There is a simple relationship between K_a and K_b, 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

1. HA < = > H⁺ + A^-      K = K_a of HA
2. + A^- + H₂O < = > HA + OH^-      K = K_b of A^-

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3. = H₂O < = > H⁺ + OH^-      K = K_w

K_w = (K_a of HA)*(K_b of A^-)

We know that K_w is 1.0*10^-14, so if we know K_a for an acid we can easily compute K_b for its conjugate base.

In a similar vein, it is easy to show that

pK_a + pK_b = 14

Example: The K_a of formic acid is 1.9*10^-4. What is the K_b and pK_b of the formate ion?

Solution: For the K_b, simply use the equation above

K_w = (K_a of HA)*(K_b of A^-)

K_b = K_w/K_a

K_b = 1.0*10^-14/1.9*10^-4

K_b = 5.3*10^-11

pK_b = -log₁₀(K_b)

pK_b = -log₁₀(5.3*10^-11)

pK_b = 10.3

Information provided by: http://learn.chem.vt.edu