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"_2xx1000)/(triangle"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"_2"FCOO"^-]["H"^+])/(["CH"_2"FCOOH"])`

= `("C"α."C"α)/("C"(1 - α))`

= `("C"α^2)/(1 - α)`

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"α^2)/(1 - α)`

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

`= (0.5xx0.00567)/0.9247`

= 0.00307 (approximately)

= 3.07 × 10^-3