作者flowerkid (Never give up)
看板Chemistry
标题Re: [问题] 弱酸 + 弱硷 pH 计算？
时间Sat Dec 19 01:15:55 2015
弱酸、弱硷滴定
以NH3(弱硷) + CH3COOH(弱酸) 为例, 酸硷中和後产生-> CH3COO- + NH4+
在水溶液中
[NH4+] -> [NH3] + [H+]
[CH3COO-] -> [CH3COOH] + [OH-]
Charge balance
[H+] + [NH4+] = [CH3COO-] + [OH-]
Mass balance
[NH4+]' = [NH4+] + [NH3] (注：[NH4+]'为解离前浓度, [NH4+]为解离後浓度)
[CH3COO-]' = [CH3COO-] + [CH3COOH] (注：同上)
由Charge balance => [H+] = [CH3COO-] - [NH4+] + [OH-]
由Mass balance
=> [CH3COO-] = [CH3COO-]' - [CH3COOH]
=> [NH4+] = [NH4+]' - [NH3]
代入上式
=>[H+] = [CH3COO-]' - [CH3COOH] - [NH4+]' + [NH3] + [OH-]
又[CH3COO-]' = [NH4+]'
代入上式
=>[H+] = [NH3] - [CH3COOH] + [OH-]
又[NH3] = Ka[NH4+]/[H+]
[CH3COOH] = [CH3COO-][H+]/Ka'
[OH-] = Kw/[H+]
代入上式
=>[H+] = Ka[NH4+]/[H+] - [CH3COO-][H+]/Ka' + Kw/[H+]
解联立方程式即可求出[H+]。
大致上应该是这样, 如有误请指正。
※ 编辑: flowerkid (111.255.136.117), 12/19/2015 01:27:38
1F：推 logs: 感谢回应！ 12/19 17:21