What is the pH of a 0.19M CH3NH3+Cl- aqueous solution

pKb(CH3NH2) = 3.44
Answer to 2 decimal places.

if the pKb = 3.44
then the
Kb = 3.63 X 10^-4

CH3NH2's conjugate acid:
CH3NH3+ --> (CH3NH2) & H+
0.19 --> X & X

Ka = Kwater / Kb = (1 X 10^-14) / (3.63 X 10^-4) = 2.75 e-11

Ka = [CH3NH2] [H+] / [CH3NH3+]

2.75 e-11 = [X] [X] / [0.19]

X2 = 5.225 e-12

X = [H+] = 2.29 e-6

pH = 5.64