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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.

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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.

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It is given that,

w1 = 500 g

w2 = 19.5 g

Kf = 1.86 K kg mol-1

ΔTf = 1K

We know that

M2 = $$\frac{K_f \times w_2 \times 1000}{ΔT_f \times w_1}$$

= $$\frac{1.86 K \ kg mol^{-1} \times 19.5g \times 1000 g kg^{-1}}{500g \times 1K}$$

= 72.54 gmol-1

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

The calculated molar mass of CH2FCOOH is

(M2)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 CH2FCOOH

CH2FCOOH ↔ CH2FCOO- + 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 Ka is given as

Ka = $$\frac{[CH_2FCOO^-][H^+]}{[CH_2FCOOH]}$$

= $$\frac{C \alpha . C \alpha}{C(1 - \alpha)}$$

= $$\frac{C \alpha^2}{1 - \alpha}$$

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, Ka = $$\frac{0.5 \times (0.0753)^2}{1-0.0753}$$

= $$\frac{0.5 \times 0.00567}{0.9247}$$

= 0.00307 (approximately)

= 3.07 x 10-3

This post was modified 2 years ago by admin
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