الأجوبة
CH3COOH + H2O ↔ H3O+ + CH3COO-
(حمض1) (أساس2) (أساس مرافق1) (حمض مرافق2)
[H3O+] = 10-PH = 10-3 mol/L
[H3O+] = √Ka.Ca
10-3 = √Ka*0.05
Ka = 2*10-5
α = [H3O+]/Ca
α = 10-3/0.05 = 2*10-3
α = 2%
[H3O+]’/[H3O] = 10-5/10-3 = 10-2
[H3O+]’ = [H3O+]/100
أسئلة مشابهة
القوائم الدراسية التي ينتمي لها السؤال