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

Asked by Akshaya Parthasarathy | 10th Jul, 2011, 12:00: AM

Expert Answer:

 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

Answered by  | 11th Jul, 2011, 09:52: AM

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