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To measure the quantity of `MnCl_(2)` dissolved in an queous solution, it was completely converted to `KMnO_(4)` using the reaction `MnCl_(2)+K_(2)S_(2)O_(8)+H_(2)O to KMnO_(4)+K_(2)SO_(4)+HCl`(equation not balanced). Few drops of concentrated HCl were added to this solution and gently warmed. Further , oxalic acid (225 mg) was added in portions till the colour of the permanganate ion disappeared. Calculate the quantity of `MnCl_(2)` (in mg) presence in the initial solution. ( Atomic weights in g `mol^(-1)`: Mn=55,Cl=35.5) |
Answer» The redox equations involved are given : `2MnCl_(2)+5K_(2)S_(2)O_(8)+8H_(2)O ot 2KMnO_(4)+4K_(2)SO_(4)+6H_(2)SO_(4)+4HCl` ……(i) `2KMnO_(4)+5H_(2)C_(2)O_(4)+3H_(2)SO_(4) to K_(2)SO_(4)+2MnSO_(4)+8H_(2)O+10CO_(2)`……(ii) Mass of oxalic acid added =225 mg Millimoles of oxalic acid added `=(225)/(90)=2.5` Millimoles of `KMnO_(4)` use to react with oxalic acid `=(5)/(2)xx(2)/(5)=1` Millimoles of `MnCl_(2)` present in the initial solution =1 Mass of `MnCl_(2)` presen in the initial solution =55+71=126 mg |
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