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Class 11
Chemistry
Equilibrium
the ionization constant of dimethylamine is 5 4 x...

Question:
The ionization constant of dimethylamine is 5.4 x 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?

Answer:

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

(CH₃)₂ NH + H₂O ↔ (CH₃)₂ NH⁺₂ + O^-H

0.02-x x x

Then (CH₃)₂ NH⁺₂= x

[OH^-] = x + 0.1 ; 0.1

⇒ K_b = [(CH₃)₂ NH⁺₂ ] [OH^-] / [CH₃)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.

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Question:
The ionization constant of dimethylamine is 5.4 x 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?

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Question: The ionization constant of dimethylamine is 5.4 x 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?

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Question:
The ionization constant of dimethylamine is 5.4 x 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?