Browse Questions

# 0.6 mL of acetic acid ($CH_3COOH$) having a density of 1.06 g $mL^{−1}$ is dissolved in 1 litre of water. The depression in freezing point observed for this strength of the acid was 0.0205$^{\large\circ}$C. Calculate the van’t Hoff factor and the dissociation constant of the acid. $K_f$ for water = 1.86 K kg $mol^{−1}$.

Can you answer this question?

0.6 mL of acetic means solute ($W_2) = 0.6 \times 1.06g = 0.636g$
1 Litre of water means solvent ($W_1)$ = 1000 g
$(\Delta T_f)_{observed}=0.0205^{\large\circ}C$
∴ Observed molar mass ($M_2$)observed =$\large\frac{1000\times K_f\times W_2}{W_1\times \Delta T_f}$
$\Rightarrow 57.7gmol^{-1}$
Calculated molar mass of $CH_3COOH = 60g mol^{−1}$
van’t Hoff factor (i) =$\large\frac{60gmol^{-1}}{57.7gmol^{-1}}$$=1.04 i=\large\frac{1+\alpha}{1}$$=1+\alpha$
$\alpha=0.04$
Dissociation constant $(K_a)=\large\frac{C\alpha C\alpha}{C(1-\alpha)}=\frac{C\alpha^2}{1-\alpha}$
$K_a=\large\frac{(0.0106)(0.04)^2}{1-0.04}$$=1.76\times 10^{-5}$
Hence (A) is the correct answer.
answered Jun 13, 2014