Answer:
HNO₂(aq) + H₂O(l) ⇄ H₃O⁺(aq) + NO₂⁻(aq)
Ka = [ H₃O⁺] . [NO₂⁻] / [HNO₂]
CH₃COOH(aq) + H₂O(l) ⇄ CHCOO⁻ (aq) + H₃O⁺(aq) Ka
Ka = [CHCOO⁻] . [H₃O⁺] / [CH₃COOH]
Remember to add the autoinization of water:
2H₂O(l) ⇄ OH⁻ (aq) + H₃O⁺(aq) Kw
Explanation:
Nitrous acid → HNO₂ (weak acid)
The ionization in water is:
HNO₂(aq) → H⁺(aq) + NO₂⁻(aq)
Let's write the equilibrium:
HNO₂(aq) + H₂O(l) ⇄ H₃O⁺(aq) + NO₂⁻(aq) Ka
Ka = [ H₃O⁺] . [NO₂⁻] / [HNO₂]
Acetic acid → CH₃COOH (weak acid)
Ionization in water: CH₃COOH(aq) → CHCOO⁻ (aq) + H⁺(aq)
Equilibrium: CH₃COOH(aq) + H₂O(l) ⇄ CHCOO⁻ (aq) + H₃O⁺(aq) Ka
Ka = [CHCOO⁻] . [H₃O⁺] / [CH₃COOH]
