Question

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

Answer 1

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 0.00132
c(H+) = 0.00132 mol/L
Ka = 0.00132*0.00132/0.09868 = 1.76x10⁻⁵

Answer 2

擦！忘了啊！大一学得，现在都忘了。。。真是学完了忘完了。