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 $(t)$ $(\min)$	$0$	$30$	$60$	$90$
Conc. of ester $(C)$ $(M)$	$0.850$	$0.800$	$0.754$	$0.710$

Show that it follows a pseudo first order reaction as the concentration of $H_{2}O$ remains nearly constant $(55 \ mol \ L^{-1})$ during the course of the reaction. What is the value of $k^{\prime}$ in this reaction?

(N/A) For a pseudo first order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{C_{0}}{C}$.
At $t = 30 \ min$: $k_{1} = \frac{2.303}{30} \log \frac{0.850}{0.800} = 0.07677 \times 0.02633 = 2.02 \times 10^{-3} \ min^{-1}$.
At $t = 60 \ min$: $k_{2} = \frac{2.303}{60} \log \frac{0.850}{0.754} = 0.03838 \times 0.0520 = 1.996 \times 10^{-3} \ min^{-1}$.
At $t = 90 \ min$: $k_{3} = \frac{2.303}{90} \log \frac{0.850}{0.710} = 0.02559 \times 0.0781 = 2.00 \times 10^{-3} \ min^{-1}$.
Since $k$ is constant,it follows a pseudo first order reaction. The average $k \approx 2.00 \times 10^{-3} \ min^{-1}$.
The rate law is $Rate = k^{\prime} [Ester][H_{2}O]$. Since $[H_{2}O]$ is constant,$k = k^{\prime} [H_{2}O]$.
$k^{\prime} = \frac{k}{[H_{2}O]} = \frac{2.00 \times 10^{-3} \ min^{-1}}{55 \ mol \ L^{-1}} = 3.636 \times 10^{-5} \ L \ mol^{-1} \ min^{-1}$.