Question

# 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 mol−1

Solution

## Step I: Calculation of van't Hoff factor of acid ΔTf=1∘C=1K;Kf=1.86 K kg mol−1ΔTf=iKfm or f=ΔTfKfm……(i)m=WBMB×WA=(19.5g)(78g mol−1)×(0.5 kg)=0.5 mol kg−1 Placing the values in Eq. (i), we find the value of van't Hoff factor (i) i=1(1.86 K kg mol−1)×(0.5 mol kg−1) =1.0753 Step II: Calculation of degree of dissociation of the acid Suppose degree of dissociation at the given concentration is \alpha CH2FCOOHaq⇌CH2FCOO−+H+Initial conc. C mol kg−100At equilibriumC(1−α)CαCα Total =C(1+α) ∴i=C(1+α)C=1+αα=i−1=1.0753−1=0.0753 Step III: Calculation of dissociation constant for the acid (Molal)C = 0.5m    (From Eq. (ii)) Ka=[CH2FCOO−][H+][CH2FCOOH]=Cα.CαC(1−α)=Cα2(1−α)Kα=(0.5)(0.0753)2(1−0.0753)=(0.5)×(0.0753)2(0.9247)=3.07×10−3

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