Question

19.5 g of \(CH_2 FCOOH\) is dissolved in 500 g of water. The depression in the freezing point of water observed is \(1.00\degree C\). Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid.

Answer 1

The correct answer is: \(3.07 \times 10^{-3}\)
It is given that:
\(w_₁ = 500 g\)
\(w_₂ =19.5g\)
\(K_f= 1.86 K\, kg mol^{-1}\)
\(ΔT_f=1K\)
We know that:
\(M_2= \frac{K_f \times w_2 \times 1000}{ ΔT_f \times w_1}\)
\(=\frac{1.86K\, kg mol^{-1} \times 19.5g \times 1000 g kg^{-1}}{500g \times 1K}\)
\(= 72.54 g mol^{-1}\)
Therefore, observed molar mass of \(CH_2FCOOH,(M_2)obs=72.54gmol\)
The calculated molar mass of \(CH_2FCOOH\) is
\((M_2)cal=14+19+12+16 +16 +1\)
\(= 78 g mol^{-1}\)
Therefore, van't Hoff factor. \(i=\frac{(M_2)cal}{(M_2)obs}\)
\(\frac{78 g mol^{-1}}{72.54 g mol^{-1}}\)
= 1.0753
Let abe the degree of dissociation of \(CH_2FCOOH\)
\(CH_2FCOOH↔CH_2FCOO^- +H^+\)
Initial conc.         \(Cmol^{-1} \)         0          0
At equilibrium   \(C(1-α)\)      \(Cα\)       \(Cα\)    Total\(=C(1+α)\)
\(∴i=\frac{C(1+α) }{C}\)
\(⇒i=1+α\)
\(⇒α=i-1\)
=1.0753-1
=0.0753
Now, the value of \(K_a\) is given as:
\(K_a=\frac{[CH_2FCOO^-][H^+]}{[CH^2FCOOH]}\)
\(=\frac{Cα.Cα}{C(1-α)}\) 
\(=\frac{Cα^2}{1-α}\)
Taking the volume of the solution as 500 mL, we have the concentration:
\(C=\frac{\frac{19.5}{78}}{500} \times 1000M\)
=0.5M
Therefore, \(K_a=\frac{Cα^2}{1-α}\)
\(=\frac{0.5 \times (0.0753)^2}{1-0.0753}\)
\(=\frac{0.5 \times 0.00567}{0.9247}\)
=0.00307(approximately)
\(=3.07 \times 10^{-3}\)