A level chen問題 acid base equilibria－貨運百科

標題:

A level chen問題 acid base equilibria

發問:

calculate the pH of a 0.01M aqueous solution of sodium ethanoate. the Ka for ethanoic acid is 1.8 * 10^-5. [Kw for water at 298K = 1.0 * 10^-14] 請列明詳細計算步驟 thx

最佳解答:

CH3COO-(aq) + H2O(l) ≒ CH3COOH(aq) + OH-(aq) First of all, calculate the equilibrium constant, Kh, for the above reaction. Kh = [CH3COOH][OH-]/CH3COO-] Kh = ([OH-][H+])/([CH3COOH]/[CH3COO-][H+]) Kh = Kw/Ka Kh = (1 x 10-14)/(1.8 x 10-5) Kh = 5.56 x 10-10 M startgeaa CH3COO-(aq) + H2O(l) ≒ CH3COOH(aq) + OH-(aq) start(M)eaaaa0.01 OO-(aq) H2O(l) a≒ C0OH(aq) + O0 change(M)aaaa-y OO-(aq) +2O(l) ≒ CH+yOH(aq) + O+y at eqm(M)aa0.01-y O-(aq) +2O(la) ≒ CHyOH(aq)a + Oy Kh = y2/(0.01 - y) = 5.56 x 10-10 M Hence y = 2.36 x 10-6 pOH = -log(y) = 5.63 pH = 14 - pOH = 8.37