no of moles of NH3 = molarity * volume in L
                              Â
= 0.48*0.108Â Â = 0.05184 moles
no of moles of NH4Cl  = molarity * volume in L
                                  Â
= 0.3*0.143Â Â = 0.0429moles
Pkb of NH3 = 4.75
  POH = Pkb + log[NH4NO3]/[NH3]
          =
4.75 + log0.0429/0.05184
         =
4.75-0.0822Â Â = 4.6678
PHÂ Â = 14-POH
       =
14-4.6678Â Â = 9.3322 >>>>answer