Question

Acid hydrolysis of ester in 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. $V_{\infty }=(V_t-V_0)$
• C. $V_{\infty }=({2V}_t-V_0)$
• D. $V_{\infty }=({2V}_t+V_0)$

Answer 1

Answer: (C)

$V_{\infty }=({2V}_t-V_0)$

$RCOO{R}^{\prime }+H_{2}O\stackrel{H^{+}}{\to }RCOOH+R^{\prime }OH$

$t=0$ $a$ $0$ $0$

$t$ $a-x$ $x$ $x$

$t=\infty$ $a-a$ $a$ $a$

Let $V_0=$ Volume of $NaOH$

$(att=o),$

$V_t=x+V_0$

$V_{\infty }=a+V_0$

If ester is $50\%$ hydrolyzed, $x=\frac{a}{2}$

$\therefore V_t=\frac{a}{2}+V_0$

$a=2V_t-2V_0$

$\therefore V_{\infty }=2V_t-2V_0+V_0$

$=2V_t-V_0$

Hence, the answer is $2V_t-V_0.$