Question

0.02mol/L的NH4Cl溶液的pH＝5.47 怎么算的?

Answer 1

Kb(NH3.H2O)=1.8*10^-5

Ka(NH4+)=Kw/1.8*10^-5=5.56*10^-10

[H+]={0.02*Ka}^1/2=3.33*10^-6

pH=-lg[H+]=5.47