Chemical Equilibrium

2010-03-06 7:47 pm
1. 50cm3 of 1M acetic acid, CH3COOH is titrated with 1M of NaOH. Find the pH of the solution after the following volumes of 1M NaOH has been added
(i) 0.00 cm3
(ii) 25.00 cm3
(iii) 50.00 cm3
(iv) 75.00 cm3
Given the Ka for acetic acid is 1.8*10^-5 mol dm^-3

回答 (1)

2010-03-06 10:31 pm
✔ 最佳答案
(i)
CH3COOH(aq) ≒ CH3COO^-(aq) + H^+(aq)
At equilibrium :
Let [CH3COO^-] = [H^+] = y M
[CH3COOH] = (1 - y) M

Ka = y^2/(1 - y) = 1.8 x 10^-5
y^2 + (1.8 x 10^-5)y - (1.8 x 10^-5) = 0
y = 0.00423
pH = -log(0.00423) = 2.37


(ii)
CH3COOH(aq) + OH^-(aq) → CH3COO^-(aq) + H2O(l)
OH^- is completely reacted.
Neglect the dissociation of CH3COOH:
[CH3COO^-]o = 1 x (25/75) = 0.333 M
[CH3COOH]o = 1 x (50/75) - 1 x (25/75) = 0.333 M

pH
= pKa - log([CH3COOH]o/[CH3COO^-]o)
= -log(1.8 x 10^-5) - log(0.333/0.333)
= 4.74


(iii)
CH3COOH(aq) + OH^-(aq) → CH3COO^-(aq) + H2O(l)
OH^‑ and CH3COOH are completely reacted.
Neglect the hydrolysis of CH3COO^-:
[CH3COO^-]o = 1 x (50/100) = 0.5 M

Consider the hydrolysis of CH3COO^- :
CH3COO^-(aq) + H2O(l) ≒ CH3COOH(aq) + OH^-(aq) .. (Kh)
At equilibrium :
Let [CH3COOH] = [OH^-] = z M
[CH3COO^-] = (1 - z) M

Kh = (1 x 10^-14)/(1.8 x 10^-5) = 5.56 x 10^-10 M
Kh = z^2/(1 - z) = 5.56 x 10^-10
z^2 + (5.56 x 10^-10)z - (5.56 x 10^-10) = 0
z = 2.36 x 10^-5
pOH = -log(2.36 x 10^-5) = 4.63
pH = 14 - 4.63 = 9.37

(iv)
CH3COOH(aq) + OH^-(aq) → CH3COO^-(aq) + H2O(l)
OH^- is in excess.
[CH3COO^-]o = 1 x (50/125) = 0.4 M
[OH^-]o = 1 x (75/125) - 1 x (50/125) = 0.2 M

OH^- is a strong base:
pOH = -log(0.2) = 0.7
pH = 14 - pOH = 14 - 0.7 = 13.3
參考: 老爺子


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