Correct Answer - D
`pK_(a)(NH_(4)^(+))=9.26`
`pK_(b)(NH_(3))=14-9.26=4.74`
`[NH_(4)]=2xx[(NH_(4))_(2)SO_(4)]`
`=2xx0.01=0.02 M`
`pOH =pK_(b)+log . ([NH_(4)^(+)])/([NH_(4)OH])`
`pOH =4.74 +log . (0.02)/(0.02)=4.74-log 1`
`pH =14-pOH =1404.74+log 1`
`=9.26+log 1`