This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
ScanSolve AI Answer
Another one ✨Fave — let's solve it. The question asks to identify which substance acts as a self-indicator during volumetric analysis. Step 1: Understand what a self-indicator is. 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. Step 2: Evaluate the given options. A: 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: KMnO_4 (Potassium permanganate) is a strong oxidizing agent. The permanganate ion (MnO_4^-) is intensely purple. In acidic solutions, it is reduced to the nearly colorless Mn^2+ ion. During a titration, as long as the reducing agent is present, the 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 KMnO_4 will impart a persistent pale pink/purple color to the solution, indicating the endpoint. Thus, KMnO_4 acts as a self-indicator. C: H_2SO_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: NH_4Cl (Ammonium chloride) is a salt. It is colorless in solution and does not change color during a titration. Step 3: Conclude the correct option. Potassium permanganate (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 MnO_4^- ion and the nearly colorless Mn^2+ ion. The final answer is B Got more? Send 'em.