[VISUAL] A particle view of a sample of H2O2(aq) is shown above. The H2O2(aq) is titrated with KMnO4 — Solutions Chemistry Question

STEPS:

1. Count the initial reactant particles: Count the number of $\text{H}_2\text{O}_2$ molecules represented in the initial reactant container. There are exactly 10 $\text{H}_2\text{O}_2$ molecules initially present.
2. Understand the definition of the equivalence point: The equivalence point is reached when the moles of added titrant ($\text{MnO}_4^-$) are stoichiometrically sufficient to react completely with all of the analyte ($\text{H}_2\text{O}_2$) originally present.
3. Calculate the titrant required for the equivalence point: Using the stoichiometric coefficients from the balanced equation:
$$2\ \text{MnO}_4^-(aq) + 5\ \text{H}_2\text{O}_2(aq) + 6\ \text{H}^+(aq) \rightarrow 2\ \text{Mn}^{2+}(aq) + 5\ \text{O}_2(g) + 8\ \text{H}_2\text{O}(l)$$
The ratio is $2\ \text{MnO}_4^- : 5\ \text{H}_2\text{O}_2$. For all 10 $\text{H}_2\text{O}_2$ molecules to react completely at the equivalence point:
$$10\ \text{H}_2\text{O}_2 \times \frac{2\ \text{MnO}_4^-}{5\ \text{H}_2\text{O}_2} = \mathbf{4\ \text{MnO}_4^- \text{ ions required}}$$
4. Determine the species present halfway to the equivalence point: Halfway to the equivalence point means that exactly half of the required titrant ($\text{MnO}_4^-$) has been added:
* $\text{MnO}_4^-$ added: $\frac{1}{2} \times 4 = \mathbf{2\ \text{MnO}_4^- \text{ ions}}$
* $\text{H}_2\text{O}_2$ consumed: These $2\ \text{MnO}_4^-$ ions react completely with:
$$2\ \text{MnO}_4^- \times \frac{5\ \text{H}_2\text{O}_2}{2\ \text{MnO}_4^-} = \mathbf{5\ \text{H}_2\text{O}_2 \text{ molecules consumed}}$$
* Remaining $\text{H}_2\text{O}_2$: Subtract the consumed molecules from the starting amount:
$$10 - 5 = \mathbf{5\ \text{H}_2\text{O}_2 \text{ molecules remaining}}$$
* $\text{Mn}^{2+}$ produced: According to the $1:1$ ratio of $\text{MnO}_4^-$ to $\text{Mn}^{2+}$ in the balanced equation, adding $2\ \text{MnO}_4^-$ ions produces:
$$2\ \text{MnO}_4^- \times \frac{2\ \text{Mn}^{2+}}{2\ \text{MnO}_4^-} = \mathbf{2\ \text{Mn}^{2+} \text{ ions formed}}$$
* Remaining $\text{MnO}_4^-$: Because we are prior to the equivalence point, the analyte ($\text{H}_2\text{O}_2$) is in excess. This means all added titrant ($\text{MnO}_4^-$) reacts completely and none is left over (0 $\text{MnO}_4^-$ ions).
5. Select the correct particle view: The correct container must depict 5 $\text{H}_2\text{O}_2$ molecules, 2 $\text{Mn}^{2+}$ ions, and 0 $\text{MnO}_4^-$ ions. This matches Option C.

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WHY_OTHERS_WRONG:

• Option A is incorrect: This diagram shows 4 $\text{Mn}^{2+}$ ions and 4 $\text{H}_2\text{O}_2$ molecules. For 4 $\text{Mn}^{2+}$ ions to be produced, 4 $\text{MnO}_4^-$ ions would have to be added, which would consume all 10 $\text{H}_2\text{O}_2$ molecules, leaving 0 $\text{H}_2\text{O}_2$ behind. It does not represent any stoichiometrically valid state for this reaction.
• Option B is incorrect: This diagram shows unreacted $\text{MnO}_4^-$ ions coexisting in the same container as excess $\text{H}_2\text{O}_2$ molecules. Because permanganate is a strong oxidizing agent that reacts rapidly and completely with hydrogen peroxide in acidic conditions, they cannot coexist in solution.
• Option D is incorrect: This diagram shows 1 unreacted $\text{MnO}_4^-$ ion present alongside 4 unreacted $\text{H}_2\text{O}_2$ molecules. Because $\text{H}_2\text{O}_2$ is in excess at this stage of the titration, any added $\text{MnO}_4^-$ must react completely to form $\text{Mn}^{2+}$, leaving no unreacted titrant.