Question

$50mL$ of a $0.1MC{H}_{3}COONa\left(aq\right)$ is titrated against a $0.1MHCl\left(aq\right)$. Calculate the $pH$ of solution when $50mLHCl$ was added.
Given: $p{K}_a\left(C{H}_{3}COOH\right)=4.74at{25}^{o}C$

• A. 4.74
• B. 3.02
• C. 5.25
• D. 4.24

Answer 1

The correct option is B 3.02

mmol of $C{H}_{3}COONa=50\times 0.1=5$
mmol of $HCl=50\times 0.1=5$

$C{H}_{3}COONa\left(aq\right)+HCl\left(aq\right)\to C{H}_{3}COOH\left(aq\right)+NaCl\left(l\right)Initial:5500Final:0055$

After titration no $C{H}_{3}COONa$ and $HCl$ are left in solution. Only weak acid, $C{H}_{3}COOH$ is present.
Final mmol of $C{H}_{3}COOH$ = $5$
Total Volume $=50+50=100mL$
$\text{Final concentration}=\frac{\text{Moles}}{\text{Total Volume}}$
$[C{H}_{3}COOH{]}_{final}=\frac{5}{100}=\frac{1}{20}M$

The formula to calculate the $pH$ of a weak acid is :
$pH=\frac{1}{2}(p{K}_a-\log C)pH=\frac{1}{2}(4.74-\log \frac{1}{20})pH=\frac{1}{2}(4.74+\log 20)pH=3.02$