Question

A $19.5g$ of $C{H_2}FCOOH$ is dissolved in $500g$ of water. The depression in the freezing point of water observed is $1.00^\circ C$ . Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid.

Answer 1

Depression in freezing point is the lowering of the freezing point of the pure solvent when there is addition of some non-volatile solute in the solvent. Here, the pure solvent is water and non-volatile solute is fluoroacetic acid.
Formula used
Freezing point depression of a pure solvent:
$\Delta {T_f} = i{K_f}m$
Where, $\Delta {T_f}$ is depression in freezing point, $i$ is van’t Hoff factor, ${K_f}$ is freezing point constant and $m$ is the molality.
Molality:
$m = \dfrac{n}{W}$
Where, $n$ is the number of moles of solute, $W$ is the weight of the solvent and $m$ is molality.
Van’t Hoff factor relation with degree of dissociation,
$i = 1 + \alpha$
Where $i$ is the Van't Hoff factor and $\alpha$ is the degree of dissociation.
Dissociation constant, ${K_a} = \dfrac{{{{\text{Concentration of products}}}}}{{{{\text{Concentration of reactants}}}}}$

Complete step-by-step answer
We start with the definition of Freezing point depression:
Freezing point depression: The freezing point depression is the lowering of the freezing point of the pure solvent when there is addition of some non-volatile solute in the solvent.
Given data in the question:
Weight of solute $= 19.5g$
Weight of solvent $= 500g$
Depression in freezing point, $\Delta {T_f} = 1.00^\circ C$
First we have to find the $i$ and then dissociation constant.
The molecular mass of $C{H_2}FCOOH$ is $78$
For finding the $i$ we have to find the molality of the solution. As the weight of solute and the solvent is given we can find out the molality, but before that we have to find the number of moles.
So, now calculating the number of moles,
$n = \dfrac{{{{\text{Given mass}}}}}{{{{\text{Molecular mass}}}}} = \dfrac{{19.5}}{{78}} = 0.25$
Hence, the number of moles present in the solution is $0.25$ .
So, now we can find out the molality of the solution.
Molality, $m = \dfrac{n}{W}$
Putting the number of moles of solute and weight of solvent.
$m = \dfrac{{0.25 \times 1000}}{{500}} = 0.5$
Hence, the molality of the given solution will be $0.5$ .
Now, we are able to find the van’t Hoff factor using the formula of freezing point depression,
$\Delta {T_f} = i{K_f}m$
As we know, ${K_f}$ of water is $1.86$
$\Rightarrow i = \dfrac{{\Delta {T_f}}}{{{K_f}m}} = \dfrac{{1.00}}{{1.86 \times 0.5}} = 1.075$
So, the van’t Hoff factor is $1.075$ .
Now, we have to calculate the degree of dissociation to find the dissociation constant,
Let the initial concentration be $c$ and degree of dissociation be $\alpha$
$C{H_2}FCOOH \to C{H_2}FCO{O^ - } + {H^ + }$
So, the concentration of the reactants after the dissociation,
$c(1 - \alpha ) \to c\alpha + c\alpha$
The total number of moles are $c - c\alpha + c\alpha + c\alpha = c + c\alpha$
Now, finding the value of $\alpha$
$i = 1 + \alpha$
$\Rightarrow \alpha = i - 1 = 1.075 - 1 = 0.075$
So, now, finding the dissociation constant,
${K_a} = \dfrac{{{{\left[ {C{H_2}FCOO} \right]}^ - }{{\left[ H \right]}^ + }}}{{\left[ {C{H_2}FCCOH} \right]}} = \dfrac{{c\alpha \times c\alpha }}{{c(1 - \alpha )}} = \dfrac{{0.5 \times 0.075 \times 0.075}}{{1 - 0.075}}$
$\Rightarrow {K_a} = \dfrac{{0.0028}}{{0.925}} = 3.04 \times {10^{ - 3}}$
Hence, the dissociation constant is $3.04 \times {10^{ - 3}}$ .
Hence, the Van’t Hoff factor of the solution is $1.075$ and the dissociation constant of the solution is $3.04 \times {10^{ - 3}}$ .

Note
The dissociation constant is the ratio of the concentration of the products to the concentrations of the reactants at that time and the concentration at that time can be calculated by multiplying the initial concentration with the degree of dissociation.