500mL水溶液中含有0.61g苯甲酸(C6H5COOH)和2.16g苯甲酸(C6H5COONa)，求此溶液中之[H+]。以知苯甲酸之Ka=6.6*10^-5。(C6H5COOH=122，C6H5COONa=144)
高中化學1題(急需)
2012-05-02 9:50 am
回答 (1)
2012-05-02 11:11 am
✔ 最佳答案
[C6H5COOH]o = (0.61/122)/(500/1000) =0.01 M[C6H5COONa]o = (2.16/144)/(500/1000) = 0.03 M
C6H5COOH(aq) ⇌ C6H5COO^-(aq)+ H^+(aq)
平衡時，由於 [H^+] 濃度很低：
[C6H5COOH] = 0.01 M - [H^+] ≈ 0.01 M
[C6H5COO^-] = 0.01 M + [H^+] ≈ 0.03 M
Ka = [C5H5COO^-][H^+]/[C6H5COOH]
(0.03)[H^+]/(0.01) = 6.6 x 10^-5
[H^+] = 2.2 x 10^-5 M
參考: fooks
收錄日期: 2021-04-13 18:40:07
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20120502000010KK00660