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298K时,HAc电离常数为1.76*10的负五次方,则0.10mol/LHAc溶液中[H+]=
人气:485 ℃ 时间:2019-10-19 04:02:53
解答
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
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