盐酸苯胺的水解常数理论值是多少
发布网友
共1个回答
热心网友
令苯胺为B,那么盐酸苯胺为BH+
苯胺的电离:B+ H2OBH+ + OH-
Kb=c(BH+)*c(OH-)/c(B)
查表电离常数：Kb=4.2X10^-10
BH+ +H2OB+ H3O+
盐酸苯胺的水解常数Kh=c(B)*c(H+) /c(BH+)
=[c(B)*c(H+)*c(OH-)] /[c(BH+)*c(OH-)]
=[c(H+)*c(OH-)] /[c(BH+)*c(OH-)/ c(B)]
=Kw /Kb
=10^-14 /4.2X10^-10
=2.4*10^-5
