Question

partb a)A 50.0-mL volume of 0.15 M HBr is titrated with 0.25 M KOH. Calculate the pH after the addition of 13.0 mL of KOH. Express your answer numerically. Part b) A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is titrated with 0.500 M HNO3. Calculate the pH after the addition of 15.0 mL of HNO3. Express your answer numerically.Part c) A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 33.0 mL of NaOH. Express your answer numerically.

Answer 1

a.

   HBr                                                                                          KOH

MA = 0.15M                                                                             MB = 0.25M

VA   = 50ml                                                                              VB   = 13ml

           M     =   MAVA -MBVB/VA+VB

                   = 0.15*50-0.25*13/50+13    = 4.25/63 = 0.0674M

   M            = [H+]

    PH = -log[H+]

          = -log0.067 = 1.1739

B.

no of moles of NH3 = molarity * volume in L

                                = 0.2*0.075 = 0.015moles

no of moles of HNO3 = molarity * volume in L

                                   = 0.5*0.015 = 0.0075 moles

             NH3    + HNO3 --------------> NH4NO3

I            0.015         0.0075                       0

C          -0.0075      -0.0075                     0.0075

E          0.0075         0                              0.0075

   POH = Pkb + log[NH4NO3]/[NH3]

          = 4.75 + log0.0075/0.0075

          = 4.75

PH    = 14-POH

        = 14-4.75 = 9.25