Question

已知在某温度下,醋酸的电离常数为4×10-5次方,0.1mol/l的醋酸溶液的PH=

Answer 1

CH3COOH ==可逆== CH3COO- + H+

初始 0.1 0 0

反应 x x x

平衡 0.1-x x x

所以 K = c(H+)c(CH3COO-)/c(CH3COOH) = x^2/(0.1-x) = 4×10^(-5)

由于x很小,所以 0.1-x≈0.1,即x^2 = 4×10^(-6),x = 2×10^(-3)

pH = - lg[2×10^(-3)] = 2.7