Ka與PH值問題

1.
首先考慮 CH3COOH 與 NaOH 的滴定：
CH3COOH(aq) + OH^-(aq) → CH3COO^-(aq) + H2O(l)
CH3COOH 的莫耳數 = 1 x (50/1000) = 0.05 mol
OH^- 的莫耳數 = 1 x (50/1000) = 0.05 mol
因此酸鹼剛好完全中和。
生成 CH3COO^- 離子的莫耳數 = 0.05 mol
溶液的總體積 = 50 + 50 = 100 ml
生成 CH3COO^- 離子的濃度 = 0.05/(100/1000) = 0.5 M

考慮 CH3COO^- 離子的水解：
CH3COO^-(aq) + H2O(l) ⇌ CH3COOH(aq) + OH^-(aq) .. (Kh)
起始時： [CH3COO^-]o = 0.5 M, [CH3COOH]o= [OH^-]o = 0 M
平衡時： [CH3COO^-] = (0.5 - y) M ≈ 0.5 M, [CH3COOH] = [OH^-] = y M

Kh = [CH3COOH][OH^-]/[CH3COO^-]
y²/0.5 = (1 x 10^-14)/(1 x 10^-5)
y = 2.24 x 10^-5

pOH = -log[OH^-] = -log(2.24 x 10^-5) = 4.65
pH = 14 - pOH = 14 - 4.65 = 9.35


2.
考慮 CH3COO^- 離子的水解：
CH3COO^-(aq) + H2O(l) ⇌ CH3COOH(aq) + OH^-(aq) .. (Kh)
起始時： [CH3COO^-]o = 0.015 M,[CH3COOH]o = [OH^-]o = 0 M
平衡時： [CH3COO^-] = (0.015 - y) M ≈ 0.015 M, [CH3COOH] = [OH^-] = y M

Kh = [CH3COOH][OH^-]/[CH3COO^-]
y²/0.015 = (1 x 10^-14)/(1.8 x 10^-5)
y = 2.89 x 10^-6

pOH = -log[OH^-] = -log(2.89 x 10^-6) = 5.54
pH = 14 - pOH = 14 - 5.54 = 8.46