# When HCl and NH_3 are mixed why is the buffer NH_3-NH_4Cl?

Apr 2, 2016

Because a buffer solution is composed of a solution of a weak base, and its conjugate base (or vice versa) in appreciable quantities.

#### Explanation:

So how does this help us? You have an solution that is stoichiometric in ammonia; immediately you add hydrochloric acid, this protonates the ammonia, and now you both $N {H}_{3}$ and $N {H}_{4}^{+}$ in some quantity. The point is that $H C l$ does not stick around, and it gives ammonium chloride quantitatively:

$N {H}_{3} \left(a q\right) + H C l \left(a q\right) \rightarrow N {H}_{4} C l \left(a q\right)$

Given ammonia and ammonium chloride, you have a buffer system. On the other hand, I could form a buffer from ammonium chloride, if I added potassium hydroxide:

$N {H}_{4} C l \left(a q\right) + K O H \left(a q\right) \rightarrow N {H}_{3} \left(a q\right) + K C l \left(a q\right) + {H}_{2} O \left(l\right)$

In the second example, the strong base potassium hydroxide immediately deprotonates ammonium chloride, but it delivers ammonia in solution to act as a buffer with the unreacted ammonium chloride.

When you perform these weak base/strong acid or strong acid/weak base titrations, and plot a curve, the first region of the titration is often known as the buffer region because, from the above, that is precisely what it is, and the $p H$ profile remains quite stable for a while, until equivalence is reached. You know that at half equivalence the $p H$ of the solution is the $p {K}_{a}$ or $p {K}_{b}$ of the weak acid/base.