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#...
