Question

A solution of weak acid $HA$ was titrated with base $NaOH$. The equivalence point was reached when $36.12mL$ of $0.1M$ $NaOH$ has been added. Now $18.06mL$ of $0.1M$ $HCl$ 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 $0.2M$ $NaOH$ and $10mL$ of $0.2M$ $HA$?

• A. $pH=3$
• B. $pH=4$
• C. $pH=9$
• D. $pH=11$

Answer 1

Answer: (C)

$pH=9$

At equivalence point, meq. of $HA=$ meq. of $NaOH=3.612$
$NaA+HCl⟶HA+HCl$
$3.612$ $1.806$
$1.356$ $-$ $1.806$

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

$4.92=p{K}_a+\log \frac{1.356}{1.806}$

$p{K}_a=5.044$

Now,
$NaOH+HA⟶NaA$
$2$ $2$
$-$ $-$ $2$

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

$pH$ of buffer solution: $pH=7+\frac{1}{2}p{K}_a+\frac{1}{2}\log C=7+\frac{5.044}{2}+\frac{1}{2}\log 0.1$

$pH=9$