官方解釋：水解是有限的,c(CH3COOH)約為c(CH3COO－)的百分之一左右.
這個解釋很難理解.通過計算可以很好的理解
pH＝7,所以[OH-]=[H+]
由電荷守恒,[Na+] + [H+] = [OH-]+ [CH3COO-]
所以 [CH3COO-] = [Na+] =0.1 mol/L
醋酸的電離常數(shù)Ka = 1.8×10^-5
CH3COOH = CH3COO-+H+
x0.1 mol/L 10^-7 mol/L
[ H+][CH3COO- ] / x = Ka
即10^-7 × 0.1=1.8×10^-5
得到[CH3COOH] =x = 5.6×10^-4 mol/L
所以溶液中,
[H+] = 10^-7 mol/L
[CH3COOH] = 5.6×10^-4 mol/L
[CH3COO-] = [Na+] =0.1 mol/L
很顯然 [Na+] = [CH3COO－]＞[CH3COOH]＞[H+] =[OH-]