What is the equilibrium constant of CH3COOH?

1 Answer
Stefan V.
Jan 9, 2015

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#.

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