0.005mol/L的醋酸pH为多少

问题描述:

0.005mol/L的醋酸pH为多少

[H][Ac]/[HAc]=1.8*10(^-5);[H]={1.8*10(^-5)*0.005}^(1/2)=3*10(^-4); PH=4-0.48=3.62

c(H+)=√cKa=√0.005×1.75×10^-5=2.96×10^-4
PH=-lg2.96×10^-4=3.53