A buffer is prepared by dissolving HONH_ and $\mathrm{HONH}_{3} \mathrm{NO}_{3}$ in some water. Write equations to show how this buffer neutralizes added \(\mathrm{H}^{+}\) and \(\mathrm{OH}^{-}\)

First, let us write down the ionization reaction for HONH3+ (conjugate acid) and the reaction for HONH2 (conjugate base) with water: \[HONH_3^+ + H_2O \rightleftharpoons HONH_2 + H_3O^+\] \[HONH_2 + H_2O \rightleftharpoons HONH_3^+ + OH^-\]

When H+ ions are added to the buffer solution, the following reaction will occur: \[HONH_2 + H^+ \rightarrow HONH_3^+\] This reaction shows that the buffer will neutralize the added H+ ions by using the conjugate base (HONH2) to form the conjugate acid (HONH3+), thereby preventing any significant change in the pH of the solution. Now let us see how the buffer neutralizes added OH- ions: \[HONH_3^+ + OH^- \rightarrow HONH_2 + H_2O\] In this case, the buffer uses the conjugate acid (HONH3+) to react with the added OH- ions, forming the conjugate base (HONH2) and water. This process neutralizes the added OH- ions and prevents significant pH changes in the solution. So, the two equations that show how this buffer solution neutralizes added H+ and OH- ions are: \[HONH_2 + H^+ \rightarrow HONH_3^+\] \[HONH_3^+ + OH^- \rightarrow HONH_2 + H_2O\]