Hydrolysis of methyl acetate in aqueous solution has been studied by titrating the liberated acetic acid against sodium hydroxide. The concentration of the ester at different times is given below:

Time $(t)$ $\text{min}$	$0$	$30$	$60$	$90$
Conc. of ester $(C)$ $\text{mol L}^{-1}$	$0.850$	$0.800$	$0.754$	$0.710$

Show that it follows a pseudo first-order reaction as the concentration of $H_2O$ remains nearly constant $(54.2 \text{ mol L}^{-1})$ during the course of the reaction. What is the value of $k'$ in this reaction?

(N/A) For a pseudo first-order reaction,the rate constant $k'$ is given by the formula:
$k' = \frac{2.303}{t} \log \frac{C_0}{C_t}$
Given $C_0 = 0.850 \text{ mol L}^{-1}$.
At $t = 30 \text{ min}$,$C_t = 0.800 \text{ mol L}^{-1}$:
$k'_1 = \frac{2.303}{30} \log \frac{0.850}{0.800} = 0.07677 \times \log(1.0625) \approx 0.07677 \times 0.0263 = 0.00202 \text{ min}^{-1}$.
At $t = 60 \text{ min}$,$C_t = 0.754 \text{ mol L}^{-1}$:
$k'_2 = \frac{2.303}{60} \log \frac{0.850}{0.754} = 0.03838 \times \log(1.1273) \approx 0.03838 \times 0.0520 = 0.00200 \text{ min}^{-1}$.
At $t = 90 \text{ min}$,$C_t = 0.710 \text{ mol L}^{-1}$:
$k'_3 = \frac{2.303}{90} \log \frac{0.850}{0.710} = 0.02559 \times \log(1.1972) \approx 0.02559 \times 0.0781 = 0.00200 \text{ min}^{-1}$.
Since the values of $k'$ are nearly constant,the reaction follows pseudo first-order kinetics.
The average value of $k' \approx 2.00 \times 10^{-3} \text{ min}^{-1}$.