And acid in aqueous solution is conceived to undergo a protonolysis reaction...
HX(aq) + H_2O(l) rightleftharpoonsH_3O^+ + X^(-)
And as for any equilibrium, we can measure and quantify it in the usual way...
K_a=([H_3O^+][X^-])/([HX(aq)])
Note that H_2O DOES NOT appear in the equilibrium expression because it is present in such high concentration that it is effectively constant..
For strong acids, i.e. HI, HBr, HCl, H_2SO_4 protonolysis is effectively quantitative: the given equilibrium lies entirely to the right as we face the page, and the acid solution is quantitative in H_3O^+. For weaker acids, HF, H_3C-CO_2H, the equilibrium lies somewhat to the left...and concentrations of the parent acid remain at equilibrium.
And likewise, we can formalize the performance of a base by an equivalent equilibrium...we use ammonia, because this is a WEAK base in aqueous solution...
NH_3(aq) + H_2O(l) rightleftharpoonsNH_4^+ + HO^-
And K_b is defined in an equivalent way to K_a...
K_b=([NH_4^+][HO^-])/([NH_3(aq)]), K_b"(ammonia)"=1.74xx10^-5...
Confused yet....?
Well, note that NECESSARILY....for a given acid/conjugate pair, say NH_4^+"/"NH_3...
K_aK_b=10^-14...or perhaps more usefully...
pK_a+pK_b=14...
