Questions on CH3COOH(aq) + H2O(l) <--> CH3COO^(-)(aq) + H3O^(+)(aq) ?

[匿名]

2016-06-10 1:35 am

The equilibrium constant for the reaction is 1.8 × 10^(-5).

1. Which substance, at equilibrium, is likely to have the highest concentration?

a. CH3COOH
b. CH3COO-
c. H2O+
d. both 2 and 3

2. When sodium acetate, NaCH3COO, is added to the system, _____.

a. both [H3O+] and [CH3COOH] increase
b. both [H3O+] and [CH3COOH] decrease
c. [H3O+] increases and [CH3COOH] decreases
d. [H3O+] decreases and [CH3COOH] increases

3. If [H3O+] and [CH3COO–] are both 0.0010 M at equilibrium, what is [CH3COOH]?

a. 56 M
b. 5.6 × 10^(-2) M
c. 1.8 × 10^(-2) M
d. 0.18 M

I'm not sure if my answers are correct, so any help would be appreciated!!

Thank you!!

回答 (1)

fooks

2016-06-10 1:58 am

✔ 最佳答案

CH₃COOH(aq) + H₂O(l) ⇌ CH3COO⁻(aq) + H₃O⁺(aq)

1.
As the equilibrium constant for the reaction is small, most of the CH₃COOH molecules do not dissociate. Therefore, The concentration of CH₃COOH is much more concentrated than those of CH3COO⁻ and H₃O⁺.

...... The answer is : a. CH₃COOH

2.
The addition of sodium acetate would increase the concentration of CH3COO⁻. According to Le Chatelier's principle, the equilibrium position would shift to the left. Therefore, [H₃O⁺] decreases and [CH₃COOH] increases.

...... The answer is : d. [H₃O⁺] decreases and [CH₃COOH] increases

3.
K = [CH3COO⁻] [H₃O⁺] / [CH₃COOH]
1.8 × 10⁻⁵ = (0.0010) (0.0010) / [CH₃COOH]
[CH₃COOH] = (0.0010) (0.0010) / (1.8 × 10⁻⁵) M
[CH₃COOH] = 5.6 × 10⁻² M

...... The answer is : b. 5.6 × 10⁻² M

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