A 1000mL buffer solution is 0180 molL1 in HClOand 0135 molL1

A 100.0-mL buffer solution is 0.180 molL1 in HClOand 0.135 molL1 in NaClO.
Ka(HClO)=4.0×108

Part A

What is the initial pH of this solution?

Express your answer using two decimal places.

Part B

What is the pH after addition of 115.0 mg of HBr?

Express your answer using two decimal places.

Part C

What is the pH after addition of 85.0 mg of NaOH?

Express your answer using two decimal places.

Solution

no of moles of HClO   = molarity * volume in L

                                  = 0.18*0.1 = 0.018 moles

no of moles of NaClO   = molarity * volume in L

                                      = 0.135*0.1 = 0.0135 moles

Pka = -logKa

         = -log4*10^-8

        = 7.3979

PH     = PKa + log[NaClO]/[HClO]

       = 7.3979 + log0.0135/0.018

      = 7.3979-0.1249   = 7.273

part-B

no of moles of HBr = W/G.M.Wt

                              = 0.115/81 = 0.00142moles

no of moles of HClO after addition of 0.00142 moles of HBr = 0.018 +0.00142   = 0.01942moles

no of moles of NaClO after addition 0.00142 moles of HBr    = 0.0135-0.00142   = 0.01208 moles

PH     = PKa + log[NaClO]/[HClO]

          = 7.3979+ log0.01208/0.01942

           = 7.3979-0.2061

           = 7.1918

part-C

no of moles of NaOH = W/G.M.Wt

                                    = 0.085/40 = 0.002125 moles

no of moles of HClO after addition of 0.002125 moles of NaOH = 0.018 -0.002125 = 0.015875moles

no of moles of NaClO after addition 0.002125 moles of NaOH = 0.0135+0.002125 = 0.015625 moles

PH     = PKa + log[NaClO]/[HClO]

          = 7.3979+ log0.015875/0.015625

       = 7.3979 + 0.006894   = 7.4048