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)
$=3.07 \times 10^{-3}$