When 0.05 M dimethylamine is dissolved in 0. | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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^

Vedclass · Accepted Answer

$5 	imes 10^{-1}$

Vedclass · Answer

(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: $(K_b)_{(CH_3)_2NH} = 5 \times 10^{-4}$

Similar Questions

If $S_1, S_2, S_3$ and $S_4$ are the solubilities of $AgCl$ in water,in $0.01 \ M \ CaCl_2$,in $0.01 \ M \ NaCl$ and in $0.05 \ M \ AgNO_3$ respectively at a certain temperature,the correct order of solubilities is

The ionization constant of phenol is $1.0 \times 10^{-10}.$ What is the concentration of phenolate ion in $0.05 \ M$ solution of phenol? What will be its degree of ionization if the solution is also $0.01 \ M$ in sodium phenolate?

Explain the common ion effect on the solubility of salts.

The solubility (in $mol \ L^{-1}$) of $AgCl$ $(K_{sp} = 1.0 \times 10^{-10})$ in a $0.1 \ M \ KCl$ solution will be

When $NH_4Cl$ is added to $NH_4OH$ solution,the dissociation of $NH_4OH$ is reduced. This is due to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module