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?

Short Answer

Expert verified
The cations that could be present in the unknown soluble ionic compound are \(Ba^{2+}\) (barium), \(Pb^{2+}\) (lead), and \(Sr^{2+}\) (strontium), based on the reactions with \(Na_2SO_4\), \(KCl\), and \(NaOH\).

Step by step solution

01

Analyze the reaction with KCl

We're given that no precipitate forms when KCl is added, which means none of the cations in the unknown ionic compound form an insoluble compound with Cl-. Therefore, the cation present in the unknown compound cannot create an insoluble chloride.
02

Analyze the reaction with Na2SO4

A precipitate is formed when Na2SO4 is added. So, our cation must form an insoluble compound with SO4^2-. We can list some possible cations that form insoluble sulfates, such as: - Ba^2+ (barium) - Pb^2+ (lead) - Sr^2+ (strontium) - Ag^+ (silver)
03

Analyze the reaction with NaOH

There is no reaction when NaOH is added to the unknown compound. Therefore, the cation in the unknown compound will not form an insoluble compound with OH- ions. This fact eliminates Ag^+ from the possibilities, since silver forms an insoluble hydroxide.
04

Conclusion

Based on the given information, the cations that could be present in the unknown soluble ionic compound are Ba^2+ (barium), Pb^2+ (lead), and Sr^2+ (strontium).

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