Solubility of Cations/precipitation question

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Discussion Overview

The discussion revolves around the solubility of cations, specifically focusing on the use of sodium sulfate in precipitation reactions to identify the presence of Li+ or Sr2+ ions in a solution. Participants explore the rationale behind using sodium as a counterion in these tests and the implications for determining precipitation.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Conceptual clarification

Main Points Raised

  • One participant questions why sodium is used with sulfate instead of any group 1 metal, suggesting that any group 1 metal would be soluble with Li+.
  • Another participant reflects on the necessity of including a soluble ion alongside the sulfate to facilitate precipitation testing.
  • A participant provides an example from a textbook where sodium is added to a compound, questioning the need for the additional sodium when testing for ions.
  • One participant expresses confusion about the requirement for an ionic compound with sulfate and acknowledges a lack of memory regarding earlier lessons on ionic compounds.
  • Another participant suggests that while any sulfate could be used, the inclusion of sodium simplifies the analysis by reducing the number of possible precipitates and avoids complications from other cations or acidification.

Areas of Agreement / Disagreement

Participants express differing views on the necessity and implications of using sodium in precipitation reactions. There is no consensus on whether sodium is the only suitable choice or if other group 1 metals could suffice.

Contextual Notes

Participants mention limitations in their understanding of ionic compounds and the complexities introduced by using different cations or acids in precipitation tests. There are references to specific examples from textbooks that may contain errors or unclear explanations.

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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