Qualitative Analysis Homework - Inference from Q Mixture

[chikis]

Homework Statement

Suppose you have Q as a mixure of two inorganic salts. The mixture was dissolved in water and filtered.
Test were carried out on both the filtrate and residue.
(1)When a portion of the filtrate was poured into dilute HCL + BaCL₂(aq), there was no visible reaction. What would be the inference?

(2) When portion of filtrate was poured into dilute HNO₃ + AgNO₃(aq) + excess dilute. NH₃(aq), there was no visible reaction. What would be the inference?

Homework Equations

The Attempt at a Solution

.Suppose there was white precipate insoluble in excess of dilute HCl+BaCl₂(aq), I would have suggested that SO4^2- ion is present but in this case, there is nothing like that, so what would the inference be?


The same apply to (2). Suppose, there was a white precipitate soluble in excess as the portion of filtrate was added to dilute HNO₃+ AgNO₃(aq). NH₃(aq) then I would have suggested that Cl^- ion but there was not reaction.

 

[Borek]

I would have suggested that SO4^- ion is present but in this case, there is nothing like that, so what would the inference be?

That there is no SO₄^2-?

 

[chikis]

That there is no SO₄^2-?

Sorry that is a typo. I actually wanted to write SO₄^2-

 

[Borek]

And I answered assuming it was a typo.

 

[chikis]

And I answered assuming it was a typo.

The specific question have not been answered. You only corrected the typo error that I made. Thank you for that anyway.

 

[Borek]

You didn't get it. Lack of any reaction suggests the inference is "no sulfates present in the solution".

 

[chikis]

You didn't get it. Lack of any reaction suggests the inference is "no sulfates present in the solution".

What about the second part of the question,
When portion of filtrate was poured into dilute HNO₃ + AgNO₃(aq) + excess dilute. NH₃(aq), there was no visible reaction. What would be the inference?

 

[Borek]

What about the second part of the question[/b]

What about you trying to analyze the situation and come to some conclusions?

 

[chikis]

What about you trying to analyze the situation and come to some conclusions?

Then I take that Cl^- is absent.

 

[Borek]

What about ammonia presence?

 

[chikis]

What about ammonia presence?

No I don't think so!

 

[chikis]

So what is our final conclusion?

 

[AGNuke]

As far as I can see, the cations present in the salts given do not precipitate with Cl^-. Can you find out which ions do not precipitate with Cl^-? And the anions do not precipitate with silver ion.

Silver and Barium have very few soluble salts. Try to figure out them and you may get your answer.

I suspect the presence of Chlorate ions, which can form soluble salts with both Barium and Silver.

 

[chikis]

As far as I can see, the cations present in the salts given do not precipitate with Cl^-. Can you find out which ions do not precipitate with Cl^-? And the anions do not precipitate with silver ion.

Silver and Barium have very few soluble salts. Try to figure out them and you may get your answer.

I suspect the presence of Chlorate ions, which can form soluble salts with both Barium and Silver.

Please which of the test are you referring to?

Is it the one:
(1)When a portion of the filtrate was poured into dilute HCL + BaCL₂(aq), there was no visible reaction. What would be the inference??

OR

Is it the one:
(2) When portion of filtrate was poured into dilute HNO₃ + AgNO₃(aq) + excess dilute. NH₃(aq), there was no visible reaction. What would be the inference?
?

 

[AGNuke]

Both. When our salt is added to HCl + BaCl₂, the salt's anion will form "soluble" salt with Barium. When our salt is added to HNO₃ + AgNO₃ + excess NH₃, the salt's anion will form "soluble" salt with Silver as well.

Then I checked the solubility chart - The only anion making soluble salt with both Barium and Silver was Chlorate.

There's a line written in the question - "Test were carried out on both the filtrate and residue.". Care to elaborate what's residue. From the looks of it, the question is incomplete, since it is impossible to accurately predict the other anion and the two cations.