C=1mol
先建立质子平衡条件
HF
得:H+
参考;NH4+F-
失;H+
NH3
[H+]=[NH3]-[HF](1)
平衡方程1:NH4+==NH3+H+电离常数Ka1=5.5*E-10
Ka1=[NH3]*[H+]/[NH4+][NH3]=Ka1*[NH4+]/[H+](2)
平衡方程2;HF==H++F-电离常数Ka2=6.6*E-4
Ka2=[H+]*[F-]/[HF][HF]=[H+]*[F-]/Ka2(3)
把(2),(3)带入(1)得
[H+]=Ka1*[NH4+]/[H+]-[H+]*[F-]/Ka2
化简Ka2*[H+]^2=Ka1*Ka2*[NH4+]-[H+]2*[F-]
[H+]=√((Ka1*Ka2*[NH4+])/(Ka2+[F-]))
如果,Ka2