Question

What will be the pH of a solution obtained by mixing $800mL$ of $0.05N$ sodium hydroxide and $200mL$ of $0.1N$ $HCl$, assuming the complete ionisation of the acid and the base?

Answer 1

Number of milli-equivalents of $NaOH=800\times 0.05=40$
Number of milli-equivalents of $HCl=200\times 0.1=20$
Number of milli-equivalents of $NaOH$ left after the addition of $HCl$ $=(40-20)=20$
Total volume $=(200+800)mL=1000mL=1litre$
$20$ milli-equivalents or $0.02$ equivalents of $NaOH$ are present in one litre
$0.02N$ $NaOH=0.02M$ $NaOH$ (Mono-acidic) and the base is completely ionised
So, $\left[{OH}^{-}\right]=0.02M$
or $\left[{OH}^{-}\right]=2\times {10}^{-2}M$
$pOH=-\log 2\times {10}^{-2}=1.7$
We know, $pH+pOH=14$
So, $pH=(14-1.7)=12.3$