Question

$19.5g$ of $C{H}_{2}FCOOH$ is dissolved in $500g$ of water. The depression in the freezing point of water observed is ${1.00}^{0}C$. Calculate the van't Hoff factor and dissociation constant of fluoroacetic acid.

Answer 1

Fluoroacetic acid has a molecular mass of $78$ g/mol.

Number of moles of fluoroacetic acid is $\frac{19.5}{78}=0.25$.

Molality is the number of moles of solute in 1 kg of solvent.

$\text{Molality}=\frac{0.25}{\frac{500}{1000}}=0.50m$

Calculated depression in the freezing point.

$\Delta T_f=K_f\times m=1.86\times 0.50=0.93$ K

Van't Hoff factor is the ratio of observed freezing point depression to calculated freezing point depression.

$i=\frac{1.0}{0.93}=1.0753$

Let $c$ be the initial concentration of fluoro acetic acid and $\alpha$ be its degree of dissociation.

$\underset{c(1-\alpha )}{C{H}_{2}FCOOH}\to \underset{c\alpha }{C{H}_{3}FCO{O}^{-}}+\underset{c\alpha }{H^{+}}$

Total number of moles $=c(1-\alpha )+c\alpha +c\alpha =c(1+\alpha )$

$i=\frac{c(1+\alpha )}{c}=1+\alpha =1.0753$

$\alpha =0.0753$

$\left[C{H}_{2}FCO{O}^{-}\right]=\left[H^{+}\right]=c\alpha =0.50\times 0.0753=0.03765$

$\left[C{H}_{2}FCOOH\right]=c(1-\alpha )=0.50(1-0.0753)=0.462$

$K_a=\frac{\left[C{H}_{2}FCO{O}^{-}\right]\left[H^{+}\right]}{\left[C{H}_{2}FCOOH\right]}$

$K_a=\frac{0.03765\times 0.03765}{0.462}$

$K_a=3.07\times {10}^{-3}$

Hence, the van't Hoff factor is 1.0753 and dissociation constant is $3.07\times {10}^{-3}$ for fluoroacetic acid.