问题描述: 25度时0.1mol/lL醋酸的电离度为1.32%,求该醋酸溶液中c(H+)及Ka 1个回答 分类:化学 2014-12-15 问题解答: 我来补答 CH3COOH H+ + CH3COO-0.1 0 0 (未电离时)0.1(1-1.32%) 0.1*1.32% 0.1*1.32% (电离后)=0.09868 0.00132 0.00132c(H+) = 0.00132 mol/LKa = 0.00132*0.00132/0.09868 = 1.76x10⁻⁵ 展开全文阅读