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?

Given:

Kb = 5.4 × 10–4.


c=0.02M


To calculate the degree of ionization, we apply the formula


α=


Where Kb is the ionization constant


c is the concentration


∴ α =√


⇒α =0.1643


Thus, the degree of ionization is 0.1643


Ionization of 0.1 M NaOH;



Since, (CH3)2 NH+2]=x


[OH– ]=x+0.1 ≈ 0.1 M


Using the formula,


Kb =


⇒Kb =


As Kb = 5.4 × 10–4.(given)


[(CH3)2 NH] = 0.02 M (given)


∴ 5.4 × 10–4 =


⇒x = 0.054


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


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