Need some help with qualitative analysis/ Net Ionic Equations I need Net Ionic E

Question

Need some help with qualitative analysis/ Net Ionic Equations

I need Net Ionic Equations for the following:

1. HCl + Na2CO3

2. HNO3 + K2CrO4 + H2O2

3. H2SO4 + NaC2H3O2

4. NH4+ + NaOH (doesnt matter which NH4 compound is used here)

And I also need help with these two questions:

5. If you are given a solution which contains only one of the Group 1 cations (Ag+, Pb 2+, or Hg2 2+) and no other metallic ions, what two chemicals could you use to determine which cation is present? Describe the observations that you would expect to see with the addition of this chemical. There are several possibilities.

6. A solution may contain Ag+, Pb 2+, Hg2 2+. A white precipitate forms on addition of HCl. The precipitate is partially soluble in hot water. The residue turns black on addition of ammonia, NH3. Which of the ions are present, which are absent, and which remain undetermined? State your reasoning for all three ions.

Thanks alot!

Explanation / Answer

(1). both HCl and Na2CO3 are completely soluble in water, and they react to form NaCl, water and carbon dioxide, out of which NaCl is completely dissolved in water.

2HCl + Na2CO3 -------> 2NaCl + H2O + CO2

Now the net ionic equation can be written as

2H+ (aq) + 2Cl- (aq) +2Na+ (aq) + CO32- (aq) --------> 2Na+ (aq)+ 2Cl- (aq) H2O(l) + CO2 (g)

In the above ionic equation Na+ (aq) and Cl- (aq) ions are spectator ions as they exist as both reactant and product. Hence they can be removed from the net ionic equation. Hence the final net ionic equation is

2H+ (aq) + CO32- (aq) --------> H2O(l) + CO2 (g)

(5) Group 1 cations like Ag+, Pb2+, and Hg22+ cannot be distingiused by adding HCl, because they all  form AgCl, PbCl2,and Hg2Cl2 rspectively which are all white in colour.Hence they are distinguished with the help of NH3 and HNO3.

PbCl2 being soluble in hot water can easily be distinguished by heating the solution.

Now  AgCl and Hg2Cl2 can be distinguished by adding NH3 first. Hg2Cl2 gives a black precipitate with NH3 that contains Hg and HgNH2Cl. However AgCl dissolves in NH3 giving [Ag(NH3)2]+ . The presence of Ag+ can be conformed by reacting with HNO3 that gives white coloured AgCl.

(6) On addition of HCl they form AgCl, PbCl2,and Hg2Cl2 rspectively which are all white in colour. Since PbCl2 is completely solube in hot water and it is given that precipitate is partially soluble in hot water, hence Pb2+ is absent in the solution.

Since the residue turmns black on addition of NH3 , it must be Hg2Cl2, because Hg2Cl2 reacts with NH3 to give black coloured Hg. Hence the ion must be Hg22+.

AgCl is partially suluble in hot water, and when it reacts with NH3 it gives Ag(NH3)2+. To conform the presence of Ag it needs to be treated with HNO3 that will give a white precipitate of AgCl. Since this is not done here, the presence of Ag+ cannot be determined.

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