2.33 19.5 g of CH2FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.00 C. Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid.
It is given that:
w1=500 g
w2=19.5 g
Kf=1.86 K kg mol-1
∆Tf=1K
We know that:
M2=Kf×w2×1000∆Tf×w1
=1.86 K kg mol-1×19.5 g×1000 g kg-1500 g×1K
=72.54 g mol-1
Therefore, observed molar mass of CH2FCOOH, (M
The calculated molar mass of CH2FCOOH
(M2)cal=14+19+12+16+16+1=78 g mol-1
Therefore, van’t Hoff factor, i=(M2)cal (M
=78 g mol-172.54 g mol-1
=1.0753
Let α be the degree of dissociation of CH2FCOOH
CH2FCOOH↔CH2FCOOH-+H+
Initial conc. C mol L-1 0 0
At equilibrium C(1-α) Cα Cα Total=C(1+α)
∴i=C(1+α)C
⇒i=1+α
⇒α=i-1
=1.0753-1
=0.0753
Now, the value of Ka is given as:
Ka=[CH2FCOOH-][H+][CH2FCOOH]
=Cα.CαC(1-α)
=Cα21-α
Taking the volume of the solution as 500 mL, we have the concentration:
C=19.578500×1000 M
=0.5 M
Therefore, Ka=Cα21-α
=0.5×(0.0753)21-0.0753
=0.5×0.005670.9247
=0.00307 (approximately)
=3.07×10-3
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