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 

Therefore, 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

Therefore, 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` = 1 + α Or α = i − 1 = 1.0753 − 1 = 0.0753

Now, the value of K_a 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

Answer 2

Calculated molar mass of CH₂FCOOH (M'_cal)

= (2 × 12) + (3 × 1) + (1 × 19) + (2 × 16)

= 78 g mol⁻¹

In the present case, w = 19.5 g, W = 500 g, ΔT_f = l.0°C,

K_f = 1.86 K kg mol⁻¹

∴ Observed molecular mass

`M'_"obs" = (1000 xx K_f xx w)/(W xx Delta T_f)`

= `(1000 xx 1.86 xx 19.5)/(500 xx 1.0)`

= 72.54 g mol⁻¹

Hence, van’t Hoff factor (i) = `(M'_(cal))/(M'_(obs))`

= `78/72.54`

= 1.0753

Suppose the degree of dissociation of fluoroacetic acid is α.

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

∴ Initial number of moles ∝ C

Number of moles at equilibrium ∝ [C(1− α) + Cα + Cα] ∝ [C(1 + α)]

∴ `i = "Number of particles in solution"/"Normal number of particles"`

= `(C(1 + alpha))/C`

= 1 + α

or α = i − 1 

= 1.0753 − 1

= 0.0753

Dissociation constant (K_a) of fluoroacetic acid is given by

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

= `(C alpha * C α)/(C (1 - alpha))` ...(∵ α is very small, 1 − α ≈ 1)

= Cα² 

Concentration (C) of the acid = `19.5/78 xx 1000/500`

= 0.5 m

∴ K_a = Cα²

= 0.5 × (0.0753)²

= 0.5 m