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`.

Explanation

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)`