`19.5g` of `CH_(2)FCOOH` is dissolved in `500g ` 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

It is given that:
`w_(1) = 500g`
`w_(2) = 19.5 g`
`K_(f) = 1.86 K kg mol^(-1)`
`DeltaT_(f) = 1K`
We know that:
`M_(2) = (K_(f) xx w_(2) xx 1000)/(DeltaT_(f) xx w_(1)) = (1.86 K kg mol^(-1) xx 19.5 g xx 1000g kg^(-1))/(500g xx 1K) = 72.54 mol^(-1)`
Therefore observed molar mass of `CH_(2)FCOOH,(M_(2))_("obs") = 72.54 mol`
The calculated molar mass of `CH_(2)FCOOH` is:
`(M_(2))_(ce) = 14 +19 +12 +16 +16 +1 = 78 g 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 `alpha` be the degree of dissociation of `CH_(2)FCOOH`
`{:(,CH_(2)FCOOH,harrCH_(2)FCOO^(-),+H^(+)),("Initial conc.",C mol L^(-1),0,0),("At equilibrium",C(1-alpha),Calpha,Calpha Total =C(1+alpha)):}`
`:. i= (C(1+alpha))/(C)`
`rArr i=1 +alpha rArr 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)FCOOH]) =(Calpha.Calpha)/(C(1-alpha)) =(Calpha^(2))/(1-alpha)`
Taking the volume of the solution as `500mL`. We have the concentration: `19.5M`
`C = (78)/(500) xx 1000M = 0.5M`
Therefore, `K_(a) = (Calpha^(2))/(1-alpha)`
`= (0.5xx(0.0753)^(2))/(1-0.0753) = (0.5 xx 0.00567)/(0.9247) = 0.00307` (approximately) `= 3.07 xx 10^(-3)`