An acid catalysed hydrolysis of ester is carried out in presence of HCl. Small but same volume of samples were taken out at various instant and were titrated with standardised solution of NaOH and following data was obtained t=Ot = 120 sec.t = ∞VNaOH required for neutralisation20 ml55ml60mlIf initially 6 M of ester was taken then calculate abed, where ab ab=t1/2 of the hydrolysis of ester cd = Molar concentration of HCl taken in the experiment.

Initially, at t = 0, m mol of HCl = 20 x N, where (N = normality of NaOH) at t = ∞, m mol of HCl + m mol of acid formed = 60 x N ⇒ m mol of acid formed = 60 x N - 20 x N = 40 N ⇒ m mol of ester taken =40N∝aAt t == 120 sec, m mol of acid formed =55×N−20×N=35N∝xa−x∝5Nk×t=ln⁡aa−xk×120=ln⁡40N5Nk×120=3ln⁡2t1/2=40Also, N1v1=N2v26×v1=40×NFor HCl,N′×v1=20×N=3v1⇒N′=3

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An acid catalysed hydrolysis of ester is carried out in presence of HCl. Small but same volume of samples were taken out at various instant and were titrated with standardised solution of NaOH and following data was obtained
t=O t = 120 sec. $t = \infty$
$V_{NaOH}$ required for neutralisation 20 ml 55ml 60ml
If initially 6 M of ester was taken then calculate abed, where ab
$a b = t_{1 / 2}$ of the hydrolysis of ester
cd = Molar concentration of HCl taken in the experiment.

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Detailed Solution

Initially, at t = 0, m mol of HCl = 20 x N, where (N = normality of NaOH) at $t = \infty$ , m mol of HCl + m mol of acid formed = 60 x N

$\Rightarrow$ m mol of acid formed $= 60 x N - 20 x N = 40 N$

$\Rightarrow$ m mol of ester taken $= 40 N \propto a$

At t == 120 sec, m mol of acid formed $= 55 \times N - 20 \times N$

$= 35 N \propto x$

$\begin{aligned} a - x \propto 5 N \\ k \times t = \ln (\frac{a}{a - x}) \\ k \times 120 = \ln \frac{40 N}{5 N} \\ k \times 120 = 3 \ln 2 \\ t_{1 / 2} = 40 \end{aligned}$