19.5 g of CH_{2}FCOOH 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.

#### Solution

It is given that:

w_{1} = 500 g

w_{2} = 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 CH_{2}FCOOH , (M_{2})_{obs} = 72.54 g mol

The calculated molar mass of CH_{2}FCOOH is

(M_{2})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 CH_{2}FCOOH

CH_{2}FCOOH ↔ CH_{2}FCOO^{-} +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}