Question

Calculate the volume (in mL) of 0.170 M NaOH that must be added to 259 mL of 0.0419 M 3-(N-Morpholino)propanesulfonic acid (MOPS) to give the solution a pH of 7.55. The pKa of MOPS = 7.18.

Answer 1

let no of moles of NaOH = x

no of moles of MOPS   = molarity * volume in L

                                     = 0.0419*0.259   = 0.01085moles

                       MOPs + NaOH ------------------> NaMOps + H2O

        I            0.0108    x                                           0

         C         -x          -x                                         +x

      E        0.0108 -x    0                                          +x

             PH   = Pka + log[NaMOPs]/[MOPs]

             7.55     = 7.18 + logx/0.0108-x

            7.55-7.18    = logx/0.0108-x

         logx/0.0108-x   = 0.37

            x/0.0108 -x    = 10^0.37

          x/0.0108-x     = 2.3442

           x                   = 2.3442*(0.0108-x)

                  x   = 0.0076

no of moles of NaOH = x   = 0.0076moles

no of moles of NaOH   = molarity * volume in L

   0.0076                        = 0.17*volume in L

volume in L                    = 0.0076/0.17    = 0.0447L

volume in ml                   = 44.7ml

volume of NaOH   = 44.7ml >>>>answer