Question

Acid hydrolysis of ester is first order reaction and rate constant is given by
$k=\frac{2.303}{t} log \frac{V_{\infty}-V_{0}}{V_{\infty}-V_{t}}$ where, $V_{0}, V_{t} $ and $V_{\infty}$
are the volume of standard $NaOH$ required to neutralise acid present at a given time, if ester is 50% neutralised then

• A. $V_{\infty} = V_{t}$
• B. $V_{\infty} = \left(V_{t}-V_{0}\right)$
• C. $V_{\infty} = 2V_{t}-V_{0}$
• D. $V_{\infty} = 2V_{t}+V_{0}$

Answer 1

Answer: (C)

At $t=0, V_{0}=$ volume of $NaOH$ due to $H ^{+}$(catalyst)
$V_{t}=x+V_{0}$
$V_{\infty}=a+V_{0}$
If ester is $50 \%$ hydrolysed then, $x=\frac{a}{2}$
or $ V_{t}=\frac{a}{2}+V_{0} $
or $ a=2 V_{t}-2 V_{0}$
$\therefore V_{\infty}=2 V_{t}-2 V_{0}+V_{0}$
$=2 V_{t}-V_{0}$