Question

0.6ml of acetic acid having a density of 1.06g/ml is dissolved in 1l of water. the depression in freezing point observed for this strength of the acid was 0.0205 degree celcius. calculate the van't hoff factor and the dissociation constant of the acid. kf for water =1.86k kg/mol

Answer 1

Number of moles of acetic acid = 0.6 ml × 1.06 g mL−1/60 g mol −1 = 0.0106 mol = n

Molality = 0.0106 mol = 0.0106 mol kg–1

Using equation
ΔTf = 1.86 K kg mol–1 × 0.0106 mol kg–1 = 0.0197 K
van’t Hoff Factor (i)= Observed freezing point/ Calculate Frezing point =0.0205 K/0.0197 K= 1.041

=>Acetic acid is a weak electrolyte and will dissociate into two ions:
Thus total moles of particles are: n(1 – x + x + x) = n(1 + x)
i = n (1 + x )/n = 1 + x = 1.041

=>Thus degree of dissociation of acetic acid = x = 1.041 – 1.000 = 0.041

=>Then

=>[CH3COOH] = n(1 – x) 0.0106 (1 – 0.041),
[CH3COO–] = nx = 0.0106 × 0.041, [H+] = nx = 0.0106 × 0.041.

=>ka=[CH3COO-] [H+] / [CH3COOH] = 0.0106 × 0.041 × 0.0106 × 0.041 / 0.0106 (1.00 − 0.041)
= 1.86 × 10^–5