CH3NH2 (aq)+ H2O(l) --> CH3NH3+(aq) + OH- (aq)
Kb= [CH3NH3+] [OH-]/ [CH3NH2]
If X mole of methylamine ionised.
Then
Kb = X × X / .36-X
4 × 10-4= X2/.36 (X<<<<.36)
X2 = 4.4 ×10-4× .36
X = 1.2586×10-2
X= .012586
pOH = - log [OH-]
= - log X= - log .012586
= 1.9
pH = 14-pOH = 14- 1.9
= 12.1
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