Question

The initial rate of hydrolysis of methyl acetate $(1M)$ by a weak acid $(HA,1M)$ is $1/100$ th of that of a strong acid $(HX,1M)$, at $25^{\circ }C.$ The $K_a(HA)$ is

• A. $1\times 10^{-4}$
• B. 1 $\times 10^{-5}$
• C. 1 $\times 10^{-6}$
• D. 1 $\times 10^{-3}$

Answer 1

Answer: (A)

$RCOO{R}^{\prime }+H_{2}O\stackrel{H^{+}}{\to }RCOOH+R^{\prime }OH$
Acid hydrolysis of ester is follows first order kinetics. For same concentration of ester in each case, rate is dependent on $\left[H^{+}\right]$from acid.
Rate $=k[RCOO{R}^{\prime }]$
Also for weak acid, $HA\rightleftharpoons H^{+}+A^{-}$
$K_a=\frac{\left[H^{+}\right]\left[A^{-}\right]}{[HA]}$
$(\text{ Rate }{)}_{HA}=k{\left[H^{+}\right]}_{HA}$
$(\text{ Rate }{)}_{Hx}=k{\left[H^{+}\right]}_{HX}$
$(\text{ Rate }{)}_{HX}=100($ Rate ${)}_{HA}$
$\therefore$ Also in strong acid, $\left[H^{+}\right]=[HX]=1M$
$\frac{(\text{ Rate }{)}_{HX}}{(\text{ Rate }{)}_{HA}}=100=\frac{{\left[H^{+}\right]}_{HX}}{{\left[H^{+}\right]}_{HA}}=\frac{1}{{\left[H^{+}\right]}_{HA}}$
$\therefore {\left[H^{+}\right]}_{HA}=\frac{1}{100}$

$\therefore K_a=\frac{\left[H^{+}\right]\left[A^{-}\right]}{[HA]}=\frac{0.01\times 0.01}{0.99}$
$1\times 10^{-4}$