Weak Bases
Similar calculations are used for weak bases.
eg.      NH3(g) + H2O(l) <---------> NH4+(aq) + OH-(aq)


This tells you that the dissociation constant for any of the conjugate bases can be obtained from the value of Kw divided by the appropriate value of Ka. (Ka can also be obtained by dividing Kw by Kb).
eg. NH3   Kw = 1.0 x 10-14                Ka for NH4+ = 5.6 x 10-10

        Kb = 1.0 x 10-14  =  1.8 x 10-5
                 5.6 x 10-10

Follow-Up Problem
What is the [OH-] of a 0.10 mol/L solution of NaCN?
When NaCN dissolves in water it dissolves completely because it is a sodium salt.
                  NaCN ------->    Na+    +    CN-
                   0.1 M             0.1 M      0.1 M
However the CN- ions that are produced then react with water and set up an equilibrium.
         CN-(aq) + H2O <-------->  HCN(aq) + OH-
         0.1 M -x                                x                x
When we set up the equilibrium equation we must set it us as a Kb equation. Ka would be used if HCN was reacting with water. This time the salt of HCN, the CN- is reacting and it is acting as a base. Therefore we must use the Kb value.

2.04 x 10-5  (x)(x)
                       0.1 - x

x2 +2.04 x 10-5 x -2.04 x 10-6 = 0

Apply the quadratic formula and solve for 'x'.

x = 1.42 x 10-3 mol/L      Therefore the [OH-] = 1.42 x 10-3 mol/L.

Follow-Up Problems
Calculate the pH of a 0.10 mol/L NaCH3COO solution?