Page 27 - Phytochemistry -1 (PG404) / Clinical Pharmacy 2nd level students ( 2019 )
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Clinical pharmacy PharmD program             Third level                        Phytochemistry-1 (PG-504)


                   • 10 ml Fehling's solution contains 0.11 gm cupric oxide which is able to

                   oxidize 0.05 gm of glucose.

                   Linge's Indicator


                   • It is an external indicator and could replace M.B.

                   • It consists of:


                       •  1 gm ferrous ammonium sulfate Fe (NH 4) 2(SO 4) 2
                       •  1.5 gm ammonium sulfocyanide

                       •  dissolved in a mix of 10 ml water + 2.5 ml conc HCI
                       •  The solution is decolorized before use by adding few pieces of zinc

                          granules.

                   • When a drop of cupric salt is brought in contact with this indicator in a

                   porcelain lab, oxidation of ferrous salt occurs with immediate production
                   of red color of ferric thiocyanate.

                   • For the use of Ling's indicator in the process of titration apply the

                   followings:

                   When the blue color of the solution has disappeared and precipitation of

                   Cu 2O is complete, a drop of sol. is brought in contact with indicator, E.P
                   is reached. When color of mixture ceases to give red color.

                                                     Calculations


                   The sugar solution taken (in ml) which reduces 10 ml Fehling's solution
                   contain 0.05 gm of any of the reducing hexoses (e.g glucose, fructose, ….

                   etc.). In case of reducing disaccharides (e.g maltose or lactose) each 10
                   ml of Fehling's solution is equivalent to 0.08 gm of maltose or lactose.


                   (i.e.):

                   A. For monosaccharide


                                                                           0.05 x 100

                    % of glucose =


                                                          ml of sugar solution taken (E.P) - 0.2
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