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標題:
calculate ph
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最佳解答:
Ionic product of water, Kw = 1 x 10-14 M2 The dissociation constant of NH3, Kb = 1.8 x 10-5 M NH4+(aq) ≒ NH3(aq) + H+(aq) Kh = Kw/Kh = (1 x 10-14)/(1.8 x 10-5) = 5.56 x 10-10 M At start (just after mixing): [NH4+]o = 0.1 x [20/(20 + 20)] = 0.05 M [H+]o = 0.18 x [20/(20 + 20)] = 0.09 M At equilibrium: Since Kh is very small, [H+] ≈ [H+]o = 0.09 M pH = -log[H+] = -log(0.09) = 1.05
其他解答:
calculate ph
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Calculate the pH of the following aqueous solutions prepared by adding 20.0 cm3 0.10M NH4Cl to 20.0 cm3 0.18M HCl. The answer is 1.05最佳解答:
Ionic product of water, Kw = 1 x 10-14 M2 The dissociation constant of NH3, Kb = 1.8 x 10-5 M NH4+(aq) ≒ NH3(aq) + H+(aq) Kh = Kw/Kh = (1 x 10-14)/(1.8 x 10-5) = 5.56 x 10-10 M At start (just after mixing): [NH4+]o = 0.1 x [20/(20 + 20)] = 0.05 M [H+]o = 0.18 x [20/(20 + 20)] = 0.09 M At equilibrium: Since Kh is very small, [H+] ≈ [H+]o = 0.09 M pH = -log[H+] = -log(0.09) = 1.05
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