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

${CH}_{3} {NH}_{2} + HCl \to C H_{3} \overset{+}{N} H_{3} {Cl}^{-}$

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.

$[{OH}^{-}] = K_{b} \times \frac{[Base]}{[Salt]}$

$= 5 \times 10^{- 4} \times \frac{0.02}{0.08} = 1.25 \times 10^{- 4}$

$∴ [H^{+}] = \frac{10^{- 14}}{[{OH}^{-}]} = \frac{10^{- 14}}{1.25 \times 10^{- 4}} = 8 \times 10^{- 11} M$

0.1 mole of ${CH}_{3} {NH}_{2} (K_{b} = 5 \times 10^{- 4})$ is mixed with 0.08 mole of HCl and diluted to one litre. The $[H^{+}]$ in solution is