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

The cell is,
`Pt H_(2(1atm)) |HA_(2)||HA_(I) |H_(2(1atm)) Pt`
At `L.H.S.: E_(H//H^(+)) = E_(OP_(H//H^(+)))^(@) +(0.059)/(1) log_(10) [H^(+)]_(2)`
`:' -log H^(+) =pH :. E_(H//H^(+)) = E_(OP_(H//H^(+)))^(@) -0.059(pH)_(2)`
At `R.H.S. : E_(H^(+)//H) =E_(RP_(H^(+)//H))^(@) +(0.059)/(1)log[H^(+)]_(1)`
`:. E_(H^(+)//H) =E_(RP_(H^(+)//H))^(@) -0.059(pH)_(1)`
For Acid `HA_(1) HA_(1) hArr H^(+) +A_(1)^(-)`
`[H^(+)] = C. alpha = sqrt(K_(a)/C)`
`:. (pH)_(1) = (1)/(2) pK_(a_(1)) -(1)/(2) log_(10)C`
Similarly, `(pH)_(2) = (1)/(2) pK_(a_(1)) -(1)/(2) log_(10)C`
`( :' C` are same)
`E_(cell) = E_(OP_(H^(+)//H))^(@) +E_(RP_(H^(+)//H))^(@)`
For II for I
`= 0.059 [(1)/(2) pK_(a_(2)) -(1)/(2) pK_(a_(1))] =(0.059)/(2) [5-3]`
`=+0.059`