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Electrolysis and Electrolytes

The pH of a...

Question

The $p H$ 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

$H A + e^{-} \to A^{-}$
$\frac{1.25 \times 241.24 \times 60}{96500} = 0.188$ moles of electrons.
Thus out of 1 mole of $A^{-}$ , 0.188 moles will convert to HA
Hence, $[A -] = 1 - 0.188 = 0.812 M$ and $[H A] = 1 + 0.188 = 1.188 M$
$p H = p K_{a} + l o g \frac{[A^{-}]}{[H A]} = 4.74 + l o g \frac{0.812}{1.188} = 4.74 - 0.165 = 4.57$
Hence, the $p H$ of the anode compartment will be 4.57.
$(4.57 \approx 5)$

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