Question

Dimethylamine ((CH₃ )₂NH) is a weak base with K_b = 5.1 × 10⁻⁴. What is the pH of a 0.60 M solution of dimethylamine and what is the percent dissociation?

Answer 1

Given that

Kb = 5.1 x 10^-4

Concentration of ((CH3 )2NH) , C = 0.6 M

[OH-]

[OH^-] = (Kb.C)^1/2

   = [(5.1x10^-4) (0.6)]^1/2

   = 0.0175 M

[OH^-]= 0.0175 M

pOH

pOH = -log [OH^-]

= -log (0.0175)

= 1.75

pOH = 1.75

pH

pH = 14- pOH

= 14 - 1.75

= 12.25

pH = 12.25

Calculation of the percent dissociation :

We know that [OH-] = c.α where α = percent of ionization

  α = [OH-]/c

= 0.0175 M / 0.6 M

= 0.029

= 2.9 %

  α = 2.9 %

Therefore,   percent dissociation = 2.9 %