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?

Kb=5.4×10-4c=0.02 MThen, α =Kbc=5.4×10-40.02=0.1643

Now, if 0.1 M of NaOH is added to the solution, then NaOH (being a strong base) undergoes complete ionization.

NaOHaqNaaq++OHaq-                     0.1 M      0.1 M

And,

CH32NH+H2O0.02-xCH32NH2+x+OHx;0.02 MThen, CH32NH2+=xOH-=x+0.1;0.1Kb=CH32NH2+OH-CH32NH5.4×10-4=x×0.10.02x=0.0054

 

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