19.5 g of `CH_(2)FCOOH` is dissolved in 500 g of water. The depression in the freezing point of water observed is `1.0^(@)C`. Calculate the van't Hoff factor and dissociation constant of fluoroacetic acid.

Explanation

`w_(1)=500 g " " w_(2)=19.5 g`
`K_(f)=1.86" K kg mol"^(-1) " " Delta T_(f)=1 K`
We known that :
`M_(2)=(K_(f)xx w_(B)xx 1000)/(Delta K_(f)xx w_(1))`
`= (1.86" K kg mol"^(-1)xx19.5 g xx 1000 g kg^(-1))/(500 g xx 1K)`
`= 72.54 mol^(-1)`
Therefore, observed molar mass of `CH_(2)FCOOH, (M_(2))_(obs)=72.54` g mol
The calculated mass of `CH_(2)FCOOH` is :
`(M_(2))_(cal)=14+19+12+16+16+1`
`= 76 mol^(-1)`
Therefore, van.t Hoff factor,
`i=((M_(2))_(cal))/((M_(2))_(obs))`
`= (78 g mol^(-1))/(72.54 g mol^(-1))=1.0753`
Let abe the degree of dissociation of
`{:("At",CH_(2)FCOOH,hArr,CH_(2)FCOO^(-),+,H^(+)),("equilibrium","C mol L"^(-1),,O,,O),(,C(1-alpha),,C alpha,,C alpha):}`
Total `= C(1-alpha)`
`i=1+alpha`
`alpha =i-1=1.0753-1`
= 0.0753
Now, the value of `K_(a)` is given as :
`K_(a)=([CH_(2)FCOO^(-)][H^(+)])/([CH_(2)FCOO])`
`=(C alpha C alpha)/(C(1-alpha))=(C alpha^(2))/(1-alpha)`
Taking the volume of the solution as 500 mL, we have the concentration :
`C=((19.5)/(78))/(500)xx1000 M`
= 0.5 M
Therefore,
`K_(a)=(C alpha^(2))/(1-alpha)`
`= (0.5(0.0753)^(2))/(1-0.0753)`
`= (0.5xx0.00567)/(0.9247)`
= 0.00307 (approximately)
`= 3.07xx10^(-3)`