Question

$40mL$ of $0.5MC{H}_{3}COOH$ is titrated with of $0.5MNaOH.$ Find the $pH$ of the solution when $39mL$ $NaOH$ has been added.
Given: $K_a\left(C{H}_{3}COOH\right)=1.9\times {10}^{-5}$

• A. 3.4
• B. 6.3
• C. 4.6
• D. 5.3

Answer 1

The correct option is B 6.3

mmol of $C{H}_{3}COOH=40\times 0.5=20$
mmol of $NaOH=39\times 0.5=19.5$

$C{H}_{3}COOH\left(aq\right)+NaOH\left(aq\right)\to C{H}_{3}COONa\left(aq\right)+H_{2}O\left(l\right)Initial:2019.500Final:0.5019.519.5$

$\text{Final concentration}=\frac{\text{Moles}}{\text{Total Volume}}$
Total Volume $=40+39=79mL$
$[C{H}_{3}COOH{]}_{final}=\frac{0.5}{79}M$
$[C{H}_{3}COONa{]}_{final}=[C{H}_{3}CO{O}^{-}{]}_{final}=\frac{19.5}{79}M$
For a buffer solution , the effective range should be : $\frac{1}{10}<\frac{\left[C{H}_{3}COONa\right]}{\left[C{H}_{3}COOH\right]}<10$
But here, $\frac{\left[C{H}_{3}COONa\right]}{\left[C{H}_{3}COOH\right]}=\frac{19.5/79}{0.5/79}=39>10$
So after titration, $C{H}_{3}COONa$ and $C{H}_{3}COOH$ is there but the mixture will not acts as a buffer.

Consider the dissociation equilibrium of $C{H}_{3}COOH$ , where $\alpha$ is the degree of dissociation.
$C{H}_{3}COOH\left(aq\right)\rightleftharpoons C{H}_{3}COOH\left(aq\right)+H^{+}\left(aq\right)Initially:0.519.5Equilibrium:0.5(1-\alpha )19.5+0.5\alpha$

Here , $C{H}_{3}COOH$ is a weak acid which is trying to dissociate , but in product $C{H}_{3}CO{O}^{-}$ concentration is very high which eventually try to reacts with $H^{+}$ and pulled the overall reaction in backward direction. Therefore degree of hydrolysis $\left(\alpha \right)$ will be very small.
$\therefore 1-\alpha \approx 119.5+0.5\alpha \approx 19.5K_a=\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[H^{+}\right]}{\left[C{H}_{3}COOH\right]}1.9\times {10}^{-5}=\frac{\left(19.5/79\right)\times \left[H^{+}\right]}{\left(0.5/79\right)}\left[H^{+}\right]=\frac{0.5\times (1.9\times {10}^{-5})}{19.5}\left[H^{+}\right]=4.9\times {10}^{-7}pH=-\log (4.9\times {10}^{-7})pH=6.3$