0.1 mole of CH3NH2  Kb = 5 × 10-4 is mixed with 0.08 mole of HCl and diluted to one litre. The H+ in solution is

CH3NH2  +  HCl    →   CH3N+H3Cl-0.1                 0.08                0         0.02               0                  0.08    [ 0.08mole HCl neutralizes 0.08 mole methylamine. Concentration of base left is (0.1-0.08) = 0.02]                                                 This is basic buffer solution.           OH- = Kb × BaseSalt=5 × 10-4 × 0.020.08 = 1.25 × 10-4∴   H+ = 10-14OH- = 10-141.25 × 10-4 = 8 × 10-11 M