Saponification of ethyl acetate $({CH}_{3} {COOC}_{2} H_{5})$ by NaOH (Saponification of ethyl acetate by NaOH is second order reaction) is studied by titration of the reaction mixture initially having 1 : 1 molar ratio of the reactants. If 10 mL of 1 N HCl is required by 5 mL of the solution at the start and 8 mL of 1 N HCl is required by another 5 mL after 10 minutes, then rate constant is :

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$k = \frac{1}{10} [\frac{1}{2} - \frac{1}{10}]$

$k = \frac{5}{10} [\frac{1}{8} - \frac{1}{10}]$

$k = \frac{2.303}{10} \log \frac{10}{2}$

$k = \frac{2.303}{10} \log \frac{10}{8}$

answer is C.

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Detailed Solution

$Ester + {OH}^{-} ⟶ Alcohol + Acid$

$\begin{aligned} O + NaOH & ⟶ () + Acid \\ a = 10 & - \\ α - x & = 8 \end{aligned}$

$\begin{aligned} \frac{1}{[A]_{t}} & = \frac{1}{[A]_{0}} + k t \\ k & = \frac{1}{t} [\frac{1}{[A]_{t}} - \frac{1}{[A]_{0}}] = \frac{5}{10} [\frac{1}{8} - \frac{1}{10}] \end{aligned}$

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