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.
Molecular mass of CH2FCOOH
14 + 19 + 12 + 16 + 16 + 1 = 78 g/mol
Now, Moles of CH2FCOOH = 19.5 / 78
= 0.25
Taking the volume of the solution as 500 mL, we have the concentration:
C = (0.25 / 500) X 1000
Therefore Molality = 0.50m
So now putting the value in the formula :
ΔTf = Kf x m
=1.86 x 0.50 = 0.93K
Van’t hoff factor = observed freezing point depression / calculated freezing point depression
= 1 / 0.93 = 1.0753
Let α be the degree of dissociation of CH2FCOOH
Now total number of moles = m(1-a) + ma +ma = m(1+a)
Or
i = α(1+α) / α = 1 +α = 1.0753
Therefore α = 1.0753- 1
= 0.0753
Now the Value of Ka is given as:
Ka = [CH2FCOO-][H+] / CH2FCOOH
= (Cα x Cα) / (C (1-α))
= Cα2 / (1-α)
Ka = 0.5 X (0.0753)2 / (1-0.0753)
= 0.5 X 0.00567 / 0.09247
= 0.00307 (approx.)
= 3 X 10-3
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Value of c will be equals to molality
Great job
Great job
What a explanation
How the value of C is calculated for Ka...???
C can be calculated by c=n/v
How the value of 'C' is calculated for Ka value.
Great hard work.