Question

A buffer is prepared by dissolving honh2 and honh3no3 in some water. write equations to show how this buffer neutralizes added h+ and oh â. (use the lowest possible coefficients. omit states-of-matter in your answer.) h+

Answer 1

Hydroxylamine in water: HONH₂(aq) + H₂O(l) ⇄ HONH₃⁺(aq) + OH⁻(aq).
Hydroxylammonium nitrate in water: HONH₃NO₃(aq) → OHNH₃⁺(aq) + NO₃⁻(aq).
1) with positive hydrogen ions (protons) react base and gives weak conjugate acid: 
H⁺(aq) + HONH₂(aq) ⇄ HONH₃⁺(aq).
2) with hydroxide anions react acid and produce weak base and weak electrolyte water:
HONH₃⁺(aq) + OH⁻(aq) ⇄ HONH₂(aq) + H₂O(l).

Answer 2

Answer : When hydroxylamine is dissolved along with hydroxylammonium nitrate to prepare a buffer into water.

The chemical equation that can represent this reaction is -

1. $HONH_{2}$ + $H_{2}O$ ⇔ $HONH_{3}^{+}$ + $OH^{-}$
2. $HONH_{3}NO_{3}$⇔ $OHNH_{3}^{+} + NO{3}^{-}$

This is the buffer which will resists the changes when an acid or base is added to this solution.

• Acid addition $H^{+}$

$HONH_{2} + H^{+}$ → $HONH_{3}^{+}$  

When an acid is added to this buffer solution the extra $H^{+}$ will be converted into hydroxylammonium ion (which is a weak conjugate acid).

• When adding $OH^{-}$

$HONH_{3}^{+} + OH^{-}$ → $HONH_{2} + H_{2}O$  

when a base it added to the buffer it stabilizes the extra $OH^{-}$ ions in the solution by converting them into hydroxylamine (which is weak base) and water molecules.