Answer
Verified
36.9k+ views
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.
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.
Recently Updated Pages
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
The path difference between two waves for constructive class 11 physics JEE_MAIN
What is the difference between solvation and hydra class 11 chemistry JEE_Main
IfFxdfrac1x2intlimits4xleft 4t22Ft rightdt then F4-class-12-maths-JEE_Main
Sodium chloride is purified by passing hydrogen chloride class 11 chemistry JEE_Main
Consider the following oxyanions PO43P2O62SO42MnO4CrO4S2O52S2O72 class 11 chemistry JEE_Main
Other Pages
The mole fraction of the solute in a 1 molal aqueous class 11 chemistry JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
Lowering in vapour pressure is highest for A 02 m urea class 11 chemistry JEE_Main
An electric bulb has a power of 500W Express it in class 11 physics JEE_Main
Which of the following Compounds does not exhibit tautomerism class 11 chemistry JEE_Main
The ratio of speed of sound in Hydrogen to that in class 11 physics JEE_MAIN