Question

0.1 mole of ethanol and 0.1 moles of butanoic acid are allowed to react. At equilibrium, the mixture is titrated with 0.85 M $NaOH$ solution and the titrated value was 100 ml. Assuming that no ester was hydrolysed by the base. Calculate $K$ for the reaction.
$C_2{H}_{5}OH+C_3{H}_{7}COOH\rightleftharpoons C_3{H}_{7}COOH{C}_2{H}_5+H_{2}O$

Answer 1

$C_2{H}_{5}OH+{CH}_3{H}_{7}COOH\rightleftharpoons C_3{H}_{7}COO{C}_2{H}_5+H_{2}O$

At $0.1$ $0.1$ $0$

$t=0$

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

$t=t_{eq}$

Given remaining acid normalized by $100ml$ of $0.85M$ $NaOH$

So $M_1{V}_1=M_2{V}_2$

$\left(0.85\right)\times \frac{100}{1000}=$ moles of $C_3{H}_{7}COOH$ remaining

$0.1-x=0.085$

$x=0.015$

$K=\frac{0.015}{0.085\times 0.085}=2.0761$