Equinormal solution of two weak acids, `HA(pK_(a) =3)` and `HB(pK_(a) =5)` are each placed in contact with standard hydrogen electrode at `25^(@)C`. When a cell is constructed by interconnecting them thorugh a salt bridge find the e.m.f. of the cell.

Answer

`E=0.059V`

Explanation

`Pt//H_(2)^(@)//H^(+)(HA)//H^(+)(HB)//H_(2)//Pt`
`H_(2)to2H^(+)+2e^(-)`
`2H^(+)+2e^(-)toH_(2)`
`2H_(HB)^(+)to2H_(HA)^(+)" "E^(@)=0`
`E=0-(0.0591)/(2)log(([H^(+)]_(HA)^(2))/([H^(+)]_(HB)^(2)))`
But `Ka=([H^(+)]^(2))/(c)`
`=-(0.0591)/(2)log((10^(-3)xxC)/(10^(-5)xxC))=-0.0591`
the cell is constructed in reversed direction.
`E_(cell)=0.0591"volt"`