已知CH3COOH 的Ka＝2×10的- 5次方，若在500mL 0.01 M CH3COOH(aq)中，加入0.02 mol CH3COONa(s)，混合均勻後，pH值為多少？(log2＝0.3)
1. 4.4
2. 5.0
3. 5.3
4. 5.7
已知CH3COOH 的Ka＝2×10- 5，若在500mL
2012-11-16 8:14 am
回答 (2)
2012-11-16 9:56 am
✔ 最佳答案
答案是： 2. 5.0CH3COOH(aq) ⇌ CH3COO⁻(aq) + H⁺(aq)
平衡時：
[CH3COOH] ≈ [CH3COOH]o = 0.01/(500/1000) =0.02 M
[CH3COO⁻] = [CH3COO⁻]o = 0.02/(500/1000)= 0.04 M
Ka = [CH3COO⁻][ H⁺]/[CH3COOH]
pH = -logKa - log([CH3COOH]/[CH3COO⁻])
pH = -log(2 x 10⁻⁵) - log(0.02/0.04)
pH = -log 2 - log 10⁻⁵ + log 2
pH = 5.0
參考: 賣女孩的火柴
2013-10-05 11:11 pm
CH3COOH---------> CH3COO^- + H^+
前 0.01 (0.02/0.5)=0.04 0
後 0.01-A 0.04+A A
Ka= (0.04+A)*A / 0.01-A = 2x10^-5
得 [ H^+ ] = 5x10^-6 M
PH= -log (5x10^-6) = 6-log 5 =6-0.7 =5.3
前 0.01 (0.02/0.5)=0.04 0
後 0.01-A 0.04+A A
Ka= (0.04+A)*A / 0.01-A = 2x10^-5
得 [ H^+ ] = 5x10^-6 M
PH= -log (5x10^-6) = 6-log 5 =6-0.7 =5.3
收錄日期: 2021-04-13 19:07:42
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20121116000016KK00074