1.687 g of N-2-acetamido-2-aminoethane-sulfonic acid potassium salt (ACES–K , MW = 220.29 g/mol) is dissolved in...

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1.687 g of N-2-acetamido-2-aminoethane-sulfonic acid potassiumsalt (ACES–K , MW = 220.29 g/mol) is dissolved in 90.76 mL ofwater. 15.18 mL of HCl is added to the solution, resulting in a pHof 7.25. Calculate the concentration of the HCl solution. The pKaof ACES is 6.85.

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4.5 Ratings (762 Votes)

no of moles of ACEs = W/G.M.Wt   = 1.687/220.29 = 0.00765 moles

                         ACES + HCl --------------> HACES

            I           0.00765     x                          0

           C           -x             -x                          +x

          E       0.00765-x      0                         +x

                     PH   = PKa + log[HACES]/[ACES]

                7.25    = 6.85 + logx/0.00765-x

               7.25-6.85    = logx/0.00765-x

                0.4         = logx/0.00765-x

         logx/0.00765-x = 0.4

             x/0.00765-x   = 100.4

           x/0.00765-x   = 2.5118

          x                      = 2.5118*(0.00765-x)

                   x            = 0.01921-2.5118x

                  x+2.5118x   = 0.01921

                  3.5118x      = 0.01921

                          x         = 0.01921/3.5118 = 0.0055moles

                no of moles oc HCl = x = 0.0055moles

                     molarity of HCl   = no of moles/volume in L

                                               = 0.0055/0.01518   = 0.362 M >>>>>> answer

            

             


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