大学无机分析化学.求解NaHN4HPO4溶液的PBE

电荷守恒:
[Na+]+ [H+]+[NH4+] =[OH-] +[H2PO4-] +2 [HPO42-] +3 [PO43-]
物料守恒:
设NaNH4HPO4的浓度为c
[Na+]=c
[NH3]+[NH4+] =c
[H3PO4]+[H2PO4-] +[HPO42-] + [PO43-]=c
因此
[Na+]+[NH3]+[NH4+] ={[H3PO4]+[H2PO4-] +[HPO42-] + [PO43-]}*2
两式相减,消除[Na+]
[Na+]+ [H+]+[NH4+] - {[Na+]+[NH3]+[NH4+] }=[OH-] +[H2PO4-] +2 [HPO42-] +3 [PO43-]-{[H3PO4]+[H2PO4-] +[HPO42-] + [PO43-]}*2
[H+]-[NH3]=[OH-] -[H2PO4-] + [PO43-]-2[H3PO4]
因此:质了守恒:
[H+]+2[H3PO4] +[H2PO4-] =[NH3]+[OH-] + [PO43-]
也可 [H+] 水 = [OH-]水
HPO42-结合一个H+,生成H2PO4-
HPO42-结合二个H+,生成H3PO4
水电离剩余的H+形成溶液的H+
因此[H+] 水=[H+] + [H2PO4-] + 2[H3PO4]
HPO42-失去一个H+(与OH-反应),生成PO43-
NH4+失去一个H+(与OH-反应),生成NH3
水电离剩余的OH-形成溶液的OH-
因此[OH-]水=[NH3]+[OH-] + [PO43-]
则:
[H+]+2[H3PO4] +[H2PO4-] =[NH3]+[OH-] + [PO43-]

[H]+2[H3PO4]+[H2PO4]=[OH]+[NH3]+[PO4]