Final answer:
A precipitate is expected to form with carbonate, sulfite, phosphate, and chromate ions when silver nitrate solution is added. Silver nitrate does not usually form a precipitate with sulfate and nitrate ions. The salt that was dissolved in the original solution was likely iodide, as indicated by the yellow precipitate and the purple solution.
Step-by-step explanation:
When silver nitrate solution is added to a solution of various salts, precipitates can form with ions that have low solubility with silver. Among the ions listed, precipitates would be expected to form with carbonate, sulfite, phosphate, and chromate. Silver nitrate does not typically form a precipitate with nitrate ions since both are components of the soluble compound silver nitrate. Nor does it usually form a precipitate with sulfate ions due to the relative solubility of silver sulfate in water. However, exceptions can occur under certain conditions, such as high concentrations or presence of other interacting substances, which might lead to the formation of precipitates with sulfate ions as well.
On the basis of the given clues, the original solution most likely contained an iodide ion. After the addition of silver nitrate, a pale yellow precipitate forms, which is indicative of silver iodide. The addition of chlorine water and carbon tetrachloride resulting in a purple solution further supports that an iodide was the source of the precipitate (the iodine liberated by the reaction with chlorine dissolves in carbon tetrachloride, forming a purple solution). The balanced equation for the reaction between iodide and silver nitrate would be:
AgNO3(aq) + KI(aq) → AgI(s) + KNO3(aq)