Which of the following reagents could be used to separate the two cations from a solution containing Lead (IV) acetate and cesium permanganate?

In this practice question it says, which of the following reagents could be used to separate the 2 cations from a solution containing lead 4 acetate and cesium permanganate? Remember, cations are positive ions. In here, what we're trying to do is we're trying to separate lead 4 ion and cesium ion. Now anything connected to acetate is soluble based on our solubility rules, so let 4 be in an aqueous state. Cesium is a group 1 a element. Anything connected to the group 1 a elements are aqueous as well. When we say that we need to separate these 2 ions, what we're kind of indirectly saying is that we need one of them to become a precipitate, a solid. Once it does, we can physically remove that solid from our solution. Now we just said that cesium is a group 1a element, so it's always going to be aqueous. So there's nothing I can combine it with to make it a solid. But there are some things that I can do with lead 4 to make it a solid. I just need to find the right negative ion, the right anion to combine with it to make my precipitate. So if we take a look here at our choices, we have nitrate ion here as a negative ion. Here we have acetate as our negative ion. Here we have sulfide as our negative ion. Here we have perchlorate and here we have nitrate. So first of all, we know that nitrate is shown there twice. So that most likely means it's not an answer. We also know that anything connected to nitrate ion will always be soluble. So a and e would be out as options. The same thing can be said for b and d. Anything connected to acetate or perchlorate would also make something soluble, so these cannot be possible answers. The answer here has to be c because if we were to have lead 4 ion combining with sulfide ion, we could make a solid based on our solubility rules. Remember, when the numbers are different in the charges they crisscross, so 4 comes here and 2 comes here. So that'll give me p b 2 s 4, which I can reduce further, but to p b s 2. So here p b s 2 would be my solid. We'd be able to remove physically remove all the lead 4 ions from the solution by forming this precipitate of lead 4sulfide. So just remember, this requires previous knowledge on the solubility rules. If you don't quite remember them, make sure you go back and look at those videos where I talk about the different rules to make an ionic compound soluble or insoluble.

Which of the following reagents could be used to separate the two...

1. Matter and Measurements4h 29m

2. Atoms and the Periodic Table5h 23m

3. Ionic Compounds2h 18m

4. Molecular Compounds2h 18m

5. Classification & Balancing of Chemical Reactions3h 17m

6. Chemical Reactions & Quantities2h 35m

7. Energy, Rate and Equilibrium3h 46m

8. Gases, Liquids and Solids3h 25m

9. Solutions4h 10m

10. Acids and Bases3h 29m

11. Nuclear Chemistry56m

BONUS: Lab Techniques and Procedures1h 38m

BONUS: Mathematical Operations and Functions47m

12. Introduction to Organic Chemistry1h 34m

13. Alkenes, Alkynes, and Aromatic Compounds2h 12m

14. Compounds with Oxygen or Sulfur1h 6m

15. Aldehydes and Ketones1h 1m

16. Carboxylic Acids and Their Derivatives1h 11m

17. Amines38m

18. Amino Acids and Proteins1h 51m

19. Enzymes1h 37m

20. Carbohydrates1h 46m

21. The Generation of Biochemical Energy2h 8m

22. Carbohydrate Metabolism2h 22m

23. Lipids2h 26m

24. Lipid Metabolism1h 45m

25. Protein and Amino Acid Metabolism1h 37m

26. Nucleic Acids and Protein Synthesis2h 54m

Which of the following reagents could be used to separate the two cations from a solution containing Lead (IV) acetate and cesium permanganate?

Watch next

Complete Ionic Equations practice set