A solution containing several metal ions is treated with dilute
...

Question

A solution containing several metal ions is treated with dilute 
HCl; no precipitate forms. The pH is adjusted to about 
1, and H2S is bubbled through. Again, no precipitate forms. 
The pH of the solution is then adjusted to about 8. Again, 
H2S is bubbled through. This time a precipitate forms. The 
filtrate from this solution is treated with 1NH422HPO4. No 
precipitate forms. Which of these metal cations are either 
possibly present or definitely absent: Al3 +, Na +, Ag +, Mg2 +?

Accepted Answer

Hey everyone today we're being asked which of the following groups of re agents can be used to separate lead two plus, calcium two plus and potassium plus from an Aquarius solution. Now let's go ahead and write these out. So um for lead two plus and calcium two plus depending on what they're bonded with they can either be soluble or insoluble in a solution which will help us for separating out the Catalans. But potassium, like sodium and ammonium for example are always soluble no matter what they're bonded to at least an Aquarius solution so they're always soluble. So we need to figure out which set of re agents in which order will help us precipitate out both. Lead two plus and calcium two plus in order to get them properly. So if we are to look at answer choice one for example we first have six molar hydrochloric acid. So that will react with all three. But while chlorine usually forms soluble salts according to our soluble itty rules there's actually an exception. And that is when it reacts with lead, silver or mercury when it reacts with lead as we have Here we end up getting lead chloride which is insoluble insoluble. So so far so good. We've now precipitated out one of the reactant. Then looking at the next step which is using ammonium phosphate and ammonia. Usually when a phosphate is involved, it makes the compound insoluble. However phosphate will only or will react with all three but it will actually only form an insoluble phosphate with calcium two plus will end up getting uh huh. C. A. Three P. 04 two which is also insoluble. So that actually gives us our answer and we can just double check the rest of them to make sure that we're on the right track. So the first step for the second answer choice is the same. We have hydrochloric acid but then it's reacting with ammonium sulfide at a ph of eight. Now ammonium sulfide will form insoluble sulfides when reacting with certain elements. But when there's no asset present at a basic condition both um lead sulfide and calcium sulfide will be soluble so they won't precipitate out so we won't really have or or sorry calcium sulfide will be soluble so it won't precipitate out so we can't really separate it. So that will be out of the question and the same thing can be said for ends choice C. And D. As well due to the presence of that um sulfide sulfate group. And just to double check for answer choices C. And D. The reason why this wouldn't work. Even though there is the present of the 0.2 molar hydrochloric acid is because that would only make the product that forms the it'll be lead sulfate or sorry lead sulfide. PBS which will only be precipitating at low phs which is why the asset is there. But it won't make C. A. S. A insoluble precipitate. So therefore we cannot identify anti choice A first using six molar hydrochloric acid and then ammonium phosphate pneumonia as the re agents that can separate all three cat ions. I hope this helps, and I look forward to seeing you all in the next one.

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