How Do You Solve Ionic Equilibrium Problems in Chemistry?

AI Thread Summary
To solve ionic equilibrium problems, it's essential to clarify whether the given concentration of ammonia is the initial or equilibrium concentration. The discussion highlights that using the initial concentration is crucial for accurate calculations. An ICE table is recommended to account for reaction stoichiometry effectively. The method of finding the concentration of Ag(NH3)2+ from Kc is valid if the correct concentrations are used. Ensuring clarity on the concentrations involved leads to a more accurate solution.
vijayramakrishnan
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Homework Statement



please see the attached image

1998 ,b part question
[/B]

Homework Equations



none

The Attempt at a Solution



i thought that Agcl would give Ag+ and cl- ions whose concentration can be found from Ksp as √Ksp,this Ag+ reacts with NH3 to give Ag(NH3)2+ (reverse of the given reaction) since Kc of the reaction is given we can find concentration of Ag(NH3)2+ from it which is Kc= [Ag(NH3)2+]/[Ag+][NH3]2

but answer obtained is not matching, is the method correct ?,please help.
 

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It is not clear to me whether 1M ammonia is the final concentration, or added concentration (in which case some of the ammonia would be consumed).

Your approach is OK in the first case, can be wrong in the latter.
 
Borek said:
It is not clear to me whether 1M ammonia is the final concentration, or added concentration (in which case some of the ammonia would be consumed).

Your approach is OK in the first case, can be wrong in the latter.
sir it is the initial concentration i suppose
 
Then you need to take the reaction stoichiometry into account (simplest approach: an ICE table).
 
Borek said:
Then you need to take the reaction stoichiometry into account (simplest approach: an ICE table).
oh i think i made a mistake concentration of ammonia in initial state is 1 M not the equilbrium concentration,sir are the rest of steps correct?
 
As I told you:

Borek said:
Your approach is OK in the first case, can be wrong in the latter.
 
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