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At equilibrium, the concentration of CH3COOH is 2.0 × 10–1 M, the concentration of CH3COO– is 1.9 × 10–3 M, and the concentration of H3O+ is 1.9 × 10–3 M. What is the value of Keq for this reaction?

User Sprutex
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2 Answers

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Acetic acid (CH3COOH) reacts with water to form the acetate ion and the hydronium ion:



At equilibrium, the concentration of CH3COOH is 2.0 × 10–1 M, the concentration of CH3COO– is
1.9 × 10–3 M, and the concentration of H3O+ is 1.9 × 10–3 M. What is the value of Keq for this reaction?
1.8 * 10^-5
User HiddenUser
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Answer:
1.8* 10^(-5)

Step-by-step explanation:

Concentration of
CH_3COOH =
2.0* 10^(-1)

Concentration of
CH_3COO^- =
1.9* 10^(-3)

Concentration of
H_3O^+ =
1.9* 10^(-3)

The balanced equilibrium reaction will be,


CH_3COOH+H_2O\rightleftharpoons CH_3COO^-+H_3O^+

The expression for equilibrium reaction will be,


K=([H_3O^+]* [CH_3COO^-])/([CH_3COOH])

Now put all the given values in this expression


K=((1.9* 10^(-3))* (1.9* 10^(-3)))/((2.0* 10^(-1)))


K=1.8* 10^(-5)

Therefore, the value of Keq for this reaction is
1.8* 10^(-5)

User Ted Betz
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