Question

What is the pH after the addition of 20.0 mL of 0.10 M NaOH to 80.0 mL of a buffer solution that is 0.25 M in NH3 and 0.25 M in NH4Cl, (K b =1.8 x 10 −5 for NH3 )?

Answer 1

no of moles of NH3 = molarity * volume in L

                                 = 0.25*0.08 = 0.02 moles

no of moles of NH4Cl = molarity * volume in L

                                    = 0.25*0.08 = 0.02 moles

no of moles of NaOH = molarity * volume in L

                                  = 0.1*0.02 = 0.002 moles

By the addition of 0.002 moles of NaOH

no of moles of NH3 by the addition of NaOH = 0.02+0.002 = 0.022 moles

no of moles of NH4Cl by the addition of NaOH = 0.02-0.002 = 0.018 moles

PKb = -logKb

         = -log1.8*10^-5 = 4.75

POH = PKb + log[NH4Cl]/[NH3]

           = 4.75 + log0.018/0.022

            = 4.75-0.08715 = 4.6628

PH = 14-POH

       = 14-4.6628   = 9.3372 >>>>answer