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.
Solution
It is given that:
w1 = 500 g
w2 = 19.5 g
Kf = 1.86 Kkg mol-1
ΔTf = 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 Ka 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