常温下,CH3COONa和CH3COOH混合溶液[pH=7,c(Na+)=0.1 mol•L-1],为什么c(CH3COO-)>c(CH3COOH)

问题描述:

常温下,CH3COONa和CH3COOH混合溶液[pH=7,c(Na+)=0.1 mol•L-1],为什么c(CH3COO-)>c(CH3COOH)

官方解释:水解是有限的,c(CH3COOH)约为c(CH3COO-)的百分之一左右.
这个解释很难理解.通过计算可以很好的理解
pH=7,所以[OH-]=[H+]
由电荷守恒,[Na+] + [H+] = [OH-]+ [CH3COO-]
所以 [CH3COO-] = [Na+] =0.1 mol/L
醋酸的电离常数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
很显然 [Na+] = [CH3COO-]>[CH3COOH]>[H+] =[OH-]