Class 11 Chemistry - Chapter Equilibrium NCERT Solutions | The ionization constant of dimethylamine

Welcome to the NCERT Solutions for Class 11th Chemistry - Chapter Equilibrium. This page offers a step-by-step solution to the specific question from Exercise 1, Question 54: the ionization constant of dimethylamine is 5 4 x....
Question 54

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

⇒ Kb = [(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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