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_化学解答

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

人气：261 ℃　时间：2020-01-28 05:54:12

解答

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)=0.098680.001320.00132c(H+) = 0.00132 mol/LK...