Qualitative Analysis Help

[4ORCE]

Question # 1

When a solution containing an unknown number of 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 and treated with (NH4)2S. This time a precipitate forms. The filtrate from this solution is not tested. Which groups of metal ions are either known to be present or cannot be excluded?

 

 

A. Group III and Group IV

B. Group III only

C. Group II and Group IV

D. Group IV only

E. Group I and Group IV

 

Question # 2

In the course of various qualitative analysis procedures, the following mixture is encountered: Na+ and K+. Suggest how this mixture might be separated. You may need to refer to Table 4.1 from Zumdahl.

 

 

A. add dilute HCl

B. add 0.2 M HCl and H2S

C. add (NH4)2S at pH 8

D. add (NH4)2HPO4 to a basic solution

E. It is not possible to separate the ions using any reaction scheme found in either Zumdahl or your lab manual.

 

 

I got D for #1 because the rest will precipitate but then I'm thinking E because only some of the group I ions will precipitate.

 

For #2 I got E, its not possible because they are always soluble.

 

Can anyone help?

[UC]

I have a copy of Zumdahl's chemical principles 5th edition. I take it that you are using a similar book.

 

Page 329-330 in the 5th edition. (I hope you know where it is in your text. Otherwise, use the index) Qualitative analysis. Are you using periodic table "groups" or the "groups" given in the book? I-insoluble chlorides, II-sulfides insoluble in acid solution, III-sulfides insoluble in basic solution, IV-insoluble carbonates, V-alkali metal and ammonium ions.

 

The two failed tests in question #1 exclude which two groups? The precipitate formed in the problem identifies which group? Are the remaining groups even able to form this kind of precipitate?

 

Your answer to #2 looks right. There are ways to seperate the two, but those are outside of what you need to know for your course, I believe. Adding perchloric acid, and chilling the solution will precipitate almost all the K+, for examaple. The difference in solubility is quite large. Oxalic acid will do the same, but with less selectivity. The potassium salt is about 10 times more soluble than the sodium salt. If you used one and then the other, you could achieve clean seperation with only a small amount of the original salts lost.