✔ 最佳答案
In both (a) and (b), it is assumed that the concentration of H3O+(aq) ions and OH-(aq) ions formed from the self-ionization of water are negligible when compared with the dissociation of C2H5COOH.
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(a)
start aaaC2H5COOH(aq) + H2O(l) ≒ C2H5COO-(aq) + H3O+(aq)
start aaaaa 0.1 MH(aq) + H2O(l) ≒ C2H0 MOO-(aq) + H0 M
change aaa -y MOH(aq) + H2O(l) ≒ 2H+y M O-(aq) +H+y M
eqm aaa(0.1 - y) M(aq) + H2O(l) ≒ aC2y M O-(aq) + H3y M
Ka = y2 / (0.1 - y) = 1.3 x 10-5 mol dm-3
y2 + (1.3 x 10-5)y - (1.3 x 10-6) = 0
[H3O+] = y = 1.13 x 10-3 M
pH = -log(1.13 x 10-3) = 2.95
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(b)
start aaaC2H5COOH(aq) + H2O(l) ≒ C2H5COO-(aq) + H3O+(aq)
start aaaaa 0.1 MH(aq) + H2O(l) ≒ C2H0.05 MOO-( + H0 M
change aaa -w MOH(aq) + H2O(l) ≒ 2H+w M O-(aq) +H+w M
eqm aaa(0.1 - w) M(aq) + H2O(l) aC2(0.05 + w) M O-(aHw M
Assumption:
Since Ka is small and the presence of CH3COO-(aq) would shift the equilibrium position to the left,
it is assumed that w is very small.
Therefore: [C2H5COOH] = (0.1 - w) M ≈ 0.1 M
Therefore: [C2H5COO-] = (0.05 + w) M ≈ 0.05 M
Ka = (0.05 + w)w/(0.1 - w) ≈ 0.05w/0.1 = 1.3 x 10-5 mol dm-3
[H3O+] = w = 2.6 x 10-5 M
pH = -log(2.6 x 10-5) = 4.59
