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.
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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.
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