Question

1：计算0.1mol/LHAc的pH值

Answer 1

0.1mol/lHAc:

[H+]=(0.1*1.76×10-5)^0.5=1.33*10^-3

PH=2.87

0.05mol/LNaAc:

[OH-]=(0.05*10^-14/1.76×10-5)^0.5=5.33*10^-6

POH=5.27

PH=14-10.55=8.73

HAc-NaAc缓冲溶液:

[H+]=1.76×10-5 *0.1/999=1.76*10^-8

PH=7.75