equilibrium

[匿名]

2008-04-30 5:27 am

A solution is formed by mixing equal volumes of 0.20M CH3COOH(aq) and 0.20M CH3COONa(aq) Given: Ka of CH3COOH = 1.76*10^-5 mol dm^-3 at 298K)
Calculate the concentration of each species, excluding H2O, presents in the solution at 298K.(Beware the DILUTION EFFECT!!)

回答 (1)

Uncle Michael

2008-04-30 7:33 am

✔ 最佳答案

Just after mixing (both solutions are diluted):
[CH3COOH]o = 0.2 x (1/2) = 0.1 mol dm-3
[CH3COO-]o = 0.2 x (1/2) = 0.1 mol dm-3

start a CH3COOH + H2O ≒ CH3COO- + H3O+
start aaaa 0.1 aaaaaaaaaaaa 0.1 aaaaaa 0 aaa mol dm-3
change aa -y aaaaaaaaaaaaa +y aaaaaa +y aa mol dm-3
at eqm a 0.1-y aaaaaaaaaa 0.1+y aaaaa y aaa mol dm-3

At equilibrium:
Since Ka is very small and the presence of CH3COO- ions shifts the equilibrium to the left, y is very small and thus 0.1 >> y.
[CH3COOH] = (0.1 - y) M ≈ 0.1 mol dm-3
[CH3COO-] = (0.1 + y) M ≈ 0.1 mol dm-3

Ka = [CH3COO-]y/[CH3COOH] = 1.76 x 10-5 mol dm-3
0.1y/0.1 = 1.76 x 10-5
Hence, [H3O+] = 1.76 x 10-5 mol dm-3
[OH-] = Kw/[H3O+] = (1 x 10-14)/( 1.76 x 10-5) = 5.68 x 10-10 mol dm-3

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