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Mansoura National University
            Pharm D-Clinical Pharmacy Program                 Level 1             Pharm. Anal. Chem. 1 (PC 101)


                                                          Example

             How many grams of silver nitrate AgNO3 are needed to prepare 500 mL 0.3 M AgNO
                                                                                                                3
             solution? (Formula mass of AgNO  = 170 g)
                                                       3




                •  No of liters (V) =  mL of solution/1000                             V= 500 mL
                                  = 500/1000 = 0.5 liter                               Molarity= 0.3 M
                •  Molarity = No of moles of solute/No of liters of solution           Molecular wt. = 170 g
                    0.3   = No of moles /0.5                                           Weight (g)??

                •  No of moles   =0.3 X 0.5= 0.15 mol
                •  No of moles =    Weight  of solute (g) / Molecular wt. of solute

                    0.15= Weight (g) / 170
                                                  Weight = 0.15 X 170 = 25.5 g


                                           Dilution and Molarity


                   ▪  The molarity of a solution and its volume are inversely proportional.

                   ▪  Therefore, adding water makes the
                       solution less concentrated.
                   ▪  This  inverse  relationship  takes  the

                       form of:

                                      ×     =     ×    
                                                           
                                           
                                                   
                                    
                   ▪  So, as water is added, increasing the
                       final volume, V , the final molarity,
                                         f
                       M , decreases.
                         f

         Example : What is the volume of 6.0 M HCl that can be made from 5.0 mL of 12.0 M HCl?


                                                    ×     =     ×                                  M = 12.0 M
                                                             
                                                                            
                                                                     
                                                     
                                                                                                     i
                                                  12.0 ×  5.0 = 6.0 ×  V                           V  = 5.0 mL

                                                                          f                         i
                                                   V   =       ×     = 10 mL                       M  = 6.0 M
                                                     f                                               f
                                                             12                                      V  ???
                                                                                                       f
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