Question

A $50mL$ solution of $0.2MC{H}_{3}COOH$ is titrated with $0.2MNaOH$. Find the $pH$ of solution when:
(a) $0mLNaOH$ is added ($p{H}_{initial}$)
(b) $10mL$ of $NaOH$ has been added ($p{H}_{10mL}$)

Given: $K_a\left(C{H}_{3}COOH\right)=1.8\times {10}^{-5}$

• A. $p{H}_{initial}=2.72,p{H}_{10mL}=4.14$
• B. $p{H}_{initial}=5.7,p{H}_{10mL}=6.72$
• C. $p{H}_{initial}=1.72,p{H}_{10mL}=7.28$
• D. $p{H}_{initial}=4.57,p{H}_{10mL}=2.17$

Answer 1

The correct option is A $p{H}_{initial}=2.72,p{H}_{10mL}=4.14$

Case(a)
mmol of $C{H}_{3}COOH=50\times 0.2=10$
mmol of $NaOH=0$
$C{H}_{3}COOH\left(aq\right)+NaOH\left(aq\right)\to C{H}_{3}COONa\left(aq\right)+H_{2}O\left(l\right)10000$
Since only weak acid $C{H}_{3}COOH$ is present initially.
$p{H}_{initial}=-log\sqrt{K_{a}C}p{H}_{initial}=-\frac{1}{2}\left(log\right(1.8\times {10}^{-5}\times 0.2\left)\right)p{H}_{initial}=-\frac{1}{2}\left(log\right(3.6\times {10}^{-6}\left)\right)p{H}_{initial}=\frac{1}{2}(6-0.56)=\frac{5.44}{2}p{H}_{initial}=2.72$

Case (b)
mmol of $C{H}_{3}COOH=50\times 0.2=10$
mmol of $NaOH=10\times 0.2=2$

$C{H}_{3}COOH\left(aq\right)+NaOH\left(aq\right)\to C{H}_{3}COONa\left(aq\right)+H_{2}O\left(l\right)Initially:10200Final:8022$

After titration we can see that $8mmol$ of $C{H}_{3}COOH$ and $2mmol$ of $C{H}_{3}COONa$ are present.
$\text{Final Concentration}=\frac{\text{Moles}}{\text{Total Volume}}$
Total Volume $=50+10=60mL$
$\left[C{H}_{3}COOH\right]=\frac{8}{60}M$
$\left[C{H}_{3}COONa\right]=\frac{2}{60}M$
An acidic buffer is formed. So,
$pH=p{K}_a+log\left(\frac{\left[C{H}_{3}COONa\right]}{\left[C{H}_{3}COOH\right]}\right)pH=4.74+log\left(\frac{\frac{2}{60}}{\frac{8}{60}}\right)pH=4.74+log\left(\frac{1}{4}\right)=4.74-0.6p{H}_{10mL}=4.14$