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  • 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?

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?

Answers (1)

best_answer

We have,

(Degree of ionization) K_b = 5.4\times 10^{-4}
Concentration of dimethylamine = 0.02 M

\therefore \alpha = \sqrt{\frac{K_b}{C}}= \sqrt{\frac{5.4\times 10^{-4}}{0.02}}
     \alpha = 0.1643

If we add 0.1 M of sodium hydroxide. It is a strong base so, it goes complete ionization

                     NaOH\rightleftharpoons Na^++OH^-
                                                (0.1 M)      (0.1 M)

and also,

(CH_{3})_2NH+H_{2}O\rightleftharpoons (CH_{3})_2NH^+_2+OH^-
    0.02- x                                           x                       x   

[OH^-]=x+0.1\approx 0.1 (since the dissociation is very small)

Therefore, 

K_b = \frac{x(0.1)}{0.02}
x=5.4\times 10^{-4}(\frac{0.02}{0.1})
     =0.0054

Hence in the presence of 0.1 M of sodium hydroxide , 0.54% of dimethylamine get dissociated.

Posted by

manish

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