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Q.
The pH of a solution obtained by mixing equal volumes of $\frac{N}{10}NaOH$ and $ \frac{N}{20}HCl$
Equilibrium
Solution:
Let the volume of $NaOH$ used $= V = $ volume of $HCl$ used.
Concentration of mixture $[OH^-] $
$ \frac{0.1 V -0.05 V}{(V+V)}$
$[\because $ Milliequivalent of $NaOH$ is in excess.$]$
$ \frac{V(0.1 +0.05)}{2V}$
$ =\frac{0.05}{2}=0.025$
$pOH=-\log[OH^-]=-\log(0.025)$
$=1.602$
$\therefore pH=14-1.602 =12.4$