Question

$100mL$ of a $0.1MC{H}_{3}COONa\left(aq\right)$ is titrated against a $0.1MHCl\left(aq\right)$. Calculate the $pH$ of the solution when $50mL$ of $HCl$ was added.
Given: $p{K}_a\left(C{H}_{3}COOH\right)=4.74at{25}^{o}C$

• A. 6.21
• B. 5.85
• C. 5.24
• D. 4.74

Answer 1

The correct option is D 4.74

mmol of $C{H}_{3}COONa=100\times 0.1=10$
mmol of $HCl=50\times 0.1=5$

$C{H}_{3}COONa\left(aq\right)+HCl\left(aq\right)\to C{H}_{3}COOH\left(aq\right)+NaCl\left(l\right)Initial:10500Final:5055$

After titration equal milimoles of $C{H}_{3}COOH$ and $C{H}_{3}COONa$ will form. The resultant solution will acts as an acidic buffer.
$\text{Final concentration}=\frac{\text{Moles}}{\text{Total Volume}}$
Total Volume $=100+50=150mL$
$[C{H}_{3}COOH{]}_{final}=\frac{5}{150}M$
$[C{H}_{3}COONa{]}_{final}=\frac{5}{150}M$
The formula to calculate the $pH$ of an acidic buffer is :
$pH=p{K}_a+\log \left(\frac{\left[C{H}_{3}COONa\right]}{\left[C{H}_{3}COOH\right]}\right)pH=4.74+\log \left(\frac{\frac{5}{150}}{\frac{5}{150}}\right)pH=4.74$