- pisgahchemistLv 71 decade agoFavorite Answer
2HNO3 + Ag(NH3)2^+ --> 2NH4+ + Ag+ + 2NO3^-
The nitric acid will cause the decomposition of the diamine complex, freeing up the silver ion to potentially react with some other ion to form an insoluble compoud, much like the next reaction:
This should be:
2HNO3(aq) + Ag(NH3)2^+ + Cl- --> AgCl(s) + 2NH4+ + 2NO3^-
The addition of ammonia solution to silver chloride will dissolve the silver chloride and produce silver diamine ion and chloride ion in a basic solution. When acid is added the solution is made acidic and the complex ion dissociates, freeing up the silver ion to recombine with chloride ion. This is the confirmatory test for silver in the "group 1" qualitative analysis scheme.
KI(aq) + Ag(NH3)2^+ --> No Reaction
This will not react, since the ammonia is a "stronger" ligand than iodide.
============= Follow up ============
With all due respect, I beg to differ.
"the second one (sic, the third one) makes K(NH3)2 + AgI"
Not even close.
"i think the first one is: --> HCl + AgNO3 + 2NH3"
Here's the reason why. As long as the solution of Ag(NH3)2^+ remains basic, then the complex ion will remain intact.
As far as having HCl and AgNO3 present together, that would immediately go to AgCl(s) and H+ and NO3^-.
- JessicaLv 44 years ago
for a, H has to be H2, because hydrogen is diatomic. then balance it. it will come out to 2Na + 2HNO3 ----> 2NaNO3 + H2 B is right C i think is right, but I'm not sure. check to make sure Mg is more reactive than Zn. if it is, you're correct.