Ionic EquilibriumHard
Question
The dissociation constant of acetic acid at a given temperature is 1.69 × 10-5. The degree of dissociation of 0.01 M acetic acid in the presence of 0.01 M HCl is equal to
Options
A.0.41
B.0.13
C.1.69 × 10-3
D.0.013.
Solution
CH3COOH (aq) ⇋ H+ (aq) + CH3COO- (aq)
t = 0 0.01
t = eq 0.01 - x x x
[H+] = x + 0.01 ≈ 0.01 M
∴ Ka =
⇒ 1.69 × 10-5 = 
∴ [CH3COO-] = 1.69 × 10-5 M
So, degree of dissociation of CH3COOH =
= 1.69 × 10-3
t = 0 0.01
t = eq 0.01 - x x x
[H+] = x + 0.01 ≈ 0.01 M
∴ Ka =
⇒ 1.69 × 10-5 = ∴ [CH3COO-] = 1.69 × 10-5 M
So, degree of dissociation of CH3COOH =
= 1.69 × 10-3Create a free account to view solution
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