溶液中解離度α疑問

若 0.1 M 之 CH3COOH 之解離度α＝1%，求[H+] 與 Ka＝？

CH3COOH(aq) ⇌ CH3COO^-(aq) + H^+(aq)

平衡時：
[CH3COOH] = 0.1 x (1 - 1%) = 0.099 M
[H^+] = [CH3COO^-] = 0.1 x (1%) = 0.001 M

Ka = [H^+][CH3COO^-]/[CH3COOH] = 0.001^2/0.099 = 1.01 x 10^-5


= = = = =
HF的K­＝7.0×10-4，求0.200 M的HF溶液中，[H^+]＝？

HF(aq) ⇌ H^+(aq) + F^-(aq)

平衡時：
設 [H^+] = [F^-] = y M
[HF] = (0.2 - y) M

Ka = [H^+][F^-]/[HF] = y²/(0.2 - y) = 7 x 10^-4
y² + (7 x 10^-4)y - (1.4 x 10^-4) = 0
y = {-(7 x 10^-4) + √[(7 x 10^-4)² - 4(1)(-1.4 x 10^-4)]} / 2
y = 0.0115

[H^+] = 0.0115 M


= = = = =
若0.1M之NH4OH之α＝0.014，求平衡後之[NH4OH]、[OH-]、Kb＝？

NH4OH(aq) ⇌ NH4^+(aq) + OH^-(aq)

平衡時：
[NH4OH] = 0.1 x (1 - 0.014) = 0.0986 M
[OH^-] = [NH4^+] = 0.1 x 0.014 = 0.0014 M

Kb = [OH^-][NH4^+]/[NH4OH] = (0.0014)²/0.0986 = 1.99 x 10^-5


= = = = =
若0.001M之NH4Cl之α＝0.076%，求NH3之Kb＝？

NH4^+(aq) ⇌ NH3(aq) + H^+(aq)

平衡時：
[NH4^+] = 0.001 x (1 - 0.076%) ≈ 0.001 M
[NH3] = [H^+] = 0.001 x 0.076% = 7.6 x 10^-7 M

Kh = [NH3][H^+]/[NH4^+] = (7.6 x 10^-7)^2/0.001 = 5.78 x 10^-10
Kb = Kw/Kh = (1 x 10^-14)/( 5.78 x 10^-10) = 1.73 x 10^-5