c(HAc)=0.001mol/lHAc溶液的PH值;(Ka=0.00018)

问题描述:

c(HAc)=0.001mol/lHAc溶液的PH值;(Ka=0.00018)

[H+] = (cKa)^(1/2) = (0.001*0.00018)^(1/2) = 4.24*10^(-4) mol/L
pH = -log[H+] =3.37