Page 85 - PC 101 practical notes 24-25..
P. 85

MANSOURA NATIONAL UNIVERSIY
         PHARM D- CLINICAL PHARMACY                        LEVEL I                        PHARM. ANAL. CHEM. I (PC 101)



          C Ca al lc cu ul la at ti io on ns s    : :

                                                                                                   st
           In this exp., the equivalence factor (F) can be calculated on 0.05 N  AgNO3  (1  standard
                                                nd
          "25 ml") OR on 0.01 N NH4SCN (2  standard "titrant") but it is preferred to be calculated
                  st
          on the 1  standard.
          If (F) is calculated on the 1  standard (0.05 N AgNO3):
                                                 st
          Equivalence factor (F):

                                                    1 AgNO 3 ≡ 1 KBr
                                                  standard      sample

                                                   1 x M.W. of KBr x 0.05
              each ml of 0.05 N AgNO3 ≡                                             ≡ ..........  g KBr
                                                            1 x 1000


          Calculation of concentration:


                                0.05 N     0.01 N            0.05 N

                                 [25  – E.P. (      . 0  01 )]  x  F  x  1000
            Concn. =                                . 0  05                             = ..........  g/L
                                                      10






          If (F) is calculated on the 2 standard (0.01 N NH4SCN):
                                                 nd
          Equivalence factor (F):

                                                  1 NH 4SCN  ≡ 1 KBr
                                                  standard        sample

                                                    1 x M.W. of KBr x 0.01
             each ml of 0.01 N NH 4SCN ≡                                             ≡ ..........  g KBr
                                                             1 x 1000

          Calculation of concentration:

                                0.05 N          0.01 N       0.01 N


                                 [25 (    . 0  05 ) – E.P.]   x   F   x 1000
            Concn. =                      . 0  01                                       = ..........  g/L
                                                      10








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