Separate samples of a solution of an unknown ionic compound are treated with dilute \(\mathrm{AgNO}_{3}, \mathrm{~Pb}\left(\mathrm{NO}_{3}\right)_{2},\) and \(\mathrm{BaCl}_{2}\) Precipitates form in all three cases. Which of the following could be the anion of the unknown salt: $\mathrm{Br}^{-}, \mathrm{CO}_{3}^{2-}, \mathrm{NO}_{3}^{-} ?$

We will write down the possible products that can be formed when the unknown salt (which contains one of the anions given - \(\mathrm{Br}^{-}, \mathrm{CO}_{3}^{2-}, \mathrm{NO}_{3}^{-}\)) encounters each of the other ions present in the added solutions. For AgNO3: - \(\mathrm{AgBr}\) - \(\mathrm{Ag}_{2}\mathrm{CO}_{3}\) - \(\mathrm{AgNO}_{3}\) For Pb(NO3)2: - \(\mathrm{PbBr}_{2}\) - \(\mathrm{PbCO}_{3}\) - \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) For BaCl2: - \(\mathrm{BaBr}_{2}\) - \(\mathrm{BaCO}_{3}\) - \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\)

Now, we will use solubility rules to determine which of these possible products form a precipitate. Common solubility rules include: - All nitrates are soluble. - All compounds with group 1 metals are soluble. - All chlorides, bromides, and iodides are soluble, except Ag, Pb, and Hg compounds. - Most carbonates are insoluble, except group 1 metals and ammonium compounds. Using these solubility rules: For AgNO3: - \(\mathrm{AgBr}\) forms a precipitate - \(\mathrm{Ag}_{2}\mathrm{CO}_{3}\) forms a precipitate - \(\mathrm{AgNO}_{3}\) is soluble For Pb(NO3)2: - \(\mathrm{PbBr}_{2}\) forms a precipitate - \(\mathrm{PbCO}_{3}\) forms a precipitate - \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) is soluble For BaCl2: - \(\mathrm{BaBr}_{2}\) forms a precipitate - \(\mathrm{BaCO}_{3}\) forms a precipitate - \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) is soluble

According to the given information, when the unknown salt is treated with each of the three solutions, a precipitate is formed in all cases. We can use this information to determine which anion is likely present in the unknown salt. Based on our results from Step 2, we see that when anions \(\mathrm{Br}^{-}\) and \(\mathrm{CO}_{3}^{2-}\) are present, precipitates form in all cases with each of the three solutions. However, when the \(\mathrm{NO}_{3}^{-}\) anion is present, it does not form a precipitate with any of the solutions. Therefore, it is not possible for the unknown salt to contain the \(\mathrm{NO}_{3}^{-}\) anion.

Based on the analysis, the anion present in the unknown salt must be either \(\mathrm{Br}^{-}\) or \(\mathrm{CO}_{3}^{2-}\).