Question

Consider the titration of 60.0 mL of 0.0400 M C6H5NH2 (a weak base; Kb = 4.30e-10) with 0.100 M HCl. Calculate the pH after the following volumes of titrant have been added: a) 45.6 mL

Answer 1

no of moles of C6H5NH2 = molarity * volume in L

                                         = 0.04*0.06 = 0.0024moles

no of moles of HCl             = molarity * volume in L

                                           = 0.1*0.0456 = 0.00456moles

         C6H5NH2 + HCl ----------------> C6H5NH3Cl

I       0.0024          0.00456                     0

excess no of moles of HCl = 0.00456-0.0024 = 0.00216moles

[H+]     = no of moles of HCl/total volume in L

               = 0.00216/0.1056 = 0.02045 M

PH   = -log[H+]

        = -log0.02045 = 1.6893