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Question
Sun July 10, 2011

# 19.5g of fluoroacetic acid is dissolved in 500g of water the observed depression in freezing point is i degree celsius . calculate the vant hoff factor and dissociation constant of the acid given kf=1.86kkg/mol

Mon July 11, 2011
Number of moles of  fluoroacetic acid (CH2FCOOH)  = 19.5/78 =0.25
Therefore,  molality  = (0.25 x1000)/500
= 0.50 mole/kg

?T = Kf x m

= 1.86 x 0.50 = 0.93 K

Van't Hoff factor (i) = 1.0/0.93  =1.0753

Since fluoroacetic acid undergo dissociation as

CH2FCOOH  ---> CH2FCOO+ H+

So, ?  = (i -1) / (n-1)  where ? is the degree of dissociation

Here n = 2

?= 1.0753 - 1 =0.0753

Now

Ka = [ CH2FCOO] [ H+] / CH2FCOOH

where,

[ CH2FCOO] =  0.50 x 0.0753 =0.03765

[ H+] = 0.50 x 0.0753 =0.03765

[CH2FCOOH] = 0.50 (1 - 0.0753) = 0.462

Thus, Ka = 0.03765 x 0.03765/0.462

= 3.07 x 10-3

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# A solution X is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.8, then depression in freezing point of solution X is. ( Kf water = 1.86 and Kf ethanol = 2.0 ) 1. 4.34 K 2. 10.86 K 3. 5.434 K 4. 1.35 K

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