Calculate the [OH ‾ ], [H+], and the pH of 0.35 M solutions of each of the following amines.?
(a) triethylamine [(C2H5)3N, Kb = 4.0 10-4]
[OH ‾ ]
[H+]
pH
(b) hydroxylamine (HONH2, Kb = 1.1 10-8)
[OH ‾ ]
[H+]
pH
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(a) (C2H5)3N(aq) + H2O ⇌ (C2H5)3NH^+(aq) + OH^-(aq)
Kb = [(C2H5)3NH^+][OH^-] / [C2H5)3N] = x² / 0.35 - x = 4.0x10^-4
We approximate x (omit ''-x'' in the denominator) as K << 0.35 M
x = [OH^-] = 0.0118 M
[H^+] = Kw / [OH^-] = 10^-14 / 0.0118 = 8.45x10^-13 M
pH = -log[H^+] = -log(8.45x10^-13) = 12.1
(b) HONH2(aq) + H2O(l) ⇌ HONH3^+(aq) + OH^-(aq)
Kb = [HONH3^+][OH^-] / [HONH2] = x² / 0.35 - x = 1.1x10^-8
Again we can approximate x
x = [OH^-] = 6.20x10^-5 M
[H^+] = 10^-14 / 6.20x10^-5 = 1.61x10^-10 M
pH = -log(1.61x10^-10) = 9.79