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

1 Answer
Apr 2, 2016

Answer:

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 #NH_3# and #NH_4^+# in some quantity. The point is that #HCl# does not stick around, and it gives ammonium chloride quantitatively:

#NH_3(aq) + HCl(aq) rarr NH_4Cl(aq)#

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:

#NH_4Cl(aq) + KOH(aq) rarr NH_3(aq) + KCl(aq) + H_2O(l)#

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 #pH# profile remains quite stable for a while, until equivalence is reached. You know that at half equivalence the #pH# of the solution is the #pK_a# or #pK_b# of the weak acid/base.