19.5 g of CH2FCOOH 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.
It is given that:
$w_{1}=500 \mathrm{~g}$
$w_{2}=19.5 \mathrm{~g}$
$K_{f}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$
$\Delta T_{f}=1 \mathrm{~K}$
We know that:
$M_{2}=\frac{K_{f} \times w_{2} \times 1000}{\Delta T_{f} \times w_{1}}$
$=\frac{1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \times 19.5 \mathrm{~g} \times 1000 \mathrm{~g} \mathrm{~kg}^{-1}}{500 \mathrm{~g} \times 1 \mathrm{~K}}$
$=72.54 \mathrm{~g} \mathrm{~mol}^{-1}$
Therefore, observed molar mass of $\mathrm{CH}_{2} \mathrm{FCOOH},\left(M_{2}\right)_{\text {obs }}=72.54 \mathrm{~g} \mathrm{~mol}$
The calculated molar mass of $\mathrm{CH}_{2} \mathrm{FCOOH}$ is:
$\left(M_{2}\right)_{\text {cal }}=14+19+12+16+16+1$
$=78 \mathrm{~g} \mathrm{~mol}^{-1}$
Therefore, van't Hoff factor, $i=\frac{\left(M_{2}\right)_{\mathrm{cal}}}{\left(M_{2}\right)_{\mathrm{obs}}}$
$=\frac{78 \mathrm{~g} \mathrm{~mol}^{-1}}{72.54 \mathrm{~g} \mathrm{~mol}^{-1}}$
$=1.0753$
Let α be the degree of dissociation of
$\therefore i=\frac{C(1+\alpha)}{C}$
$\Rightarrow i=1+\alpha$
$\Rightarrow \alpha=i-1$
$=1.0753-1$
$=0.0753$
Now, the value of Ka is given as:
$K_{a}=\frac{\left[\mathrm{CH}_{2} \mathrm{FCOO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_{2} \mathrm{FCOOH}\right]}$
$=\frac{C \alpha \cdot C \alpha}{C(1-\alpha)}$
$=\frac{C \alpha^{2}}{1-\alpha}$
Taking the volume of the solution as 500 mL, we have the concentration:
$C=\frac{\frac{19.5}{78}}{500} \times 1000 \mathrm{M}$
$=0.5 \mathrm{M}$
Therefore, $K_{a}=\frac{C \alpha^{2}}{1-\alpha}$
$=\frac{0.5 \times(0.0753)^{2}}{1-0.0753}$
$=\frac{0.5 \times 0.00567}{0.9247}$
$=0.00307$ (approximately)
$=3.07 \times 10^{-3}$
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