Question

Calculate the change in pH when 5.00 mL of 0.100 M HCl(aq) is added to 100.0 mL of a buffer solution that is 0.100 M in NH3(aq) and 0.100 M in NH4Cl(aq). Consult the table of ionization constants as needed.

Calculate the change in pH when 5.00 mL of 0.100 M NaOH is added to the original buffer solution.

Answer 1

no of moles of NH3 = molarity * volume in L

                               = 0.1*0.1 = 0.01 moles

no of moles of NH4Cl = molarity * volume in L

                                    = 0.1*0.1 = 0.01 moles

PKb of NH3 = 4.75

POH   = PKb + log[NH4Cl]/[NH3]

           = 4.75 + log0.01/0.01

           = 4.75+ log1

           = 4.75 + 0

           = 4.75

PH   = 14-POH

           = 14-4.75

           = 9.25

no of moles of HCl = molarity *volume in L

                               = 0.1*0.005   = 0.0005 moles

no of moles of NH3 after addition of 0.0005 moles of HCl = 0.01-0.0005   = 0.0095 moles

no of moles of NH4Cl after addition of 0.0005 moles of HCl = 0.01+0.0005   = 0.0105moles

POH   = PKb + log[NH4Cl]/[NH3]

           = 4.75 + log0.0105/0.0095

           = 4.75+ 0.04346

           = 4.79346

PH   = 14-POH

         = 14-4.79346

          = 9.20654

change in PH = 9.20654-9.25 = -0.04346>>>>answer

no of moles of NaOh = molarity *volume in L

                               = 0.1*0.005   = 0.0005 moles

no of moles of NH3 after addition of 0.0005 moles of NaOH = 0.01+0.0005   = 0.0105 moles

no of moles of NH4Cl after addition of 0.0005 moles of NaOH = 0.01-0.0005   = 0.0095moles

POH   = PKb + log[NH4Cl]/[NH3]

           = 4.75 + log0.0095/0.0105

           = 4.75-0.04346

           = 4.70654

PH   = 14-POH

         = 14-4.70654

          = 9.29346

change in pH = 9.29346-9.25 = 0.04346 >>>>answer