The ionization constant of dimethylamine is $$5.4 \times 10^{-4}$$ 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 in NaOH?

Asked by Pragya Singh | 11 months ago |  138

##### Solution :-

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

$$c=0.02M\\ \\$$

$$Then,\alpha$$

$$=\sqrt{\dfrac{K_{b}}{c}}\\ \\ =\sqrt{\dfrac{5.4\times 10^{-4}}{0.02}}=0.1643$$

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)

0.1M        0.1M

And,

(CH3)2 NH   +  H2$$\leftrightarrow$$ (CH3)2 NH2+  + OH

(0.02-x),     x,    x

0.02M        0.1M

Then,[ (CH3)2 NH+2] = x

[OH ] = x + 0.1;0.1

$$\Rightarrow K_{b}=\dfrac{[(CH_{3})_{2}NH_{2}^{+}][OH^{-}]}{[(CH_{3})_{2}NH]}\\$$

$$= 5.4\times 10^{-4}$$

$$=\dfrac{x\times 0.1}{0.02}\\ \\$$

$$x=0.0054$$

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

Answered by Pragya Singh | 11 months ago

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