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+

Respuesta :

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).


PBCHEM

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. [tex]HONH_{2}[/tex] + [tex]H_{2}O[/tex] ⇔ [tex]HONH_{3}^{+}[/tex] + [tex]OH^{-}[/tex]
  2. [tex]HONH_{3}NO_{3}[/tex]⇔ [tex] OHNH_{3}^{+} + NO{3}^{-}[/tex]

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

  • Acid addition [tex]H^{+}[/tex]

[tex]HONH_{2} + H^{+} [/tex] → [tex] HONH_{3}^{+} [/tex]  

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

  • When adding [tex]OH^{-}[/tex]

[tex]HONH_{3}^{+} + OH^{-}[/tex] → [tex] HONH_{2} + H_{2}O[/tex]  

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

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