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 :

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

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

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

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

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

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$Ester + {OH}^{-} ⟶ Alcohol + Acid$

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

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

Watch 3-min video & get full concept clarity

tricks from toppers of Infinity Learn

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

🔥 Start Your JEE/NEET Prep at Just ₹1999 / month - Limited Offer! Check Now!

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET