I have doubt in derivation of the relation between Ka and Kb in ionic equilibrium topic

Question

While deriving the relation between Ka and Kb in ionic equilibrium topic, say I take an example of NH4+ + H2O ----> NH3 + H3O+ ......(1)

Ka of this reaction is : [NH3][H3O+]/[NH4+][H2O]

Now NH3 is a conjugate base of NH4+.

Its Kb can be written as: NH3 + H2O ----> NH4+ + OH- .....(2)

Kb of the reaction: [NH4+][OH-]/[NH3][H2O]

In my textbooks while deriving relation between Ka and Kb, it has added the two equation (hence multiplies the K's). In the addition of two equations we get:

NH4+ + 2H2O + NH3 ----> NH3 + NH4+ + H3O+ + OH-

My textbook cancelled out [NH4+] concentration terms from LHS and RHS aswell as [NH3] concentration terms from LHS and RHS. [NH4+] on LHS is from equation (1) and [NH4+] on RHS is from equation (2).

My question is how can we cancel them out. I mean how can you say [NH4+] formed from hydrolysis of NH3 in reaction (2) is equal to that of [NH4+] initially taken in reaction (1) ? Even if you initially take same concentration of [NH3] for reaction 2 as formed from reaction 1, that concentration of NH3 on hydrolysis can never give same concentration of [NH4+] as initially taken in reaction 1. Then how can you cancel these different [NH4+] concentrations when you are adding reaction 1 and 2 ?

I know its a small question. But please don't underestimate the question.

Answer 1

In the third equation that you wrote (i.e. the one in which you added the two earlier equations), the effective reaction is the ionization of water to form a hydronium ion (or if you cancel one of the H2O molecules, a proton) and an OH- ion. NH4+ and NH3 are not reactants or products in this reaction. Since the reaction constant for the ionization of water is well-known, it can be used to find Kb if you know Ka and vice versa.

 

best,

Ram