Start with the dissociation of the ethanoic acid, CH_3COOHCH3COOH, in aqueous solution to give hydronium cations and ethanoate anions, CH_3COO^(-)CH3COO−.
CH_3COOH_((aq)) + H_2O_((l)) rightleftharpoons H_3O_((aq))^(+) + CH_3COO_((aq))^(-)CH3COOH(aq)+H2O(l)⇌H3O+(aq)+CH3COO−(aq)
The acid dissociation constant, K_aKa, is equal to
K_a = ([H_3O^(+)] * [CH_3COO^(-)])/([CH_3COOH])Ka=[H3O+]⋅[CH3COO−][CH3COOH]
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))^(-)CH3COO−(aq)+H2O(l)⇌CH3COOH(aq)+OH−(aq)
The base dissociation constant, K_bKb, is equal to
K_b = ([CH_3COOH] * [OH^(-)])/([CH_3COO^(-)])Kb=[CH3COOH]⋅[OH−][CH3COO−]
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^(+)])Ka=[H3O+]⋅[CH3COO−][CH3COOH]⇒[CH3COO−][CH3COOH]=Ka[H3O+]
Since K_bKb uses the reciprocal of this ratio, you can write
([CH_3COOH])/([CH_3COO^(-)]) = ([H_3O^(+)])/K_a[CH3COOH][CH3COO−]=[H3O+]Ka
Plug this into the equation for K_bKb to get
K_b = ([H_3O^(+)])/K_a * [OH^(-)] = ([H_3O^(+)] * [OH^(-)])/K_aKb=[H3O+]Ka⋅[OH−]=[H3O+]⋅[OH−]Ka " "color(blue)((!)) (!)
Now take a look at the equation for the self-ionization of water
2H_2O_((l)) rightleftharpoons H_3O_((aq))^(+) + OH_((aq))^(-)2H2O(l)⇌H3O+(aq)+OH−(aq)
The ionization constant of water is known to be
K_W = [H_3O^(+)] * [OH^(-)]KW=[H3O+]⋅[OH−]
Plug this into equation color(blue)((!))(!) to get
K_b = K_W/K_a <=> K_a * K_b = K_WKb=KWKa⇔Ka⋅Kb=KW
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)Ka⋅Kb=10−14
