Separate samples of a solution of an unknown soluble ionic compound are treated with \(\mathrm{KCl}, \mathrm{Na}_{2} \mathrm{SO}_{4},\) and \(\mathrm{NaOH}\). A precipitate forms only when \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is added. Which cations could be present in the unknown soluble ionic compound?

Begin by analyzing the reagent, \(\mathrm{Na}_{2}\mathrm{SO}_{4}\), and its solubility with different cations. Based on solubility rules, we know that some cations will form a precipitate with sulfate ions (\(\mathrm{SO}_{4}^{2-}\)). Some common insoluble sulfate salts are: 1. \(\mathrm{Ag}_{2}\mathrm{SO}_{4}\) 2. \(\mathrm{Ba}\mathrm{SO}_{4}\) 3. \(\mathrm{Ca}\mathrm{SO}_{4}\) 4. \(\mathrm{Pb}\mathrm{SO}_{4}\)

We are given that the unknown ionic compound forms precipitates when treated with \(\mathrm{Na}_{2}\mathrm{SO}_{4}\) but not with \(\mathrm{KCl}\) and \(\mathrm{NaOH}\). This information helps us eliminate some possibilities. For example, \(\mathrm{Ag}^{+}\) would react with both \(\mathrm{KCl}\) and \(\mathrm{Na}_{2}\mathrm{SO}_{4}\), forming \(\mathrm{AgCl}\) and \(\mathrm{Ag}_{2}\mathrm{SO}_{4}\) respectively. Since no precipitate is formed with \(\mathrm{KCl}\), we can eliminate \(\mathrm{Ag}^{+}\) from the possibilities. Similarly, \(\mathrm{Ca}^{2+}\) would react with both \(\mathrm{Na}_{2}\mathrm{SO}_{4}\) and \(\mathrm{NaOH}\), forming \(\mathrm{Ca}\mathrm{SO}_{4}\) and \(\mathrm{Ca(OH)}_{2}\) respectively. Since no precipitate is formed with \(\mathrm{NaOH}\), we can eliminate \(\mathrm{Ca}^{2+}\) from the possibilities.

Based on our analysis, we are left with two possible cations from our original list, which only form precipitates with \(\mathrm{Na}_{2}\mathrm{SO}_{4}\): 1. \(\mathrm{Ba}^{2+}\), forming \(\mathrm{Ba}\mathrm{SO}_{4}\) 2. \(\mathrm{Pb}^{2+}\), forming \(\mathrm{Pb}\mathrm{SO}_{4}\) Therefore, the cations that could be present in the unknown soluble ionic compound are \(\mathrm{Ba}^{2+}\) and \(\mathrm{Pb}^{2+}\).