Preparation of 0.1 Molar and 1 Molar Hydrochloric Acid solution

 



Preparation of 0.1 Molar and 1 Molar Hydrochloric Acid solution

Preparation of 0.1 Molar Hydrochloric Acid solution

¨  For the preparation of 0.1 M Hydrochloric Acid (HCl), take 8.3ml of HCl and dilute it with carbon-dioxide free water up to 1000 ml.

¨  Molecular Weight of HCl = 1+35.5= 36.5

¨  Mass in gram= Molarity x molecular weight x required volume

                                                             1000

¨  Mass in gram= 0.1 x 36.5x 1000  = 3.65 gram

                                                    1000

¨  Now convert mass into volume by using following formula

¨  Density = Mass  

                              Volume

              Re arrange the formula, Volume = Mass

                                                                          density

               Density of HCl = 1.19

               Volume = 3.65/ 1.19 = 3.06 ml

¨  The above said volume is for 100% pure HCl, however the available HCl is 37% pure, hence the volume required will become

¨   3.06 x 100  = 8.3 ml

                 37

¨  Hence to prepare 1 liter, 0.1 molar solution of HCl we will take 8.3 ml of HCl in carbon-dioxide free 1 liter water.

Preparation of 1M Hydrochloric Acid Solution

¨  For the preparation of 1 M Hydrochloric Acid (HCl), take 83ml of HCl and dilute it with carbon-dioxide free water up to 1000 ml.

¨  Molecular Weight of HCl = 1+35.5= 36.5

¨  Mass in gram= Molarity x molecular weight x required volume

                                                             1000

¨  Mass in gram= 1 x 36.5x 1000  = 36.5 gram

                                                 1000

¨  Now convert mass into volume by using following formula

¨  Density = Mass  

                               Volume

             Re arrange the formula, Volume = Mass

                                                                         density

             Density of HCl = 1.19

             Volume = 36.5/ 1.19 = 30.67 ml

¨  The above said volume is for 100% pure HCl, however the available HCl is 37% pure, hence the volume required will become

¨   30.67 x 100  = 82.89 (83) ml

                 37

¨  Hence to prepare 1 liter, 1 molar solution of HCl we will take 83 ml of HCl in carbon-dioxide free 1 liter water.

 Presented by

Muhammad Atif

PhD Scholar Pharmaceutical Sciences

Abdul Wali Khan University Mardan, Pakistan

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