Question

$20mL$ of $0.2MC{H}_{3}COONa$ solution is titrated against $0.1MHCl$. Calculate the $pH$ of solution when $10mLHCl$ has been added.
Given: $p{K}_a\left(C{H}_{3}COOH\right)=4.74$

• A. 5.22
• B. 6.44
• C. 3.55
• D. 7.64

Answer 1

The correct option is A 5.22

mmol of $C{H}_{3}COONa=20\times 0.2=4$
mmol of $HCl=10\times 0.1=1$

$C{H}_{3}COONa\left(aq\right)+HCl\left(aq\right)\to C{H}_{3}COOH\left(aq\right)+NaCl\left(l\right)Initial:4100Final:3.0011$

After titration we can see that $3.0mmol$ of $C{H}_{3}COONa$ and $1mmol$ of $C{H}_{3}COOH$ are present. So , the resultant solution will act as an buffer.
$\text{Final concentration}=\frac{\text{Moles}}{\text{Total Volume}}$
Total Volume $=20+10=30mL$
$[C{H}_{3}COOH{]}_{final}=\frac{1}{30}M$
$[C{H}_{3}CONa{]}_{final}=\frac{3}{30}M$
The formula to calculate the $pH$ of an acidic buffer is :
$pH=p{K}_a+\log \left(\frac{\left[C{H}_{3}COONa\right]}{\left[C{H}_{3}COOH\right]}\right)pH=4.74+\log \left(\frac{\frac{3}{30}}{\frac{1}{30}}\right)pH=4.74+log\left(3\right)=4.74+0.48pH=5.22$