Question

The initial rate of hydrolysis of methyl acetate (1M) by a weak acid (HA, 1M) is \[\left( \dfrac{1}{100}th \right)~\] of that of a strong acid (HX,1M) at \[25{}^\circ C\], the \[{{K}_{a}}\]of HA is:
(A) $1.00\times {{10}^{-4}}$
(B) $1.00\times {{10}^{-5}}$
(C) $1.00\times {{10}^{-6}}$
(D) $1.00\times {{10}^{-3}}$

Answer 1

Hint: To find out the correct answer of this question, we should know about the rate of hydrolysis. We should know that acid hydrolysis is a process in which we use a protic acid to catalyze the cleavage of a chemical bond via a nucleophilic substitution reaction, with the addition of the elements of water.
Step by step answer:
We have to find Ka for this question. First of all, we will write the reaction.
\[RCOO{{R}^{'}}+{{H}_{2}}O\xrightarrow{{{H}^{+}}}RCOOH+{{R}^{'}}OH\]
The above reaction is an acid hydrolysis reaction. We should know that this is a first order kinetics. We should know that acid hydrolysis is a process in which a protic acid is used to catalyze the cleavage of a chemical bond via a nucleophilic substitution reaction, with the addition of the elements of water (H2O). and we should also understand first order kinetics. We should know that it is an order of chemical reaction in which the rate of the reaction depends on the concentration of only one reactant, and is proportional to the amount of the reactant. It may be represented by the equation, rate = kA, where k is the reaction rate constant, and A is the concentration of the reactant.
We should know that, concentration of ester in each case will depend on hydrogen ion.
Rate of reaction will depend on \[[{{H}^{+}}]\].
Rate=K[A]
Rate=[RCOOR’]
We should know that reaction for weak acid HA is:
$HA\rightleftarrows {{H}^{+}}+{{A}^{-}}$
And\[{{K}_{a}}\]for this reaction is ${{K}_{a}}=\dfrac{[{{H}^{+}}]\times [{{A}^{-}}]}{[HA]}$
\[\begin{align}
  & {{[Rate]}_{HA}}=k{{[{{H}^{+}}]}_{HA}} \\
 & {{[Rate]}_{HX}}=k{{[{{H}^{+}}]}_{HX}} \\
\end{align}\]
\[{{[Rate]}_{HX}}=100{{[Rate]}_{HA}}\]

We should note that in strong acid that:
\[[{{H}^{+}}]=[HX]=1M\to Given\]
\[\dfrac{{{[Rate]}_{HX}}}{{{[Rate]}_{HA}}}=100=\dfrac{{{[{{H}^{+}}]}_{HX}}}{{{[{{H}^{+}}]}_{HA}}}=\dfrac{1}{{{[{{H}^{+}}]}_{HA}}}\]
\[\begin{align}
  & \dfrac{1}{{{[{{H}^{+}}]}_{HA}}}=\dfrac{{{[Rate]}_{HX}}}{{{[Rate]}_{HA}}}=100 \\
 & \dfrac{1}{100}={{[{{H}^{+}}]}_{HA}} \\
\end{align}\]
\[\begin{align}
  & HA\rightleftarrows {{H}^{+}}+{{A}^{-}} \\
 & 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,0 \\
 & (1-x)\,\,\,\,x\,\,\,\,\,\,\,\,\,x \\
\end{align}\]
From this we can say that x=0.01
${{K}_{a}}=\dfrac{[{{H}^{+}}]\times [{{A}^{-}}]}{[HA]}=\dfrac{0.01\times 0.01}{0.99}\to 1.00\times {{10}^{-4}}$
So, the correct answer to this question is $1.00\times {{10}^{-4}}$ and that is option A.

Note: We should know that hydrolysis is widely used in industry to break down chemicals into smaller fractions or pieces. For example, a compound called organophosphate ester can be hydrolysed, or undergo a hydrolysis reaction. This hydrolysis aids in the production of insect killers and pesticide sprays.