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Q.
The pH of a solution obtained by mixing $50\: mL$ of $HCl$ and $430 \: mL$ of $IN\: NaOH$ is $[ \log \: 2.5 = 0.3979]$
Equilibrium
Solution:
In $HCl=1\,M\,HCl. 1\,N\,NaOH=1M\,NaOH$
Moles of $HCl=M\times V=50\times 10^{-3} \times 1$
$=5 \times 10^{-2}$
Moles of $NaOH = M \times V =30 \times 10^{-3} \times 1$
$=3 \times 10^{-2}$
As number of moles of $HCl$ are more than $NaOH$ the resulting solution $(30+ 50 = 80\, mL)$ will contain excess of $HCl$
$=5 \times 10^{-2}-3 \times 10^{-2}=2\times 10^{-2}\,mol$
$\therefore H^{+}$ (after mixing) $=\frac{2\times10^{-2}mol}{80\times10^{-3}L}=\frac{1}{4} M $
$PH =-log\,H^{+}=-log \frac{1}{4}=log\,4$
$=0.6021$