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.
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.
It is given that,
w_{1} = 500 g
w_{2} = 19.5 g
K_{f} = 1.86 K kg mol^{1}
ΔT_{f} = 1K
We know that
M_{2} = \(\frac{K_f \times w_2 \times 1000}{ΔT_f \times w_1}\)
= \(\frac{1.86 K \ kg mol^{1} \times 19.5g \times 1000 g kg^{1}}{500g \times 1K}\)
= 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 gmol^{1}
Therefore, van't Hoff factor, i
= \(\frac{(M_2)_{cal}}{(M_2)_{obs}}\)
= \(\frac{78g \ mol^{1}}{72.54 g \ mol^{1}}\)
= 1.0753
Let α be the degree of dissociation of CH_{2}FCOOH
CH_{2}FCOOH ↔ CH_{2}FCOO^{} + H^{+}
Initial conc. C mol L^{1} 0 0
At equilibrium C(1  α) Cα Cα
Total = C(1 + α)
∴ i = \(\frac{C(1  \alpha)}{C}\)
⇒ i = 1 + α
⇒ α = i  1
= 1.0753  1
= 0.0753
Now, the value of K_{a} is given as
K_{a} = \(\frac{[CH_2FCOO^][H^+]}{[CH_2FCOOH]}\)
= \(\frac{C \alpha . C \alpha}{C(1  \alpha)}\)
= \(\frac{C \alpha^2}{1  \alpha}\)
Taking the volume of the solution as 500 mL, we have the concentration
C = \(\frac{\frac{19.5}{78}}{500}\)x 1000 M
= 0.5 M
Therefore, K_{a} = \(\frac{0.5 \times (0.0753)^2}{10.0753}\)
= \(\frac{0.5 \times 0.00567}{0.9247}\)
= 0.00307 (approximately)
= 3.07 x 10^{3}

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