# 01 Mole Of CH3NH2 (Kb=5 * 10-4 Is Mixed With 0.08 Mole Of HCl And Diluted To One Litre. What Will Be The H+ Concentration In The Solution?

The reaction is as follows:

$$CH_{3}NH_{2} + HCl \rightarrow CH_{3}NH_{3}^{+} + Cl^{-}$$

0.10 0.08 0 0

0.02 0 0.08 0.08

The expression for the hydroxide ion concentration is $$[OH^{-}]$$

is

= $$K_{b} * \frac{[Base]}{[Conjugated acid]}$$

Now by substituting the values in the above expression, we get,

= $$1.25 * 10^{-4}$$

Since $$[H^{+}] [OH^{-}] = K_{w} = 10^{-14}$$

Therefore, $$[H^{+}] = \frac{10^{-14}}{[OH^{-}]}$$

= $$\frac{10^{-14}}{1.25 * 10^{-4}}$$

= $$8 * 10^{-11} M$$