You know that an unlabeled bottle contains a solution of one of the following: \(\mathrm{AgNO}_{3}, \mathrm{CaCl}_{2}\), or \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} .\) A friend suggests that you test a portion of the solution with \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) and then with \(\mathrm{NaCl}\) solutions. Explain how these two tests together would be sufficient to determine which salt is present in the solution.

Let's first analyze what will happen if the unknown solution reacts with \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\). If the unknown solution contains: 1. \(\mathrm{AgNO}_{3}\): \(\mathrm{AgNO}_{3} + \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2} \rightarrow \mathrm{Ba}\left[\mathrm{NO}_{3}\right]_{2} + \mathrm{AgNO}_{3}\) (No visible reaction) 2. \(\mathrm{CaCl}_{2}\): \(\mathrm{CaCl}_{2} + \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2} \rightarrow \mathrm{BaCl}_{2} + \left[\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\right]\) (No visible reaction) 3. \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\): \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} + 3\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2} \rightarrow 3\mathrm{Ba}\left[\mathrm{SO}_{4}\right] + 2\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\) (Production of a white precipitate of barium sulfate) If a white precipitate forms, we can conclude that the unknown solution contains \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\). Otherwise, it can be either \(\mathrm{AgNO}_{3}\) or \(\mathrm{CaCl}_{2}\).

Now let's analyze what will happen if the unknown solution reacts with \(\mathrm{NaCl}\). If the unknown solution contains: 1. \(\mathrm{AgNO}_{3}\): \(\mathrm{AgNO}_{3} + \mathrm{NaCl} \rightarrow \mathrm{AgCl} + \mathrm{NaNO}_{3}\) (Formation of a white precipitate of silver chloride) 2. \(\mathrm{CaCl}_{2}\): \(\mathrm{CaCl}_{2} + 2\mathrm{NaCl} \rightarrow 2\mathrm{NaCl} + \mathrm{CaCl}_{2}\) (No visible reaction) 3. \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) (already ruled out in the previous test as we assume that there was no precipitation): \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} + 6\mathrm{NaCl} \rightarrow 2\mathrm{AlCl}_{3} + 3\mathrm{Na}_{2}\mathrm{SO}_{4}\) (No visible reaction) If a white precipitate forms in this reaction, we can conclude that the unknown solution contains \(\mathrm{AgNO}_{3}\). If not, it must contain \(\mathrm{CaCl}_{2}\)

We have now established the reactions and outcomes for each test. To determine the salt present in the unknown solution, we can use the results of the two tests as follows: 1. If a white precipitate is formed in the first test (with \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\)), the unknown solution contains \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\). 2. If no precipitate is formed in the first test but a white precipitate is formed in the second test (with \(\mathrm{NaCl}\)), the unknown solution contains \(\mathrm{AgNO}_{3}\). 3. If no precipitate is formed in either test, the unknown solution contains \(\mathrm{CaCl}_{2}\). These two tests together are sufficient to identify the salt present in the solution.