Calculate the pH of a 0.500M sodium ethanoate solution at 298K , given that the
dissociation constant of ethanoic acid is 1.74x10^-5 mol dm^-3 and the ionic
product of water is 1.00x10^-14 mol^2 dm^-6.
Chem- equilibrium
2010-04-05 9:33 am
回答 (1)
2010-04-05 5:51 pm
✔ 最佳答案
At eqm:Let [OH^-] = y M
Then [CH3COO^-] = (0.5 - y) M
and [CH3COOH] = y M
Kh = y^2/(0.5 - y) = 5.75 x 10^-10
y^2 + (5.75 x 10^-10)y - 2.875 x 10^-10 = 0
y = 1.7 x 10^-5
pOH = -log(1.7 x 10^-5) = 4.77
pH = 14 - 4.77 = 9.23
參考: 胡說八
收錄日期: 2021-04-22 22:21:18
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