计算0.1mol/L醋酸钠溶液中醋酸根离子的浓度大小.

答：
【1】[OH-] = {KbC}1/2 = {(Kw/Ka)C}1/2 ={ (10^-14/1.75*10^-5)x0.1}1/2 = 7.56x10^-6
【2】[H+] = Kw/[OH-] = 10^-14 /7.56x10^-6 = 1.32x10^-7
【3】 [HAc] + [Ac-] = c [H+] = [OH-] - [HAc] [HAc] = [OH-] - [H+]
【4】 [Ac-] = c — [HAc] = c — [OH-] +[H+] = 0.10 - 7.56x10^-6 + 1.32x10^-7 = 0.09999 = 0.1受你的启发，我用 [Ac-]*[OH-]/[Ac-]=10^-14/Ka [Ac-] = c — [OH-] +[H+] [H+]*[OH-]=10^-14 三个方程联立得到 [Ac-] = 0.0333217 怎么和你算的差那么多？你的计算结果，不合理。方程应该没错吧你试解一下谢谢