The correct options are
A The concentration of H+ ions is 8×10−11 M
C This forms a basic buffer
CH3NH2+HCl→CH3NH+3Cl−
Initial 0.1 0.08
moles
Final (0.1-0.08) 0 0.08
All of the HCl gets neutralised since there is a larger quantity of the base present. The low Kb value of the base does not affect this.
Since weak base CH3NH2 is left along with its salt of strong base, it forms a basic buffer solution so using Henderson Hasselbalch equation:
pOH=pKb+log[salt][base]
or
[OH−]=Kb[Base][Salt]=5×10−4×0.020.08=1.25×10−4 M
and
[H+]=Kw[OH−]=10−141.25×10−4=8×10−11 M
so, pH=−log(8×10−11)
=−log(23)+11=−3×0.3+11
pH=10.1
pOH=14−10.1=3.9