A buffer is prepared by dissolving \(\mathrm{HONH}_{2}\) and \(\mathrm{HONH}_{3} \mathrm{NO}_{3}\) in some water. Write equations to show how this buffer neutralizes added \(\mathrm{H}^{+}\) and \(\mathrm{OH}^{-}\).

: In a buffer solution containing HONH₂ and HONH₃NO₃, the following acid-base reactions occur: - HONH₂ can accept an H⁺ ion and form its conjugate acid HONH₃⁺ - HONH₃NO₃ can donate an H⁺ ion and form its conjugate base HONH₂ These reactions allow the buffer to neutralize added H⁺ and OH⁻ ions.

: For the weak base HONH₂ accepting an H⁺ ion, the equilibrium reaction is: \(HONH_2 + H^+ \rightleftharpoons HONH_3^+\) Here, the HONH₂ is in equilibrium with its conjugate acid, HONH₃⁺. When an H⁺ ion is added to the solution, the equilibrium will shift to the right, and more HONH₃⁺ will be formed, neutralizing the added H⁺ ion.

: For the weak acid HONH₃NO₃ donating an H⁺ ion, the equilibrium reaction is: \(HONH_3NO_3 \rightleftharpoons H^+ + NO_3^- + HONH_2\) Here, HONH₃NO₃ is in equilibrium with its conjugate base, HONH₂. When an OH⁻ ion is added to the solution, it reacts with the H⁺ ion in the solution, reducing its concentration. The equilibrium will shift to the right, and more HONH₂ and H⁺ ions will be formed. The newly formed H⁺ ions will neutralize the added OH⁻ ions. In conclusion, the buffer solution containing HONH₂ and HONH₃NO₃ can neutralize added H⁺ and OH⁻ ions, maintaining the pH of the solution. The equilibrium reactions for these neutralization processes are: \(HONH_2 + H^+ \rightleftharpoons HONH_3^+\) (for added H⁺ ions) \(HONH_3NO_3 \rightleftharpoons H^+ + NO_3^- + HONH_2\) (for added OH⁻ ions)