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  4. Percentage ionization of 0.02 M dimethyl amine in the presence of 1 M NaOH, is ( K b =5.4 × 10-4)

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Q. Percentage ionization of $0.02\, M$ dimethyl amine in the presence of $1 \,M \,NaOH$, is $\left( K _{ b }=5.4 \times 10^{-4}\right)$

Equilibrium

Solution:

$\left( CH _{3}\right)_{2} NH + H _{2} O \rightleftharpoons\left( CH _{3}\right)_{2} NH _{2}^{+}+ OH^{-}$

image

$K_{b}=\frac{\left[\left( CH _{3}\right)_{2} NH _{2}^{+}\right]\left[ OH ^{-}\right]}{\left[\left( CH _{3}\right)_{2} NH \right]}$

$ 5.4 \times 10^{-4}=\frac{C \alpha \times 0.1}{C(1-\alpha)}=\alpha(0.1)$ (if $\alpha<<1)$

$ \therefore \alpha=5.4 \times 10^{-3}$

$ \\ \% $ of ionization $=5.4 \times 10^{-3} \times 100=0.54$