A 0.2 M acetic acid (CH3OOH) was prepared in water at 25 degrees Celsius and sta

Question

A 0.2 M acetic acid (CH3OOH) was prepared in water at 25 degrees Celsius and started to dissociate to produce acetate (CH3COO-) and hydronium (H3O+) ions as described below. The solution is then allowed to come to equilibrium. Analysis of the equilibrium mixture shows that the concentration of H30+ is 0.0019 M.

CH3COOH (aq) + H2) (l) (equilibrium symbol) CH3COO- (aq) + H3O+ (aq)

A) What's the expression of equilibrium constant (Kc) using the concentration of reactants and products?

B) What's the concentration of CH3COOH (aq) and CH3COO- (aq) at equilibrium?

C) Calculate Kc

Explanation / Answer

(a) CH3COOH (aq) + H2O (l) <=> CH3COO- (aq) + H3O+ (aq)

Kc = [CH3COO-][H3O+]/[CH3COOH]


(b)...CH3COOH..+..H2O..<=>...H+...+...CH3COO-

I........0.2..................................0..................0

C.........-a...............................+a...............+a

E.......0.2-a.............................a..................a


At equilibrium:

[H3O+] = a = 0.0019 M

[CH3COOH] = 0.2 - a = 0.1981 M

[CH3COO-] = a = 0.0019 M


(c) Kc = [CH3COO-][H3O+]/[CH3COOH]

= 0.0019 x 0.0019/0.1981

= 1.82 x 10^(-5) = 1.8 x 10^(-5)

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