# Don't understand what is exchanging with what in this acid/base equili

NH4+ + H2O (kr) = (kf) NH3 + H3O+;

the same reaction can be written as:
NH4+ + HO- (kr) = (kf) NH3 + H2O;

where (kf) - rate of forward reaction and (kr) - rate of reverse reaction.

1 Question. NH4+ or NH3 exchanging with H2O?

If NH4+ is exchanging with H2O:
NH4+ (k-1) =(k1) H2O,

where (k1) - rate of exchange from NH4+ to H2O and (k-1) - rate of exchange from H2O to NH4+.

If NH3is exchanging with H2O:
NH3(k-1) =(k1) H2O,

where (k1) - rate of exchange from NH3 to H2O and (k-1) - rate of exchange from H2O to NH3.

2 Question. What is the relationship between k1, k-1 and kf, kr?

Borek
Mentor
NH4+ + H2O (kr) = (kf) NH3 + H3O+;

the same reaction can be written as:
NH4+ + HO- (kr) = (kf) NH3 + H2O;

I don't see why you assume it is the same reaction with the same rate constants. They are definitely related to the same equilibrium, but that's all.

I don't see why you assume it is the same reaction with the same rate constants. They are definitely related to the same equilibrium, but that's all.

At pH=7, I assume that 1st one is the most likely case. I am not sure I understand NH4+ or NH3 is exchanging H+ with H2O?

Borek
Mentor
To be honest, I have no idea what the exact mechanism is - at the same time I don't think it matters. Equilibrium doesn't depend on these details, and it is equilibrium that we are typically interested in.

Please note there is no such thing as H+ nor OH- in the solution. H3O+ is only an approximation, as proton is most likely solvated by more than one water molecule. OH- is solvated as well, so it is present as - at least - H3O2-. We ignore all these and it doesn't interfere with our understanding of acid/base reactions, why should it matter for ammonia?