Question

A 100.0 ml buffer solution is 0.20 M HC7H5O2 and 0.15 M NaC7H5O2. Calculate the pH of the solution after the addition of 0.0025 moles of NaOH. Assume the volume of the buffer does not change. (HC7H5O2 = Ka 6.5x10^-5)

Answer 1

no of moles of HC7H5O2   = molarity*volume in L

                                            = 0.2*0.1 = 0.02 moles

no of moles of NaC7H5O2   = molarity*volume in L

                                            = 0.15*0.1 = 0.015 moles

no of moles of HC7H5O2 after addition of 0.0025 moles of NaOH   = 0.02-0.0025   = 0.0175 moles

no of moles of NaC7H5O2 after addition of 0.0025 moles of NaOH = 0.015+0.0025 = 0.0175 moles

PKa   = -logKa

           = -log6.5*10^-5

            = 4.187

PH    = Pka + log[NaC7H5O2]/[HC7H5O2]

          = 4.187 + log0.0175/0.0175

          = 4.187 + log1

          = 4.187 + 0

        = 4.187 >>>>>answer