Suggest how the cations in each of the following solution mixtures can be separated: (c) Pb2 + and Al3 +.

Video transcript

Hey everyone today, we're being asked to identify which of the following four groups of re agents can be used to separate zinc two plus and ammonium in solution. So one very quick thing that we need to realize that's not explicitly stated in the question. But something that is safe to assume is that both zinc two plus and ammonium are in an aqueous solution. They're dissolved in water. That's important because when considering our re agents, we need to know are solid ability rules which are most commonly or can only be really known for water. Their standard for various solutions. So right off the bat from our Celje bility rules, we know that Um NH four plus ammonium along with sodium plus or potassium plus can form soluble salts and they will always be soluble. So that means no matter what it's paired with, ammonium will always end up being soluble, it'll always be soluble. So, this means in order to separate ZN two plus and NH four plus, we need to find out which of the re agents can aid in making ZN two plus into something that is unsolvable or insoluble, sorry, not insoluble. So, taking a look at our answer choices, we can actually rule out a right off the bat. This is because we are dealing with hydrochloric acid or hcl. Now in a reaction between zinc two plus and hydrochloric acid as with other halogen, like bromine or iodine will end up forming soluble salts again, zinc two plus. Or if zinc was to react with hydrochloric acid, we would end up forming the compound zinc chloride. zN cl two, which is a soluble chloride. So it is a soluble salt, which means that we cannot use six molar hydrochloric acid to separate zinc chloride and NH four plus because they will still be Aquarius. The same can actually be said for answer choice B and C. Even though we have the presence of a different sulfide, we have hydrogen sulfide and answer choice B. And we have ammonium sulfide anti choice C. But the reason that we can't include these or that we can't use these is because we still have the presence of hydrochloric acid of two molar, it's slightly less concentration, but it's still a very strong acid. And because this strong asset is there, it will take precedence over the reaction between zinc and sulfur, which means that it will interfere and you won't necessarily get the precipitate that we need. However, finally looking at answer choice D. Which is indeed our correct option. We have ammonium sulfide at a very high ph which means we won't have the strong acid there. And when zinc sulfide or when zinc reacts with ammonium sulfide, you will end up forming the compound zinc sulfide. And from our liability rules, we know that sulfides are usually insoluble. Except for a few exceptions, such as ammonium sulfide or sodium sulfide to name a few. And this will leave us with just ammonium present in the solution. So the only re agent that can be used to separate, sinks two plus and ammonium in solution is sodium or sorry, ammonium sulfide. Answer choice D. I hope this helps, and I look forward to seeing you all in the next one.