Question #2109a

Start with the dissociation of the ethanoic acid, #CH_3COOH#, in aqueous solution to give hydronium cations and ethanoate anions, #CH_3COO^(-)#.

#CH_3COOH_((aq)) + H_2O_((l)) rightleftharpoons H_3O_((aq))^(+) + CH_3COO_((aq))^(-)#

The acid dissociation constant, #K_a#, is equal to

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

When in aqueous solution, the ethanoate ion, which is the conjugate base of ethanoic acid, reacts with water to reform ethanoic acid and produce hydroxide ions.

#CH_3COO_((aq))^(-) + H_2O_((l)) rightleftharpoons CH_3COOH_((aq)) + OH_((aq))^(-)#

The base dissociation constant, #K_b#, is equal to

#K_b = ([CH_3COOH] * [OH^(-)])/([CH_3COO^(-)])#

Notice that both equations contain a form of the weak acid - conjugate base ratio. This means that you can use this ratio to determine a relationship between the concentration of hydronium ions, that of hydroxide ions, and the two dissociation constants.

#K_a = ([H_3O^(+)] * [CH_3COO^(-)])/([CH_3COOH]) =>([CH_3COO^(-)])/([CH_3COOH]) = K_a/([H_3O^(+)])#

Since #K_b# uses the reciprocal of this ratio, you can write

#([CH_3COOH])/([CH_3COO^(-)]) = ([H_3O^(+)])/K_a#

Plug this into the equation for #K_b# to get

#K_b = ([H_3O^(+)])/K_a * [OH^(-)] = ([H_3O^(+)] * [OH^(-)])/K_a# #" "color(blue)((!))#

Now take a look at the equation for the self-ionization of water

#2H_2O_((l)) rightleftharpoons H_3O_((aq))^(+) + OH_((aq))^(-)#

The ionization constant of water is known to be

#K_W = [H_3O^(+)] * [OH^(-)]#

Plug this into equation #color(blue)((!))# to get

#K_b = K_W/K_a <=> K_a * K_b = K_W#

For conjugate acid/base pair, the product of the two dissociation constants is equal to the ionization constant of water.

#K_a * K_b = 10^(-14)#