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Question

The pH of an anode of a cell of each compartment after a passage of 1.25 A current for 241.24 minutes.(Give answer as greatest integer)

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Solution

HA+eA
1.25×241.24×6096500=0.188 moles of electrons.
Thus out of 1 mole of A, 0.188 moles will convert to HA
Hence, [A]=10.188=0.812M and [HA]=1+0.188=1.188M
pH=pKa+log[A][HA]=4.74+log0.8121.188=4.740.165=4.57
Hence, the pH of the anode compartment will be 4.57.
(4.575)

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