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Equinormal solutions of two weak acids , HA ( pKa=3) and HB ( pKa=5) are each placed in contact with standard hydrogen electrode at 298 K. When a cell is constructed by interconnecting them through a salt bridge find the emf of the cell.

Equinormal solutions of two weak acids , HA ( pKa=3) and HB ( pKa=5) are each placed in contact with standard hydrogen electrode at 298 K. When a cell is constructed by interconnecting them through a salt bridge find the emf of the cell.

Grade:12

1 Answers

Sunil Kumar FP
askIITians Faculty 183 Points
7 years ago
we have pKa =3,Ka= 10^-3 and pKb=5Kb=10^-5
Take HA as the right hand electrode and HB as the left hand electrode
The standard electrode potential for each acid is 0
Now we have.
RT/(2F) * ln ([H+][A]2 / [H+][B]2).
emf=.0596 V

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