Question

The ionisation constant of dimethylamine is 5.4 × $10^{-4}$. Calculate its degree of ionization in its 0.02 M solution. What percentage of dimethylamine is ionized if the solution is also 0.1 M NaOH ?

Answer 1

Kb = 5.4x10-4
c = 0.02M
Now, if 0.1 M of NaOH is added to the solution, then NaOH (being a strong base) undergoes complete ionization.
NaOH (aq) ↔ Na + (aq) + OH - (aq)
0.1M 0.1M
and
(CH 3 ) 2 NH + H 2 O ↔ (CH 3 ) 2 NH + 2 + O- H
0.02-x x x
Then (CH 3 ) 2 NH + 2 = x
[OH - ] = x + 0.1 ; 0.1
⇒ Kb = [(CH 3 )2 NH + 2 ] [OH - ] / [CH 3 )2 NH]
5.4x10-4 = x x 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.

Answer 2

"Let's calculate the “degree of ionization”.

Formula:$\quad \alpha \quad =\quad \sqrt {\frac {{k}_{b}}{c}}$

From the given

Ionization constant = ${ k }_{ b } = 5.4\quad \times \quad { 10 }^{ -4 }$

c = 0.02 M

Substitute the values

$=\quad \sqrt { \frac { 5.4\quad \times\quad { 10 }^{ -4 } }{ 0.02 } \quad =\quad 0.1643 }$

If 0.1 M of NaOGH is added to the solution, then NaOH undergoes complete ionization.

The ionization reaction is as follows.

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

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

${ (CH }_{ 3 }{ ) }_{ 2 }NH\quad +\quad { H }_{ 2 }O\quad \leftrightarrow \quad { (CH }_{ 3 }{ ) }_{ 2 }N{ H }^{ + }\quad +\quad { OH }^{ - }\\ (0.02\quad -\quad x)\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad x\quad \quad \quad \quad \quad \quad x\\ 0.02\quad M\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 0.1\quad M$

Then,

$[{ (CH }_{ 3 }{ ) }_{ 2 }N{ H }_{ 2 }^{ + }]\quad =\quad x$

$[{ OH }^{ - }]\quad =\quad x\quad +\quad 0.1$

${ K }_{ b }\quad =\quad \frac { [{ (CH }_{ 3 }{ ) }_{ 2 }N{ H }_{ 2 }^{ + }][{ OH }^{ - }] }{ [{ (CH }_{ 3 }{ ) }_{ 2 }NH] } \\ 5.4\quad \times \quad { 10 }^{ -4 }\quad =\quad \frac { x\quad \times \quad 0.1 }{ 0.02 }$

$x\quad =\quad 0.0054$

It indicates, In the presence of 0.1 M NaOH, 0.54% of dimethylamine will get dissociated."