常温下,CH3COONa和CH3COOH混合溶液[pH＝7,c(Na＋)＝0.1 mol•L－1],为什么c
2个回答

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+

x 0.1 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-]

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