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標題:
equilibrium
發問:
Calculate the pH of 0.2 M solution of ammonium sulphate (NH4)2SO4 at 298K (Given Kb of NH3 = 1.8*10^-5 mol dm^-3)
最佳解答:
Each mol of (NH4)2SO4 contains 2 mol of NH4+ ions. [NH4+]o = 0.2 x 2 = 0.4 M start aaaNH4+(aq) + H2O(l) ≒ NH-3(aq) + H3O+(aq) start aaa 0.4 M q) + H0 M) ≒ NH0 Mq) + H0 M change a -y Maq) + H2O(l) ≒ NH+y M) + H+y M eqm aa(0.4 - y) M+ H2O(l) ≒ NH y M) + HOy M Hydrolysis constant Kh = Kw / Kb = (1 x 10-14) / (1.8 x 10-5) = 5.56 x 10-10 mol dm-3 Kh = y2 / (0.4 - y) = 5.56 x 10-10 y2 + (5.56 x 10-10)y - (2.22 x 10-10) = 0 [H3O+] = y = 1.49 x 10-5 M pH = -log(1.49 x 10-5) = 4.83
其他解答:
equilibrium
發問:
Calculate the pH of 0.2 M solution of ammonium sulphate (NH4)2SO4 at 298K (Given Kb of NH3 = 1.8*10^-5 mol dm^-3)
最佳解答:
Each mol of (NH4)2SO4 contains 2 mol of NH4+ ions. [NH4+]o = 0.2 x 2 = 0.4 M start aaaNH4+(aq) + H2O(l) ≒ NH-3(aq) + H3O+(aq) start aaa 0.4 M q) + H0 M) ≒ NH0 Mq) + H0 M change a -y Maq) + H2O(l) ≒ NH+y M) + H+y M eqm aa(0.4 - y) M+ H2O(l) ≒ NH y M) + HOy M Hydrolysis constant Kh = Kw / Kb = (1 x 10-14) / (1.8 x 10-5) = 5.56 x 10-10 mol dm-3 Kh = y2 / (0.4 - y) = 5.56 x 10-10 y2 + (5.56 x 10-10)y - (2.22 x 10-10) = 0 [H3O+] = y = 1.49 x 10-5 M pH = -log(1.49 x 10-5) = 4.83
其他解答:
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