My tutor got the answer 5.0 E -4, and I got the answer 1.752 E -4, but they were wrong. Please give me a step-by-step answer. Thank you.

Calculate the [OH] in a solution where 0.050 mole CH3NH3Cl is added to 200.0mL of 0.50M in CH3NH2. CH3NH2+H2OCH3NH3++OHKbCH3NH2=4.38104[OH]=[?]10[7]M Exponent (yellow) Calculate the [OH]in a solution where 0.050 melo CH3NH3, is added to 200.0mL of 0.50M in CH3NH2. CH3NH2+H2OCH3NH3+OHKbCH3NH24.38E4.50.2=0.1MCHCH3NH2CH3NH3Cl=2.5MCH3NH2+H2OCH3NH3+OHRB+H2OBHOH1.1ll.25HB1Ex1xl+x.25+x+xx4.384=(.1x)(25ix)(x)x=1.7524