Question

0.1 mol each of ethyl alcohol and acetic acid are allowed to react and at equilibrium, the acid was exactly neutralised by 100 mL of 0.85 N NaOH. If no hydrolysis of ester is supposed to have undergo, find $K_C$.

• A. $K_C=0.006$
• B. $K_C=0.031$
• C. $K_C=1.34$
• D. $K_C=3.1$

Answer 1

Answer: (B)

$K_C=0.031$

${CH}_{3}COOH+C_2{H}_{5}OH\rightleftharpoons {CH}_{3}COO{C}_2{H}_5+H_{2}O$
0.1 0.1 0 0 Before reaction
$(0.1-x)$ $(0.1-x)$ $x$ $x$
mEq of of acetic aid left = mEq of NaOH used = 100 x 0.85 = 85
$\therefore$ Millimoles of aetic acid left = 85 $(\because monobasic)$
$\therefore Molesofaceticacidleft=0.085$
$(0.01-x)=0.085$
$x=0.015$
Now,
$K_C=\frac{x^2}{{(0.1-x)}^2}=\frac{{\left(0.015\right)}^2}{{\left(0.085\right)}^2}=0.031$