equilibrium_0－dvhtlbz的部落格

標題:

equilibrium

發問:

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!!)

最佳解答:

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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