Saved Bookmarks
| 1. |
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» `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` |
|