Question

$0.1$ mole each of ethyl alcohol and acetic acid are allowed to react and at equilibrium, the acid was exactly neutralized by $100mL$ of $0.85NNaOH$. If no hydrolysis of ester is supposed to have undergone and $K_c$$=y/100$, then find $y$ (nearest integer).

Answer 1

The equilibrium reaction and the initial number of moles and the equilibrium number of moles are as given below.
$C{H}_{3}COOH+C_2{H}_{5}OH\rightleftharpoons C{H}_{3}COO{C}_2{H}_5+H_{2}O$
Before reaction $0.1$ $0.1$ $0$ $0$
At equilibrium $(0.1-x)$ $(0.1-x)$ $x$ $x$
Meq. of acetic acid left = Meq. of $NaOH$ used $=100\times 0.85=85$
$\therefore$ Milli mole of acetic acid left $=85$ ($\because$ monobasic)
$\therefore$ Mole of acetic acid left $=0.085$
or $(0.1-x)=0.085$
or $x=0.015$
The equilibrium constant expression is as given below.
$K_c=\frac{x^2}{{(0.1-x)}^2}=\frac{{\left(0.015\right)}^2}{{\left(0.085\right)}^2}=0.031$

Now, according to question, $K_c=\frac{y}{100}$

$\therefore y=3$
The answer is 3.