Question

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.

Answer 1

Answer:

0.059V.

Explanation:

Since, from the question we have that for the hydrogen electrode which is at 1 atm pressure and 298K we will have reduction potential Ered as -0.059pH. If we let that the right hand electrode is HA and the left one is the HB.  

So, we know that the for weak acid pH = 1/2 pKa - 1/2 log C.

For the right hand electrode HA the E red will be = -0.059/2 [ pKa - log C]  which on solving we will get - 0.0295 [3 - log C]

Similarly, for the left hand electrode HB the E red will be

-0.059/2[pKa-log C] which on solving we will get -0.0295 [5 - log C].

Therefore, the Ecell will be E red (HA) - E red (HB) which is  - 0.0295 [3 - log C] + 0.0295 [5 - log C] = 0.059 V.