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 Kkg mol^-1 

ΔT_f = 1K

We know that:

`M_2 = (K_fxxw_2xx1000)/(triangleT_fxxw_1)`

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

= 72.54 gmol^-1

Therefore, observed molar mass of  CH₂FCOOH , (M₂)_obs = 72.54 g mol

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 gmol^(-1))/(72.54 g mol^(-1))`

= 1.0752

Let α be the degree of dissociation of CH₂FCOOH 

                                  CH₂FCOOH ↔ CH₂FCOO^- +H⁺

Initial Conc              Cmol L^-1         0                0

At equilibrium         C(1-α)            Cα              Cα         Total = C(1+α)

`:.i = (C(1+alpha))/C`

`=> i = 1+alpha`

`=>alpha = i -1`

= 1.0753 -1

 = 0.0753

Now, the value of K_a is given as:

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

` = (Calpha.Calpha)/(C(1-alpha))`

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

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

`C = (19.5/78)/500 xx 1000M`

= 0.5M

Therefore, `K_alpha = (Calpha^2)/(1-alpha)`

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

`= (0.5xx0.00567)/0.9247`

= 0.00307 (approximately)

= 3.07 x  10^-3