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

Question

Cameron Andrews · Accepted Answer

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

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)# You have a buffer system because you have ammonia and ammonium chloride. However, if I added potassium hydroxide to ammonium chloride, I could create a buffer: #NH_4Cl(aq) + KOH(aq) rarr NH_3(aq) + KCl(aq) + H_2O(l)# In the second example, ammonium chloride is instantly deprotonated by the strong base potassium hydroxide; however, ammonia in solution is delivered 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.

Avery Adams · Answer

undefined When HCl and NH₃ are mixed, the resulting buffer system is NH₃-NH₄Cl because NH₃ reacts with HCl to form NH₄Cl, which consists of ammonium ions (NH₄⁺) and chloride ions (Cl⁻). This system helps to resist changes in pH by providing both a weak base (NH₃) and its conjugate acid (NH₄⁺/NH₄Cl).