Question
The ionization constant of dimethylamine is $5.4 \times 10^{–4}$​​​​​​​. Calculate its degree of ionization in its $0.02M$ solution. What percentage of dimethylamine is ionized if the solution is also $0.1M$ in NaOH?

Answer

$\text{K}_\text{b}=5.4\times10^{-4}$
$\text{c}=0.02\text{M}$
$\text{Then, }\alpha=\sqrt{\frac{\text{K}_\text{b}}{\text{c}}}$
$=\sqrt{\frac{5.4\times10^{-4}}{0.02}}$
$=0.1643$
Now, if 0.1M of NaOH is added to the solution, then NaOH (being a strong base) undergoes complete ionization.
$\text{NaOH}_\text{(aq)}$ $\leftrightarrow$ $\text{Na}^+_\text{(aq)}$ $+$ $\text{OH}^-_\text{(aq)}$
    $0.1\text{M}$   $0.1\text{M}$
And,
$\text{(CH}_3)_2\text{NH}$ $+$ $\text{H}_2\text{O}$ $\leftrightarrow$ $(\text{CH}_3)_2\text{NH}_2^+$ $+$ $\text{OH}^-$
$(0.02-\text{x})$       $\text{x}$   $\text{x}$
$;0.02\text{M}$           $;0.1\text{M}$
Then, $[(\text{CH}_3)_2\text{NH}_2^+]=\text{x}$
$[\text{OH}^-]=\text{x}+0.1;0.1$
$\Rightarrow\text{K}_\text{b}=\frac{[(\text{CH}_3)_2\text{NH}_2^+][\text{OH}^-]}{[(\text{CH}_3)_2\text{NH]}}$
$5.4\times10^{-4}=\frac{\text{x}\times0.1}{0.02}$
$\text{x}=0.0054$
It means that in the presence of 0.1M NaOH, 0.54% of dimethylamine will get dissociated.

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