Question

A solution of weak acid $HA$ was titrated with base $NaOH$. The equivalent point was reached when $40mL$ of $0.1MNaOH$ has been added. Now $20mL$ of $0.1MHCl$ were added to titraded solution, the $pH$ was found to be $5.0$. What will be the $pH$ of the solution obtained by mixing $20mL$ of $0.2MNaOH$ and $20mL$ of $0.2MHA$?

Answer 1

$HA+NaOH\to NaA+H_{2}O$
milli-moles of salt $NaAor{A}^{-}=40\times 0.1=4$
Now $A^{-}+H^{+}\to HA$
$Initialmilli-moles42$
$Finalmilli-moles2-$
Acidic buffer solution is formed and $\left[A^{-}\right]=\left[HA\right]$
$pH=p{K}_a+log\frac{\left[A^{-}\right]}{\left[HA\right]}\Rightarrow p{H}_a=5$
Now,
$HA+NaOH\to NaA+H_{2}O$
Hydrolysis of $A^{-}$ will takes place
$\left[NaA\right]=\frac{millimolesofacid}{Totalvolume}=\frac{20\times 0.2}{20+20}=0.1$
$pH=\frac{1}{2}(p{K}_w+p{K}_a+logC)=\frac{1}{2}[14+5-1]=9$