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Q.
The $ pH $ of the solution produced by mixing equal volume of $ 2 .0 \times 10^{-3}\, M \,HClO_4 $ and $ 1.0 \times 10^{-2}\,M\,KClO_4 $ is
AMUAMU 2010Equilibrium
Solution:
$KClO_{4}$ being a neutral salt does not affect the nature of solution
$[HClO_{4}]=\frac{2.0\times10^{-3}}{2}$
$=1.0\times 10^{-3}=[H^{+}]$
$PH=-log[H^{+}]=-log\, 1.0\times 10^{-3}$
$=3$