Question

0.1 mole of ethanol and 0.1 mole of butanoic acid are allowed to react. At equilibrium mixture is titrated with 0.85 M NaOH solution and the titre value was 100 ml.Assuming that no ester was hydrolysed by the base, calculate K for the reaction.
$C_2{H}_{5}OH+C_3{H}_{2}COOH\rightleftharpoons C_3{H}_{2}COO{C}_2{H}_5+H_{2}O$

Answer 1

$C_2{H}_{5}OH+C_3{H}_{2}COOH\rightleftharpoons {CH}_2{COOC}_2{H}_5+H_{2}O$

At $0.1$ $0.1$ $0$ $0$

$t=0$

At $0.1-x$ $0.1-x$ $x$ $x$

$t=t_{eq}$

$(0.1-x)$ moles neutralised by $0.85M$, $100ml$ $NaOH$

$0.1-x=0.85\times \frac{100}{1000}=0.085$

$x=0.015$

$K=\frac{\left[C_3{H}_{2}COO{C}_2{H}_5\right]}{\left[C_2{H}_{5}OH\right]\left[C_3{H}_{2}COOH\right]}=\frac{0.085}{0.015\times 0.015}\approx 377.77$