($\% \text{ w/v}$).


Step 3: Detailed Explanation:
The balanced chemical equation for the oxidation of iodide by hydrogen peroxide in an acidic medium is:
\[ \text{H}_2\text{O}_2 + 2\text{I}^- + 2\text{H}^+ \rightarrow \text{I}_2 + 2\text{H}_2\text{O} \]
From the stoichiometry of the reaction:
\[ 1\text{ mole of }\text{H}_2\text{O}_2 \text{ produces } 1\text{ mole of }\text{I}_2 \]
Let us calculate the moles of iodine liberated:
- Molar mass of $\text{I}_2 = 2 \times 127\text{ g mol}^{-1} = 254\text{ g mol}^{-1}$.
- Moles of $\text{I}_2 = \frac{1.27\text{ g}}{254\text{ g mol}^{-1}} = 0.005\text{ mol}$.
Since the mole ratio of $\text{H}_2\text{O}_2$ to $\text{I}_2$ is $1:1$:
- Moles of $\text{H}_2\text{O}_2$ in $5.0\text{ mL}$ of solution = $0.005\text{ mol}$.
Now, let us calculate the mass of $\text{H}_2\text{O}_2$:
- Molar mass of $\text{H}_2\text{O}_2 = 34\text{ g mol}^{-1}$.
- Mass of $\text{H}_2\text{O}_2 = 0.005\text{ mol} \times 34\text{ g mol}^{-1} = 0.17\text{ g}$.
The percentage strength ($\% \text{ w/v}$) is:
\[ \% \text{ strength} = \frac{\text{mass of solute (g)}}{\text{volume of solution (mL)}} \times 100 \]
\[ \% \text{ strength} = \frac{0.17\text{ g}}{5.0\text{ mL}} \times 100 = 3.4\% \]


Step 4: Final Answer:
The correct option is (D).