Solubility of Cations/precipitation question

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The discussion centers on the solubility of cations, specifically Li+ and Sr2+, in relation to precipitation reactions involving sulfate ions (SO42-). It is established that Sr2+ is not soluble with sulfate, leading to precipitation, while Li+ remains soluble. The course material utilizes sodium sulfate (Na2SO4) to facilitate the precipitation test, as sodium compounds are preferred for their low cost and high solubility. The discussion also highlights the importance of using a group 1 or 2 metal to form ionic compounds with sulfate for effective analysis.

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supernova1203
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The question is asking us to determine weather there is a Li+ Ion or Sr+2 ion in the solution

We are using the precipitation method

Using various solubility tables I determined that Sr+2 ion is not soluble with the sulphate ion
SO-24

the course book uses the compound Na2SO4


my question is, they intentionally use sulphate ion because they know if there is Sr+2 ion in solution, there will be a precipitate, if there isn't then there will be no precipitate, but why did they use sodium? Couldnt they have used any group 1 Metal? since any group 1 metal would have been soluble with Li ion?
 
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Iv run into similar questions down the line now, you can always just put one ion that isn't soluble and find the precipitation, but why do they put the other element in there? Like they did in the problem described above, where they put in sodium along with sulphate?
 
I add a screenshot for further clarity

Here is another example, where they do this again, they just need to add

C2H3O2

but they also add a Na to each stage when testing for a certain Ion in a solution, to see wheather or not a precipitate forms


There is a typo for the H3

in the book it should be C2H3O2

but it says H5

Why do they add the extra sodium? i don't get it..
 

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How are you going to add just SO42-?
 
well i didnt know that i couldnt, i had forgetten from earlier lessons what an ionic compound was,(i have a terrible memory) anyway i figured out eventually that you need to have an element from group 1 or 2 and make an ionic compound with sulphate and then you can proceed to see if precipitation occurs or not
 
Actually you could use any sulfate, or even sulfuric acid. But that means troubles - other cations will just increase number of possible precipitates, sulfuric acid will acidify the solution, in both cases the analysis becomes more and more difficult as the number of factors one has to take into account grows. Sodium compounds are cheap, and soluble - so usually they are preferred over anything else.
 

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