Question

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

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

Answer 1

The correct option is A $1\times {10}^{-4}$

$RCOO{R}^{{}^{\prime }}+H_{2}O\stackrel{H^{+}}{\to }RCOOH+R^{{}^{\prime }}OH$
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[RCOOR']
Also for weak acid, $HA\rightleftharpoons H^{+}+A^{-}$
$K_a=\frac{\left[H^{+}\right]\left[A^{-}\right]}{\left[HA\right]}$
$(Rate{)}_{HA}=k[H^{+}{]}_{HA}$
$(Rate{)}_{HX}=k[H^{+}{]}_{HX}$
$(Rate{)}_{HX}=100(Rate{)}_{HA}$
$\therefore$ Also in strong acid, $\left[H^{+}\right]=\left[HX\right]=1M$
$\frac{(Rate{)}_{HX}}{(Rate{)}_{HA}}=100=\frac{[H^{+}{]}_{HX}}{[H^{+}{]}_{HA}}=\frac{1}{[H^{+}{]}_{HA}}$
$\therefore [H^{+}{]}_{HA}=\frac{1}{100}$
$\begin{array}{ccccc}HA& \rightleftharpoons & H^{+}& +& A^{-}\\ 1& & 0& & 0\\ (1-x)& & x& & x\end{array}$
x = 0.01
$\therefore K_a=\frac{\left[H^{+}\right]\left[A^{-}\right]}{\left[HA\right]}=\frac{0.01\times 0.01}{0.99}$
$=1.01\times {10}^{-4}$