PH = 8.58
POH = 14-PH
       = 14-8.58 = 5.42
PKb = -logKb
       =
-log1.8*10-5
      = 4.75
n0 0f moles of NH3 = molarity * volume in L
                            Â
= 0.4*2.1 = 0.84 moles
POH = PKb + log [salt]/[base]
5.42 = 4.75 + log[NH4Cl]/0.84
5.42-4.75Â Â = log[NH4Cl]/0.84
0.67Â Â Â Â Â Â Â Â Â Â
= log[NH4Cl]/0.84
[NH4Cl]/0.84 = 100.67Â Â = 4. 678
[NH4Cl]Â Â Â = 4.678*0.84 = 3.9295 moles
mass of NH4Cl = no of moles * gram molar mass
                     Â
= 3.9295*53.5 = 210.2gm >>>> answer
