Ncert Exercises & Solutions

7.54 The ionization constant of dimethylamine is 5.4 × 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?

K_{b} = 5.4 \times 10^{- 4}

c = 0.02 M

Then, α = \sqrt{\frac{K_{b}}{c}}

= \sqrt{\frac{5.4 \times 10^{- 4}}{0.02}}

= 0.1643

Now, if 0.1 M of NaOH is added to the solution, then NaOH (being a strong base) undergoes complete ionization.

${NaOH}_{(aq)} \leftrightarrow {Na}_{(aq)}^{+} + {OH}_{(aq)}^{-}$
$0.1 M 0.1 M$

And,

$\underset{(0.02 - x)}{{({CH}_{3})}_{2} N H + H_{2} O} \leftrightarrow \underset{x}{{(C H_{3})}_{2} N H_{2}^{+}} + \underset{x}{O H}$
$; 0.02 M$
$T h e n, [{(C H_{3})}_{2} N H_{2}^{+}] = x$
$[O H^{-}] = x + 0.1; 0.1$
$\Rightarrow K_{b} = \frac{[{(C H_{3})}_{2} N H_{2}^{+}] [O H^{-}]}{[{(C H_{3})}_{2} N H]}$
$5.4 \times 10^{- 4} = \frac{x \times 0.1}{0.02}$
$x = 0.0054$

It means that in the presence of 0.1 M NaOH, 0.54% of dimethylamine will get dissociated.

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