Question

19.5 g of CH₂FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.0°C. Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid.

Answer 1

It is given that:

w₁ = 500 g

w₂ = 19.5 g

K_f = 1.86 K kg mol⁻¹ 

ΔT_f = 1.0°C

We know that:

M₂ = `(K_f xx w_2 xx 1000)/(Delta T_f xx w_1)`

= `(1.86  K  kg  "mol"^(-1) xx 19.5  g xx 1000  g  kg^(-1))/(500  g xx 1.0)`

= 72.54 g mol⁻¹ 

∴ Observed molar mass of CH₂FCOOH, (M₂)_obs = 72.54 g mol⁻¹ 

The calculated molar mass of CH₂FCOOH is (M₂)_cal = 14 + 19 + 12 + 16 + 16 + 1

= 78 g mol⁻¹

∴ Van’t Hoff factor (i) = `((M_2)_(cal))/(M_2)_(obs)`

= `78/72.54`

= 1.0753

Let α be the degree of dissociation of CH₂FCOOH.

	\[\ce{CH2FCOOH ⇌ CH2FCOO^- + H^+}\]
Initial Conc.	C mol L−1                     0                0
At equilibrium	C(1− α)                      Cα              Cα

∴ i = `(C(1 + α))/C`

⇒ i = 1 + α 

⇒ α = i − 1

= 1.0753 − 1

= 0.0753

Now, the value of K_α is given as:

K_α = `([CH_2FCOO^-][H^+])/([CH_2FCOOH])`

= `(C alpha * C α)/(C (1 - alpha))`

= `(C alpha^2)/(1 - alpha)`

Taking the volume of the solution as 500 mL, we have the concentration:

C = `19.58/78 xx 1/500 xx 1000`

= 0.5 M

∴ K_α = `(C alpha^2)/(1 - alpha)`

= `(0.5 xx (0.0753)^2)/(1 - 0.0753)`

= `(0.5 xx 0.00567)/0.9247`

= 0.00307 (approximately)

= 3.07 × 10⁻³