Correct Answer - C
40 ml of 0.1 M `NH_(3)` solution `= 40xx0.1` millimole
= 4 millimoles
`NH_(4)OH+HCl rarr NH_(4)Cl + H_(2)O`
2 millimole of HCl will neutralize 2 millimoles of `NH_(4)OH` to form 2 millimoles of `NH_(4)Cl`.
`NH_(4)OH` left = 60 ml
`:. [NH_(4)OH]=(2)/(60)M, [NH_(4)Cl]=(2)/(60)M`
`pOH=pK_(b)+log .([NH_(4)Cl])/([NH_(4)OH])`
`=4.74 + log .(2//60)/(2//60)=4.74`
`:. pH=14-4.4.74=9.26`