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, \([H^{+}] = \frac{10^{-14}}{[OH^{-}]} \)

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

= \( 8 * 10^{-11} M \)

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