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Q1.33:
19.5 g of \(CH_{2}FCOOH\) is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.00 C. Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid.

Solution :

Step 1: Identifying Given Data and Physical Constants

Given parameters for fluoroacetic acid ($CH_{2}FCOOH$):

  • Mass of solute ($w_2$): $19.5\text{ g}$
  • Mass of solvent ($w_1$): $500\text{ g} = 0.5\text{ kg}$
  • Observed depression in freezing point ($\Delta T_f$): $1.00\text{ K}$
  • Molar mass of $CH_{2}FCOOH$ ($M_2$): $(12.01 \times 2) + (1.008 \times 3) + (19.00 \times 1) + (16.00 \times 2) \approx 78\text{ g mol}^{-1}$
  • Cryoscopic constant of water ($K_f$): $1.86\text{ K kg mol}^{-1}$ [Standard Physical Constant]

Step 2: Calculating Theoretical Molar Concentration (Molality)

The molality ($m$) is calculated by the ratio of moles of solute to the mass of solvent in kilograms:

$n_2 = \frac{w_2}{M_2} = \frac{19.5\text{ g}}{78\text{ g mol}^{-1}} = 0.25\text{ mol}$

$m = \frac{n_2}{w_1 (\text{in kg})} = \frac{0.25\text{ mol}}{0.5\text{ kg}} = 0.50\text{ m}$

Step 3: Determining the van’t Hoff Factor ($i$)

The relationship between observed depression and the theoretical calculation is given by $\Delta T_f = i \cdot K_f \cdot m$ [Per Colligative Property Theory].

First, calculate the theoretical freezing point depression ($\Delta T_f^{\text{theoretical}}$) assuming no dissociation:

$\Delta T_f^{\text{theoretical}} = K_f \cdot m = 1.86\text{ K kg mol}^{-1} \times 0.50\text{ mol kg}^{-1} = 0.93\text{ K}$

Now, calculate $i$:

$i = \frac{\Delta T_f^{\text{observed}}}{\Delta T_f^{\text{theoretical}}} = \frac{1.00\text{ K}}{0.93\text{ K}} \approx 1.0753$

Step 4: Calculating Degree of Dissociation ($\alpha$) and Dissociation Constant ($K_a$)

For the dissociation $CH_{2}FCOOH \rightleftharpoons CH_{2}FCOO^- + H^+$, the van't Hoff factor is $i = 1 + \alpha$ (for $n=2$ particles).

$\alpha = i - 1 = 1.0753 - 1 = 0.0753$

The dissociation constant $K_a$ is expressed as [Per Law of Mass Action]:

$K_a = \frac{C\alpha^2}{1 - \alpha}$

Where $C$ is the molarity (approximated to molality for dilute aqueous solutions, $C \approx 0.5\text{ M}$):

$K_a = \frac{0.5 \times (0.0753)^2}{1 - 0.0753} = \frac{0.5 \times 0.00567}{0.9247} \approx 3.07 \times 10^{-3}$

Summary of Variables

van't Hoff factor (i) = 1.075 Degree of dissociation (α) = 0.0753 Dissociation constant (Ka) = 3.07 × 10⁻³

Final Solution: The van’t Hoff factor is 1.075 and the dissociation constant is $3.07 \times 10^{-3}$.


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