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

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.

Answer 2

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).