# Buffers: what does the strong base react with?

I am a bit confused on buffers: If I have a solution with a weak acid (say $\ce{NH4}$) and its conjugate weak base ($\ce{NH3}$), I have a buffer solution (in $\ce{H2O}$).

$$\ce{NH4 <--> NH3 + H+}$$

But now If I add a strong base to my solution ($\ce{OH-}$ for example) I get that my pH will go up but I don't understand the dynamics of it.

1. Will my added $\ce{OH-}$ react with the weak acid $\ce{NH4}$ (more $\ce{NH3}$ produced)?

$$\ce{NH4+OH -> NH3 + H2O}$$

1. OR will my added $\ce{OH-}$ react with the $\ce{H+}$ produced by my weak acid (Less $\ce{H+}$ produced)?

$$\ce{NH3 + H+ (products of \ce{NH4}) + OH- -> NH3 + H2O}$$

I understand that both cases up end up with the pH going up but but I want to be sure to understand with what exactly my $\ce{OH-}$ molecule reacts once added to my solution.

Is it the $\mathrm{1^{st}}$ reaction which means more $\ce{NH3}$ produced?

Or is it the $\mathrm{2^{nd}}$ reaction which means less $\ce{H+}$ produced?

The reaction NH$_4^+$ $\Leftrightarrow$ NH$_3$ + H$^+$ is in a dynamic equilibrium that means the reaction will happen in both directions all the time and the acid dissociation constant will tell you how much on average you have all components in solution so by that you can also calculate the pH value. So of course since you have NH$_4^+$ in solution as well as H+ both can in principle react with the OH$^-$ in providing the proton to form water. The question is, which of the two reactions happens more. This will be determined by the two equilibrium constants
$K_{1} = \frac{[NH3][H2O]}{[NH4+][OH-]}$ $K_2 = \frac{[NH3][H2O]}{[NH3][H+]}$
You see the only difference are the concentrations of the educts, so how large are these? Suppose you prepare a 0.05M NH$_4$Cl solution. After a short time there will be thermodynamic equilibrium and by using the $pK_\mathrm{a}=9.25$ value for NH$_4$Cl you can calculate the respective concentrations of NH$_4^+$ and NH$_3$ in equilibrium:
$K_a=\frac{[H+][NH_3]}{[NH_4^+]}$
By using $[H^+]=[NH_3]$ we get
$10^{-9.25}=\frac{[x][x]}{0.05-[x]}$ gives $[x]=5 \cdot 10^{-6} M$, so actually not much has dissociated. So you can be nearly completely sure that the reaction takes place with the NH$_4$ rather than protons or hydronium ions. And that is what a buffer is for! The pH should not change even when you add $OH^-$ until you reach the stability window of the buffer, that is when the NH$_4^+$ concentration gets too low compared to the $H^+$ concentration. Free protons will also never exist in solution, they can nearly always be associated to some water molecule.