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QUESTION 2.33

19.5 g of CH2FCOOH is dissolved in 500g of water. The depression in the freezing point of water observed in 1.0 C. Calculate the van't Hoff factor and dissociation constant of fluoroacetic acid.
Kf=1.86 K kg mol1


Solution

Step I: Calculation of van't Hoff factor of acid
ΔTf=1C=1K;Kf=1.86 K kg mol1ΔTf=iKfm
or f=ΔTfKfm(i)m=WBMB×WA=(19.5g)(78g mol1)×(0.5 kg)=0.5 mol kg1
Placing the values in Eq. (i), we find the value of van't Hoff factor (i)
i=1(1.86 K kg mol1)×(0.5 mol kg1)
=1.0753
Step II: Calculation of degree of dissociation of the acid
Suppose degree of dissociation at the given concentration is \alpha
CH2FCOOHaqCH2FCOO+H+Initial conc. C mol kg100At equilibriumC(1α)CαCα
Total =C(1+α)
i=C(1+α)C=1+αα=i1=1.07531=0.0753
Step III: Calculation of dissociation constant for the acid
(Molal)C = 0.5m    (From Eq. (ii))
Ka=[CH2FCOO][H+][CH2FCOOH]=Cα.CαC(1α)=Cα2(1α)Kα=(0.5)(0.0753)2(10.0753)=(0.5)×(0.0753)2(0.9247)=3.07×103

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