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A saturated calomel electrode is coupled through a salt bridge with a quinhydrone electrode dipping in 0.1 M NH_(4)Cl. The observed EMF at 25^(@)C is 0.152V. Find the dissociated constant of NH_(4)OH. The oxidation potential of saturated calomel electrode=-0.699V at 25^(@)C. |
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Answer» Solution :For the cell`:` Saturated CALOMEL `||0.1 MNH_(4)Cl|H_(2)Q,Q|Pt` `E_(cell)=E_(qui nhydron e el etrode)-E_(saturated calomel)` `E_(cell)=0.152,,,,[{:(E_(saturated calomel(o x i d)),=,-0.242V),(E_(saturated calomel (red)),=,-0.242V):}]` `(E_(qui nhydron e)=0.6994-0.059pH)` `( :. E^(C-)._(o x i de)=-0.6994,so E^(c-)._(Red)=0.6994)` `0.152=0.6994-0.059pH-0.242` `:. PH=(0.457-0.152)/(0.059)=5.17` For `0.1 M NH_(4)Cl,` salt of `S_(A)//W_(B)` which on hydrolysis GIVES `NH_(4)OH` `pH=(1)/(2)(pK_(W)=pK_(b)-logc)` `5.17=(1)/(2)(14-pK_(b)-log0.1) Solving , we get `K_(b)=2.18xx10^(-5)` |
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