Question

Calculate the pH value of a solution obtained by mixing $50mL$ of $0.2N$ $HCl$ with $50mL$ of $0.1N$ $NaOH$.

Answer 1

Number of milli-equivalents of the acid $=50\times 0.2=10$
Number of milli-equivalents of the base $=50\times 0.1=5$
Number of milli-equivalents of the acid left after the addition of base $=(10-5)=5$
Total volume of the solution $=50+50=100mL$
thus, $5$ milli-equivalents of the acid are present in $100mL$ of solution
or $50$ milli-equivalents of the acid are present in one litre of solution
or $0.05$ equivalents of the acid are present in one litre of solution.
The acid is monobasic and completely ionised in solution
$0.05N$ $HCl=0.05M$ $HCl$
So, $\left[H^{+}\right]=0.05M$
$pH=-\log \left[H^{+}\right]=-\log 5\times {10}^{-2}=-[\log 5.0+\log {10}^{-2}]=1.3$