第1个回答 2012-08-23
用解离常数Ka计算
设原醋酸溶液体积为1 L
CH3COOH = CH3COO- + H+
10^(-3) 10^(-3)
由Ka = [H+][CH3COO-] / [CH3COOH] = 1.8*10^(-5)
解得[CH3COOH] = 1/18 mol
所以总酸c(CH3COOH) = 1/18 + 10^(-3) = 0.05656 mol/L
稀释4倍,总酸 c(CH3COOH) = 0.05656 / 4 = 0.01414
CH3COOH = CH3COO- + H+
0.01414-x x x
x^2 / (0.01414-x ) = 1.8*10^(-5)
解得x = 4.96*10^(-4)
pH = 3.3本回答被提问者采纳