Themes > Science > Chemistry > Inorganic Chemistry > Acids and Bases > Acids and Bases Index > Relationship between Ka and Kb There is a simple relationship between Ka and Kb, the equilibrium constants for a weak acid and a weak base. If you remember back to the rules for combining equations and equilibrium constants, when you add two chemical equations the resulting equilibrium constant K for the reaction is the product of the first two. Consider then two reactions, one an acid and one a base HA < = > H+ + A-      K = Ka of HA + A- + H2O < = > HA + OH-      K = Kb of A- = H2O < = > H+ + OH-      K = Kw Since the product of the Ks for the first two reactions is the third, then Kw = (Ka of HA)*(Kb of A-) We know that Kw is 1.0*10-14, so if we know Ka for an acid we can easily compute Kb for its conjugate base. In a similar vein, it is easy to show that pKa + pKb = 14 Example: The Ka of formic acid is 1.9*10-4. What is the Kb and pKb of the formate ion? Solution: For the Kb, simply use the equation above Kw = (Ka of HA)*(Kb of A-) Kb = Kw/Ka Kb = 1.0*10-14/1.9*10-4 Kb = 5.3*10-11 For the pKb, just take the negative log of the Kb pKb = -log10(Kb) pKb = -log10(5.3*10-11) pKb = 10.3 Information provided by: http://learn.chem.vt.edu