Question

$500mL$ of $0.2M$ aqueous solution of acetic acid is mixed with $500mL$ of $0.2M$ $HCl$ at ${25}^{o}C$
(a) Calculate the degree of dissociation of acetic acid in the resulting solution and pH of the solution.
(b) If $6g$ of $NaOH$ is added to the above solution, determine final pH. Assume there is no change in volume on mixing. $K_a$ for acetic acid is $1.75\times {10}^{-5}M{L}^{-1}$.

Answer 1

(a) Conc. of $HCl$ and ${CH}_{3}COOH$ after mixing will be $0.1M$.
${CH}_{3}COOH\rightleftharpoons {CH}_{3}CO{O}^{-}+H^{+}$
$t=0$ $0.1$ $0$ $0.1$ (from $HCl$)
$t_{eq}$ $0.1-x$ $x$ $(0.1+x)$
$K_a=\frac{\left[{CH}_{3}CO{O}^{-}\right]\left[H^{+}\right]}{\left[{CH}_{3}COOH\right]}$
$1.75\times {10}^{-5}=\frac{x\times (0.1+x)}{(0.1-x)}$
On approximation
$x\approx 1.75\times {10}^{-5}$
$\left[H^{+}\right]=0.1+x\approx 0.1M$
$pH=-\log \left[0.1\right]=1$
Degree of dissociation of acetic acid $=\frac{1.75\times {10}^{-5}}{0.1}=1.75\times {10}^{-4}$
(b) number of moles of $NaOH$ added $=\frac{6}{40}=0.15$
${CH}_{3}COOH+HCl+NaOH\rightleftharpoons {CH}_{3}COONa+NaCl+H_{2}O$
$t=0$ $0.1$ $0.1$ $0.15$ $0$ $0$ $0$
$t_{eq}$ $0.05$ $0$ $0$ $0.05$ $0.1$ $0$
$pH=p{K}_a+\log \frac{\left[Salt\right]}{\left[Acid\right]}$
$=-\log 1.75\times {10}^{-5}+\log \frac{0.05}{0.05}=4.757$