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Homework Help: Solubility of Cations/precipitation question

  1. Sep 19, 2011 #1
    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

    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 isnt 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????
  2. jcsd
  3. Sep 19, 2011 #2
    Iv run into similar questions down the line now, you can always just put one ion that isnt 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?
  4. Sep 19, 2011 #3
    I add a screenshot for further clarity

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


    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 dont get it..

    Attached Files:

  5. Sep 19, 2011 #4


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    Staff: Mentor

    How are you going to add just SO42-?
  6. Sep 19, 2011 #5
    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
  7. Sep 20, 2011 #6


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    Staff: Mentor

    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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