Question

QUESTION 2.33

19.5 g of $C{H}_{2}FCOOH$ is dissolved in 500g of water. The depression in the freezing point of water observed in 1.0 C. Calculate the van't Hoff factor and dissociation constant of fluoroacetic acid.
$K_f=1.86Kkgmo{l}^{-1}$

Answer 1

Step I: Calculation of van't Hoff factor of acid
$\Delta T_f=1^{\circ }C=1K;K_f=1.86Kkgmo{l}^{-1}\Delta T_f=i{K}_{f}m$
or $f=\frac{\Delta T_f}{K_{f}m}\dots \dots \left(i\right)m=\frac{W_B}{M_B\times W_A}=\frac{\left(19.5g\right)}{\left(78gmo{l}^{-1}\right)\times \left(0.5kg\right)}=0.5molk{g}^{-1}$
Placing the values in Eq. (i), we find the value of van't Hoff factor (i)
$i=\frac{1}{\left(1.86Kkgmo{l}^{-1}\right)\times \left(0.5molk{g}^{-1}\right)}$
$=1.0753$
Step II: Calculation of degree of dissociation of the acid
Suppose degree of dissociation at the given concentration is \alpha
$\begin{array}{cccc}& C{H}_{2}FCOOH\stackrel{aq}{\rightleftharpoons }& C{H}_{2}FCO{O}^{-}& +H^{+}\\ Initialconc.Cmolk{g}^{-1}& & 0& 0\\ Atequilibrium& C(1-\alpha )& C\alpha & C\alpha \end{array}$
Total $=C(1+\alpha )$
$\therefore i=\frac{C(1+\alpha )}{C}=1+\alpha \alpha =i-1=1.0753-1=0.0753$
Step III: Calculation of dissociation constant for the acid
(Molal)C = 0.5m (From Eq. (ii))
$K_a=\frac{\left[C{H}_{2}FCO{O}^{-}\right]\left[H^{+}\right]}{\left[C{H}_{2}FCOOH\right]}=\frac{C\alpha .C\alpha }{C(1-\alpha )}=\frac{C{\alpha }^2}{(1-\alpha )}{K}_{\alpha }=\frac{\left(0.5\right)(0.0753{)}^2}{(1-0.0753)}=\frac{\left(0.5\right)\times (0.0753{)}^2}{\left(0.9247\right)}=3.07\times {10}^{-3}$