Question

Acid hydrolysis of ester is first order reaction and rate constant is given by k=$\dfrac{2.303}{\text{t}}\text{log}\left( \dfrac{{{\text{V}}_{\infty }}\text{-}{{\text{V}}_{0}}}{{{\text{V}}_{\infty }}\text{-}{{\text{V}}_{\text{t}}}} \right)$where, ${{\text{V}}_{0}}\text{,}{{\text{V}}_{\text{t}}}$ and ${{\text{V}}_{\infty }}$ are the volumes of standard NaOH required to neutralise acid present at a given time, if ester is $50%$neutralised then:
A. ${{\text{V}}_{\infty }}\text{=}{{\text{V}}_{\text{t}}}$
B. ${{\text{V}}_{\infty }}\text{=}\left( {{\text{V}}_{\text{t}}}\text{-}{{\text{V}}_{0}} \right)$
C. ${{\text{V}}_{\infty }}\text{=2}{{\text{V}}_{\text{t}}}-{{\text{V}}_{0}}$
D. ${{\text{V}}_{\infty }}\text{=2}{{\text{V}}_{\text{t}}}+{{\text{V}}_{0}}$

Answer 1

Hint: The reaction of hydrolysis of ester is $\text{RCOO}{{\text{R}}^{'}}+{{\text{H}}_{2}}\text{O}\xrightarrow{{{\text{H}}^{+}}}\text{RCOOH}+{{\text{R}}^{'}}\text{OH}$. Now, we need to find the concentrations of esters, catalyst and relate it to the volumes of the NaOH given for neutralisation. The general formula for first order reaction is $\text{k=2}\text{.303log}\left( \dfrac{\text{a-x}}{\text{a}} \right)$.

Complete answer:
Let us solve this question step by step:
Step (1)- Write the reaction whose kinetic study is to be done and write its rate law, the reaction is $\text{RCOO}{{\text{R}}^{'}}+{{\text{H}}_{2}}\text{O}\xrightarrow{{{\text{H}}^{+}}}\text{RCOOH}+{{\text{R}}^{'}}\text{OH}$and the rate law is $\text{Rate = k}\left( \text{ester} \right)\text{ or k}\left( \text{ester} \right)\left( {{\text{H}}^{+}} \right)$ . The order of the reaction is 1.

Step (2)- Deal with concentrations of reactant or ester at different times, at t=0, t=t and t=$\infty $, the concentrations will be considering initial concentration as ‘a’ and at t=t, the part that dissociated or hydrolysed be ‘x’. Then, the concentrations will be-
$\begin{align}
  & \text{ RCOO}{{\text{R}}^{'}}+{{\text{H}}_{2}}\text{O}\xrightarrow{{{\text{H}}^{+}}}\text{RCOOH}+{{\text{R}}^{'}}\text{OH} \\
 & \text{t=0 a 0 } \\
 & \text{t=t a-x x } \\
 & \text{t=}\infty \text{ a-a=0 a } \\
\end{align}$
The ester completely finished when time reaches $\infty $, it means that reaction has completed and acid is completely obtained.

Step (3)- Relate these concentrations with volumes of NaOH given for neutralization. The volume of NaOH at t=0 or initially is ${{\text{V}}_{0}}$. Initially just esters and catalysts are present and NaOH will try to neutralize that catalyst or the ${{\text{H}}^{+}}$ ions. At time t=t, there is some acid present and catalyst. So, we can directly relate the volume of NaOH $\left( {{\text{V}}_{0}}\text{ and }{{\text{V}}_{\text{t}}} \right)$ with the concentration of acid, as, ${{\text{V}}_{\text{t}}}=\text{x}+{{\text{V}}_{0}}$. Similarly, when time reaches $\infty $, there is only acid present in the solution, so, ${{\text{V}}_{\infty }}$ will be related ${{\text{V}}_{\infty }}=\text{a}+{{\text{V}}_{\text{o}}}$.

Step (4)- The value of acid hydrolysed is 50%, which is represented by ‘x’. The value of x will be $\dfrac{\text{50}\times \text{a}}{100}$. The value of x is $\dfrac{\text{a}}{2}$.

Step (5)- Put the value of x in ${{\text{V}}_{\text{t}}}=\text{x}+{{\text{V}}_{0}}$, the expression is transformed to ${{\text{V}}_{\text{t}}}=\dfrac{\text{a}}{2}+{{\text{V}}_{0}}$. So, $\text{a = 2}{{\text{V}}_{\text{t}}}-2{{\text{V}}_{0}}$. Put this value of a in expression ${{\text{V}}_{\infty }}=\text{a}+{{\text{V}}_{\text{o}}}$ to replace ‘a’, the new expression is ${{\text{V}}_{\infty }}=2{{\text{V}}_{\text{t}}}-2{{\text{V}}_{0}}+{{\text{V}}_{\text{o}}}\text{ or }{{\text{V}}_{\infty }}=2{{\text{V}}_{\text{t}}}-{{\text{V}}_{\text{o}}}$.
The correct answer to this question is ${{\text{V}}_{\infty }}=2{{\text{V}}_{\text{t}}}-{{\text{V}}_{\text{o}}}$
So, the correct answer is “Option C”.

Note: The important point to note in this question is just to relate the volumes of NaOH required for neutralisation with the concentrations of the ions. The ions include only the ${{\text{H}}^{+}}$ ions from acetic acid and catalyst. As, the base can only neutralise acid so, just include those concentrations.