✔ 最佳答案
CH3COOH(aq) ⇌ CH3COO⁻(aq) + H⁺(aq)
Ka of CH3COOH = 1.8 x 10⁻⁵ M
在原 10 mL 溶液中含 CH3COOH 的莫耳數 = 1 x (10/1000) = 0.01 mol
在原 10 mL 溶液中含 CH3COO⁻ 離子的莫耳數 = 1 x (10/1000) = 0.01 mol
(1)
加入 OH⁻ 離子的莫耳數 = 0.2 x (0.5/1000) =0.0001 mol
加入 NaOH 溶液後,溶液的體積 = (10 + 0.5)/1000 = 0.0105L
加入 NaOH 溶液後的反應:
CH3COOH(aq) + OH⁻(aq) → CH3COO⁻(aq) + H2O(l)
不考慮 CH3COOH 的解離,以上反應後:
[CH3COOH]o = (0.01 - 0.0001)/0.0105 = 0.0099/0.0105 M
[CH3COO⁻]o = (0.01 +0.0001)/0.0105 = 0.0101/0.0105 M
平衡時:
pH = pKa - log ([CH3COOH]/ [CH3COO⁻] ≈ pKa - log ([CH3COOH]o/[CH3COO⁻]o
所以 pH ≈ -log(1.8 x 10⁻⁵) - log (0.0099/0.0101) = 4.75
(2)
加入 H⁺ 離子的莫耳數 = 0.2 x (0.5/1000) =0.0001 mol
加入 HCl 溶液後,溶液的體積 = (10 + 0.5)/1000 = 0.0105L
加入 HCl 溶液後的反應:
CH3COO⁻ (aq) + H⁺(aq) → CH3COOH(aq)
不考慮 CH3COOH 的解離,以上反應後:
[CH3COOH⁻]o = (0.01 -0.0001)/0.0105 = 0.0099/0.0105 M
[CH3COOH]o = (0.01 + 0.0001)/0.0105 = 0.0101/0.0105 M
平衡時:
pH = pKa - log ([CH3COOH]/ [CH3COO⁻] ≈ pKa - log ([CH3COOH]o/[CH3COO⁻]o
所以 pH ≈ -log(1.8 x 10⁻⁵) - log (0.0101/0.0099) = 4.74