Correct Answer - C
40 Ml of 0.1 M HCl `=20 xx 0.1` milli mole `=2` milli mole
`NH_(4)OH+HCl rarr NH_(4)Cl +H_(2)O`
2 milli mole of HCl will neutralize 2 milli moles of `NH_(4)OH` to form 2 milli moles of `NH_(4)Cl`.
`NH_(4)OH ` left `=2 ` milli moles
Total volume `=60 mL`
`:. [NH_(4)OH]=2//60M,[NH_(4)Cl]=2//60M`
` pOH =pK_(b)+log.([NH_(4)Cl])/([NH_(4)OH])`
`=4.74 +log. (2//60)/(2//60)=4.74`
`:. pH =14-4.74 =9.26`