貨運百科

此文章來自奇摩知識+如有不便請留言告知

標題:

equilibrium

發問:

50 cm3 of 1M NH3 is titrated with 1M HCL What is the pH value before any HCL added? What is the pH after 25cm3 of HCL ADDDED? What is the pH at equivalence pt , when 50cm3 of HCL has added? Ka for NH4+ is 5.6*10^-10

最佳解答:

50 cm^3 of 1M NH3 is titrated with 1M HCl. What is the pH value before any HCl added ? NH3(aq) + H2O(l) = NH4^+(aq) + OH^-(aq) Kb = Kw/Ka = (1 x 10^-14)/(5.6 x 10^-10) = 1.79 x 10^-5 M At equilibrium : Let [OH^-] = a M Then [NH4+] = a M and [NH3] = (1 - a) M Kb = [NH4^+][OH^-]/[NH3] a^2/(1 - a) = 1.79 x 10^-5 a^2 + (1.79 x 10^-5)a - (1.79 x 10^-5) = 0 a = 0.00422 pOH = -log(0.00422) = 2.37 pH = 14 - 2.37 = 11.63 What is the pH after 25cm^3 of HCl ADDDED? H^+(aq) + NH3(g) → NH4^+(aq) Neglected any dissociation: Total volume = 25 + 50 = 75 cm^3 [NH4^+]o = 1 x (25/75) = 0.333 M [NH3]o = 1 x (50/75) - 1 x (25/75) = 0.333 M Consider the dissociation of NH4^+(aq) : NH4^+(aq) = NH3(aq) + H^+(aq) pH ≈ pKa - log([NH4^+]o/[NH3]o) pH ≈ -log(5.6 x 10^-10) - log(0.333/0.333) = 9.25 What is the pH at equivalence pt , when 50cm^3 of HCl has added? H^+(aq) + NH3(g) → NH4^+(aq) Neglected any dissociation: Total volume = 50 + 50 = 100 cm^3 [NH4^+]o = 1 x (50/100) = 0.5 M [NH3]o = 1 x (50/100) - 1 x (50/100) = 0 M Consider the dissociation of NH4^+(aq) : NH4^+(aq) = NH3(aq) + H^+(aq) Let [H^+] = b M Then [NH3] = b M and [NH4^+] = (0.5 - b) M Ka = [NH3][H^+]/[NH4^+] b^2/(0.5 - b) = 5.6 x 10^-10 b^2 + (5.6 x 10^-10)b - (2.8 x 10^-10) = 0 b = 1.67 x 10^-5 pH = -log(1.67 x 10^-5) = 4.78

其他解答:6CC7293C79127CE5