Question

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

PLAN $ RCOOR' + H2_O \xrightarrow{ H^+ } RCOOH + R'OH $
Acid hydrolysis of ester is follows first order kinetics.
For same concentration of ester in each case, rate is dependent on $[H^+ $]from acid.
Rate = k [RCOOR' ]
Also for weak acid, HA $\rightleftharpoons H^+ A^- $
$ K_a = \frac{ [ H^+ ] \ [ A ^- ] }{ [ HA ]} $
$ ( Rate)_{ HA } = k [H^+ ]_{ H A } $
$ ( Rate)_{H x } = k [ H^+ ] _{ HX } $
$ ( Rate)_{ HX } = 100 Rate)_{ HA } $
$\therefore$ Also in strong acid, $[ H^+ ]$ = [ HX ] = 1M
$ \frac{ (Rate )_{ HX} }{ (Rate )_{ HA}} = 100 = \frac{ [ H^+ ]_{ HX }}{ [ H^+ ]_{ HA }} = \frac{1}{ [ H^+ ]_{ HA }} $
$ \therefore [ H^+ ]_{ HA } = \frac{1}{ 100 } $
HA $\rightleftharpoons H^+ + A^- $
$ \begin{array}
\ 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \ \ \ \ \ \ \ \ \ \ \ \ 0 \\
(1 - x) \ \ \ \ \ \ x \ \ \ \ \ \ \ \ \ \ \ \ x \\
\end{array}$
$\therefore K_a = \frac{ [ H^+ ] \ [A^- ] }{ [HA] } = \frac{ 0.01 \times 0.01 }{ 0.99 } $
$ 1 \times 10^{ - 4 } $