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19.5 g of CH2FCOOH is dissolved in 500 g...

19.5 g of `CH_2FCOOH` 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.

Text Solution

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The correct Answer is:
`I =1.0753; K_(a) = 3.07 xx 10^(-3)`
`W_(B) = 19 . 5 g`
`W_(A) = 500 g = 0 . 5 kg`
Molar mass of
`(FCH_(2) CO OH)`
`M_(B) = 24 + 32 + 3 + 19`
`= 78 g mol^(-1)`
`K_(f) ("water ") = 1 . 86 K kg mol^(-1)`
`Delta T_(f) = 1^(@) C = 1 K`
`Delta T_(f) = i K_(f) xx (W_(B))/( M_(B)) xx (1)/( W_(A) (kg))`
` i = (Delta T_(f) xx M_(B) xx W_(A) (kg))/( W_(B) xx K_(f)`
`= (1 K xx 78 g mol^(-1) xx 0 . 5 kg)/( 19 . 5 g xx 1 . 86 K kg mol^(-1) )`
`= (78 xx 5)/( 195 xx 1. 86) = 1 . 0753`
Fluoroacetic acid dissociates as follows ,
`CH_(2) FCO OH (aq) hArr CH_(2) F CO O ^(-) (aq) + K^(+) (aq)`
`{:("Initial Conc C __ __" ),("Eqn. Cone . C" (1 - alpha) " " C alpha" " C alpha ):}`
` i = (C (1 - alpha ) + C alpha + C alpha)= 1 + alpha `
`:. 1 + alpha = 1 . 0753 `
`alpha = 0 . 0 753`
`K_(a) = ([FCH_(2) CO O^(-)H^(+)])/([CH_(2) F CO OH])`
`K_(a) = (C alpha * C alpha)/( C (1 - alpha)) = (C alpha^(2))/(1 - alpha)`
Here, `C = (n_(B))/(V) = (W_(B))/(M_(B)) xx (1)/(V)`
`= (1 9 . 5 g)/( 78 g mol^(-1)) xx (1) /(0 . 5 L)`
`= 0 . 5 mol L^(-1)` [Vol. of solution Mass of water]
`:. K_(a) = ( 0 . 5 xx (0 . 0753)^(2))/( 1 - 0 .0753) = (0 . 5 xx (0 . 0 753)^(2))/( 0 . 9247)`
`= 3 . 07 xx 10^(-3)`
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