Which of the following substances acts as a self-indicator in volumetric analysis: NaOH, KMnO4, or H2SO4?

A self-indicator is a reactant in a titration that changes color at or near the equivalence point, signaling the end of the reaction without the need for an additional indicator. This color change is usually due to a change in its oxidation state or chemical form.

• A: $\text{NaOH}$ (Sodium hydroxide) is a strong base. It is colorless in solution and does not change color during a titration. It requires an external indicator (e.g., phenolphthalein for acid-base titrations).
• B: $\text{KMnO}_4$ (Potassium permanganate) is a strong oxidizing agent. The permanganate ion ($\text{MnO}_4^{-}$) is intensely purple. In acidic solutions, it is reduced to the nearly colorless $\text{Mn}^{2+}$ ion. During a titration, as long as the reducing agent is present, the $\text{KMnO}_4$ added is consumed and the solution remains colorless (or the color of the analyte). Once all the reducing agent has reacted, the first drop of excess $\text{KMnO}_4$ will impart a persistent pale pink/purple color to the solution, indicating the endpoint. Thus, $\text{KMnO}_4$ acts as a self-indicator.
• C: $\text{H}_2\text{SO}_4$ (Sulfuric acid) is a strong acid. It is colorless in solution and does not change color during a titration. It requires an external indicator.
• D: $\text{NH}_4\text{Cl}$ (Ammonium chloride) is a salt. It is colorless in solution and does not change color during a titration.

Potassium permanganate ($\text{KMnO}_4$) is well-known for its ability to act as a self-indicator in redox titrations due to the distinct color change between the purple $\text{MnO}_4^{-}$ ion and the nearly colorless $\text{Mn}^{2+}$ ion.

The final answer is $\text{B}$