A solution is believed to contain one or more of the following ions: \(\mathrm{Cr}^{3+}, \mathrm{Zn}^{2+}, \mathrm{Fe}^{3+}, \mathrm{Ni}^{2+} .\) When the solution is treated with excess \(\mathrm{NaOH}(\mathrm{aq}),\) a precipitate forms. The solution in contact with the precipitate is colorless. The precipitate is dissolved in \(\mathrm{HCl}(\mathrm{aq}),\) and the resulting solution is treated with \(\mathrm{NH}_{3}(\text { aq })\). No precipitation occurs. Based solely on these observations, what conclusions can you draw about the ions present in the original solution? That is, which ion(s) are likely present, which are most likely not present, and about which can we not be certain? [Hint: Refer to Appendix D for solubility product and complex-ion formation data.

Step by step solution

01

Examine the effect with NaOH

The reaction with NaOH results in precipitation. This suggests that the original solution likely contains ions that react with hydroxide ions to form insoluble compounds. According to the solubility rules, Zn^2+ and Fe^3+ could form precipitates with hydroxide ions. So the possible ions could be Zn^2+ and Fe^3+.

02

Check the colorlessness of the solution

The colorless solution after precipitation suggests that Cr^3+ is probably not present, because Cr^3+ forms a green precipitate with hydroxide. So, we could further conclude that Fe3+ is probably present as it could result colorless solution.

03

Check the reaction with NH3

Consider that Zn^2+ forms soluble complexes with ammonia. Since no precipitate forms when the solution is treated with NH3, it suggests that Zn^2+ can't be present. This further confirms the likelihood of Fe^3+ being in the original solution.

04

Summarize conclusions

Considering these three steps, the Fe3+ is likely to be present, because it forms a precipitate with hydroxide, does not give the solution color, and does not form a precipitate with NH3. Ni^2+ and Zn^2+ are not likely to be present due to the clues from the reactions, and there is no clear indication about Cr^3+ with given information.

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