Question

Saponification of ethyl acetate $(C{H_3}COO{C_2}{H_5})NaOH$ (Saponification of ethyl acetate by $NaOH$ is second order Rxn) is studied by titration of the reaction mixture initially having 1:1 molar ratio of the reactants. If 10mL of 1N $HCl$ is required by 5mL of the solution at the start and 8M of 1N $HCl$ is required by another 5ml after 10 minutes, then rate constant is:
A.$K = \dfrac{{2.303}}{{10}}\log \dfrac{{10}}{8}$
B.$K = \dfrac{{2.303}}{{10}}\log \dfrac{{10}}{2}$
C. $K= \dfrac{1}{{10}}\left( {\dfrac{1}{8} - \dfrac{1}{{10}}} \right)$
D.$K= \dfrac{1}{{10}}\left( {\dfrac{1}{2} - \dfrac{1}{{10}}} \right)$

Answer 1

The reaction occurs as follows
$C{H_3}COO{C_2}{H_5} + NaOH \to C{H_3}COONa + {C_2}{H_5}OH$
This reaction is a second order reaction because the rate of reaction is directly proportional to the both of the reactants. This is called saponification of Ethyl acetate.

Formula used: $k = \dfrac{1}{t}\dfrac{{\left( {{V_0} - {V_t}} \right)}}{{{V_0}{V_t}}}$;
k=rate constant for saponification of Ethyl acetate.
${V_0} =$Initial volume of acid
${V_t} =$Final volume of acid
t= time

Complete step by step answer:
For Saponification of Ethyl acetate, a water bath is prepared, a flask is kept into it, solution of $C{H_3}COO{C_2}{H_5},NaOH$ is mixed in the flask. We observe the concentration of reactants decreases with time and concentration of products increases with time. To find the rate of reaction we will titrate a small amount of the mixture taken at regular intervals against $HCl$ (hydrochloric acid solution), taken in burette. Now, the question arises in our mind, why did we use the water bath? Can’t this reaction be carried without its use? The simple answer to this question is that we need to maintain the temperature during this whole process, if the temperature is not maintained, our results will be faulty and will not match the general conclusions. The theoretical part ends here and now we will perform the calculations after noting the observations.
For the given reaction,
$C{H_3}COO{C_2}{H_5} + NaOH \to C{H_3}COONa + {C_2}{H_5}OH$
Volume of standard acid at the beginning of the reaction (${V_0}$) is directly proportional to Initial concentration of $NaOH$(a).
${V_0}\propto$ $a$…… (1)
After time $t$, $x$ amount of $NaOH$ is used in the reaction. $a - x$is the remaining concentration.
Volume of standard acid at time $t,{V_t}$=$a - x$which is the remaining concentration of $NaOH$.
${V_t}\propto$ $a - x$ …………. (2)
Substituting equation (1) in equation (2)
${V_t}\propto$ ${V_0} - x$
Or, x$$$$\propto $$$${V_0} - {V_t} …………. (3)
Substituting these values in the rate equation for second order reaction;

$$k = \dfrac{1}{t}\dfrac{x}{{a\left( {a - x} \right)}} \\\ \Rightarrow k = \dfrac{1}{t}\dfrac{{\left( {{V_0} - {V_t}} \right)}}{{{V_0}\left( {{V_0} - \left( {{V_0} - {V_t}} \right)} \right)}} \\\ \Rightarrow k = \dfrac{1}{t}\dfrac{{\left( {{V_0} - {V_t}} \right)}}{{{V_0}\left( {{V_t}} \right)}} \\\ \Rightarrow k = \dfrac{1}{t}\left( {\dfrac{1}{{{V_t}}} - \dfrac{1}{{{V_0}}}} \right) \\\$$

This is the required relation.
Now, using the information given in the question and substituting it,

$$t = 10minutes \\\ {V_0} = 10ml \\\ {V_t} = 8ml \\\ \Rightarrow k = \dfrac{1}{t}\left( {\dfrac{1}{{{V_t}}} - \dfrac{1}{{{V_0}}}} \right) \\\ \Rightarrow k = \dfrac{1}{{10}}\left( {\dfrac{1}{8} - \dfrac{1}{{10}}} \right) \\\$$

Thus option C. is the correct for the given question.

Additional information:
While working on this experiment in the laboratory, we will observe the sweet, fruity smell of one of the reactants. That reactant is Ethyl acetate $C{H_3}COO{C_2}{H_5}$.

Note:
$NaOH$ is commonly called caustic soda or lye. $C{H_3}COO{C_2}{H_5}$ is also used as the decaffeinating agent in tea and coffee. Whereas $NaOH$, Sodium hydroxide is basic in nature and the most important ingredient for soap making but it is absent from the final product.