Question

A solution of weak acid $HA$ was titrated with base $NaOH$. The equivalence point was reached when $36.12mL$ of $0.1MNaOH$ was added. Now $18.06mL$ of $0.1MHCl$ were added to titrated solution, the pH was found to be $4.92$. What will be the $pH$ of the solution obtained by mixing $10mL$ of $0.2MNaOH$ and $10mL$ of $0.2MHA$?

• A. $5$
• B. $9$
• C. $10$
• D. $5$

Answer 1

The correct option is B $9$

At equivalence point,
$\text{milliequivalents of}HA=\text{milliequivalents of}NaOH=3.612$

$\begin{array}{ccc}\mathrm{NaA}+& \mathrm{HCl}& {\to }^{}\mathrm{HA}+\mathrm{NaCl}\\ 3.612& 1.806& \\ 1.806& -& 1.806\end{array}$

$pH=p{K}_a+log\frac{\left[Salt\right]}{\left[Acid\right]}$

$4.92=p{K}_a+log\frac{1.806}{1.806}$

$\therefore p{K}_a=4.92$

Now,
$\begin{array}{ccc}\mathrm{NaOH}+& \mathrm{HA}& \stackrel{}{\to }\mathrm{NaA}\\ 2& 2& \\ -& -& 2\end{array}$

$\left[NaA\right]=\frac{2}{20}=0.1$

For a salt of Strong base and weak acid,
$pH=7+\frac{1}{2}p{K}_a+\frac{1}{2}logC=7+\frac{4.92}{2}+\frac{1}{2}log0.1$
$\therefore pH=9$