Question

298K时,HAc电离常数为1.76*10的负五次方,则0.10mol/LHAc溶液中[H+]=

Answer 1

HAc + H2O = Ac- + H3O+ Ka=1.76*10^-5

Ka= [H3O+][Ac-]/[HAc] 设[H3O+]=[Ac-]=x mol/L

x^2/Ka = 0.10

x = (0.10*1.76*10^-5)^0.5 = 1.33*10^-3 mol/L