Question

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?

Answer 1

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

Answer 2

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)`