When 0.05 M dimethylamine is dissolved in 0. | vedclass

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(C) Dimethylamine is a weak base: $(CH_3)_2NH + H_2O ightleftharpoons (CH_3)_2NH_2^+ + OH^-$Given: $C = 0.05 \ M$,$K_b = 5 	imes 10^{-4}$,and $[OH^

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$5 	imes 10^{-1}$

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(C) Dimethylamine is a weak base: $(CH_3)_2NH + H_2O ightleftharpoons (CH_3)_2NH_2^+ + OH^-$Given: $C = 0.05 \ M$,$K_b = 5 	imes 10^{-4}$,and $[OH^-]_{NaOH} = 0.1 \ M$.Due to the common ion effect,the concentration of $OH^-$ is dominated by $NaOH$,so $[OH^-] \approx 0.1 \ M$.The expression for $K_b$ is: $K_b = \frac{[(CH_3)_2NH_2^+][OH^-]}{[(CH_3)_2NH]} = \frac{C\alpha 	imes 0.1}{C(1-\alpha)} \approx \frac{0.05 	imes \alpha 	imes 0.1}{0.05} = 0.1 \alpha$.$5 	imes 10^{-4} = 0.1 \alpha \implies \alpha = 5 	imes 10^{-3}$.Percentage dissociation $= \alpha 	imes 100 = 5 	imes 10^{-3} 	imes 100 = 0.5 \% = 5 	imes 10^{-1} \%$.

When $0.05 \ M$ dimethylamine is dissolved in $0.1 \ M$ $NaOH$ solution,the percentage dissociation of dimethylamine is: $(K_b)_{(CH_3)_2NH} = 5 \times 10^{-4}$

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