Final answer:
The unknown salt in the solution is NaB, containing the bromide ion, as indicated by the pale yellow precipitate upon reaction with silver nitrate and the purple solution formed after adding chlorine water and carbon tetrachloride.
Step-by-step explanation:
The question is about determining the nature of the anion present in a solution that was initially colorless and then reacted with silver nitrate, forming a pale yellow precipitate, and finally turned purple upon adding chlorine water and carbon tetrachloride. The key to the solution lies in the chemical reactions involved. When a few drops of silver nitrate solution are added to a solution of a halide salt, a precipitate forms if the halide is chloride (white precipitate), bromide (pale yellow precipitate), or iodide (yellow precipitate). Since a pale yellow precipitate formed, this indicates the presence of bromide ions. Therefore, the original salt was NaB or sodium bromide. The balanced chemical equation for this reaction is AgNO3(aq) + NaB(aq) → AgB(s) + NaNO3(aq).
Adding chlorine water and carbon tetrachloride results in a purple solution, which is indicative of the formation of bromine (Br2) in the organic layer, as bromide ions (B⁻) are oxidized by chlorine (Cl2). This further confirms the presence of bromide ions in the original solution. The balanced equation for this reaction is 2Cl2(aq) + 2NaB(aq) → Br2(l) + 4NaCl(aq). The reaction demonstrates the use of chlorine as an oxidizing agent that oxidizes bromide to bromine, which is soluble in carbon tetrachloride, producing the purple color characteristic of molecular bromine.