Question

6ml of acetic acid having density 1.05 is dissolved

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