Question

If the Kb of a weak base is 4.2

Answer 1

first, write your Kb expression:
Kb = [OH-] [BH]/ [B]
we know that the concentration of B is 0.22, and that Kb = 4.2e-6. [OH-] = [BH], so

Kb = x^2/0.22-x

4.2 e-6 = x^2/0.22 (we can eliminate the -x on the bottom because 4.2e-6 is such a small number that subtracting x from 0.22 won't even make a significant difference to our answer)

thus, x = 0.000961, and [OH-] = 0.000961. we can then plug this into our pH and pOH equations:

pOH = -log [OH-], -log[0.000961] = 3.017

pOH + pH = 14,

14 - 3.017 = 10.98

pH = 10.98

Answer 2

pKb = - log( Kb)

so,
pKb = -log( 4.2*10^-6)

pKb = 5.3767

Now
pOH = 1/2 [ pKb - log C]

putting values,

pOH = 1/2*[ 5.3767- log 0.22]

solving it ,

pOH = 3.017

now,
pH = 14- pOH

so,
pH = 10.983