Question

When 9.45 g of CICH2COOH is added to 500 mL of water, its freezing point drops by 0.5°C. The dissociation constant of CICH2COOH is $x \times 10^{-3}$. The value of x is ____. [$K_{f(H_2O)} = 1.86 K kg mol^{-1}$]

Answer 1

36

Answer 2

The problem asks for the dissociation constant of chloroacetic acid (ClCH2COOH) given the freezing point depression of its aqueous solution.

First, calculate the molality of the solution. The molar mass of ClCH2COOH is $12.01 + 2(1.008) + 35.45 + 12.01 + 2(16.00) + 1.008 = 94.504$ g/mol. Using standard atomic masses (C=12, H=1, Cl=35.5, O=16), the molar mass is $12 + 2(1) + 35.5 + 12 + 2(16) + 1 = 94.5$ g/mol.

Number of moles of ClCH2COOH = $\frac{\text{mass}}{\text{molar mass}} = \frac{9.45 \text{ g}}{94.5 \text{ g/mol}} = 0.1 \text{ mol}$.

The volume of water is 500 mL. Assuming the density of water is 1 g/mL, the mass of water is 500 g = 0.5 kg.

Molality ($m$) = $\frac{\text{moles of solute}}{\text{mass of solvent (kg)}} = \frac{0.1 \text{ mol}}{0.5 \text{ kg}} = 0.2 \text{ mol/kg}$.

Next, use the freezing point depression formula to find the van't Hoff factor ($i$). $\Delta T_f = i \cdot K_f \cdot m$

Given $\Delta T_f = 0.5^\circ C$ and $K_f = 1.86 K kg mol^{-1}$.

$0.5 = i \cdot 1.86 \cdot 0.2$

$0.5 = i \cdot 0.372$

$i = \frac{0.5}{0.372} \approx 1.344086$

Chloroacetic acid is a weak acid that dissociates in water according to the equilibrium:

ClCH2COOH $\rightleftharpoons$ ClCH2COO$^-$ + H$^+$

Let $\alpha$ be the degree of dissociation. For this type of dissociation (1 molecule dissociating into 2 ions), the van't Hoff factor $i$ is related to $\alpha$ by the equation $i = 1 + \alpha$.

So, $\alpha = i - 1 = 1.344086 - 1 = 0.344086$.

Now, calculate the acid dissociation constant ($K_a$) for ClCH2COOH. The equilibrium concentrations (or more precisely, molalities) are:

[ClCH2COOH] = $m(1-\alpha)$

[ClCH2COO$^-$] = $m\alpha$

[H$^+$] = $m\alpha$

where $m$ is the initial molality (0.2 mol/kg).

The dissociation constant is given by: $K_a = \frac{[ClCH_2COO^-][H^+]}{[ClCH_2COOH]} = \frac{(m\alpha)(m\alpha)}{m(1-\alpha)} = \frac{m\alpha^2}{1-\alpha}$

Substitute the values: $m = 0.2$ mol/kg and $\alpha = 0.344086$.

$K_a = \frac{0.2 \times (0.344086)^2}{1 - 0.344086}$

$K_a = \frac{0.2 \times 0.118405}{0.655914}$

$K_a = \frac{0.023681}{0.655914} \approx 0.036100$

The question asks for the dissociation constant in the form $x \times 10^{-3}$.

$K_a \approx 0.036100 = 36.100 \times 10^{-3}$.

So, $x \approx 36.100$.

Rounding off to the nearest integer, the value of x is 36.