Question

Question #2109a

Answer 1

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)`