Question

What is the equilibrium constant of CH3COOH?

Answer 1

When placed in water, weak acids (generic `HA`) form a homogeneous equilibrium in which acid molecules react with water to form aqueous hydronium ions, `H_3^(+)O`, and aqueous anions, `A^(-)`.

In the case of acetic acid, which is a weak acid, the equilibrium can be described like this:

`CH_3COOH_((aq)) + H_2O_((l)) rightleftharpoons CH_3CHOO_((aq))^(-) + H_3^(+)O_((aq))`

The equilibrium constant is

`K_(eq) = ([H_3^(+)O]*[CH_3CHOO^(-)])/([CH_3COOH]*[H_2O])`

Since the concentration of liquid water is left out of the expression, the equilibrium constant for this reaction is called acid dissociation constant, `K_a`

`K_a = ([H_3^(+)O]*[CH_3CHOO^(-)])/([CH_3COOH])`

The values of the acid dissociation constants for various acids are usually given to you in an exam, acetic acid's equilibrium constant being `1.8*10^(-5)`; however, if the value is not given to you, you can always use the equilibrium concentrations described in the above equation to solve for `K_a`.