Question
I'm studying chemistry and I need help solving a problem involving a weak acid. My lab partner mentioned that a 0.1 molal solution of this acid lowers the freezing point of water by 0.193°C. Can you find the Ka value for this acid using the information on this webpage?
Answers
GPT-4o mini
To find the \( K_a \) value of the weak acid using the information provided, we need to follow a few steps.
**Step 1: Calculate the van't Hoff factor (i)**
The freezing point depression (\( \Delta T_f \)) is related to the molality (\( m \)) of the solution and the van't Hoff factor (\( i \)) by the equation:
\[
\Delta T_f = i \cdot K_f \cdot m
\]
Where:
- \( \Delta T_f \) is the change in freezing point (0.193°C),
- \( K_f \) is the freezing point depression constant for water (which is approximately 1.86 °C kg/mol),
- \( m \) is the molality of the solution (0.1 molal).
Now we can rearrange this formula to find \( i \):
\[
i = \frac{\Delta T_f}{K_f \cdot m}
\]
Substituting the values:
\[
i = \frac{0.193°C}{1.86°C \cdot 0.1\, \text{molal}} = \frac{0.193}{0.186} \approx 1.037
\]
**Step 2: Determine the degree of ionization (\( \alpha \))**
The van't Hoff factor can also be expressed in terms of the degree of ionization (\( \alpha \)) for a weak acid, where the base form of the acid is HA and it partially ionizes into H⁺ and A⁻:
\[
i = 1 + \alpha
\]
From our \( i \) value,
\[
1.037 = 1 + \alpha \\
\alpha = 0.037
\]
This means that approximately 3.7% of the acid has ionized.
**Step 3: Relate the degree of ionization \( \alpha \) to \( K_a \)**
For a weak acid dissociation represented as:
\[
HA \rightleftharpoons H^+ + A^-
\]
The equilibrium concentration expressions relate to \( K_a \):
\[
K_a = \frac{[H^+][A^-]}{[HA]}
\]
In a 0.1 molal solution, if \( \alpha \) is the fraction that ionizes, then at equilibrium:
\[
[H^+] = [A^-] = 0.1 \cdot \alpha = 0.1 \cdot 0.037 = 0.0037\, \text{mol/kg} \quad \text{(both products)}
\]
\[
[HA] = 0.1 - 0.1 \cdot \alpha = 0.1 - 0.0037 = 0.0963\, \text{mol/kg}
\]
Now substituting these values into the \( K_a \) expression:
\[
K_a = \frac{(0.0037)(0.0037)}{0.0963} = \frac{0.00001369}{0.0963} \approx 0.0001427
\]
Thus, the \( K_a \) value for the weak acid is approximately:
\[
K_a \approx 1.43 \times 10^{-4}
\]
This value represents the strength of the weak acid in terms of its dissociation in water.
**Step 1: Calculate the van't Hoff factor (i)**
The freezing point depression (\( \Delta T_f \)) is related to the molality (\( m \)) of the solution and the van't Hoff factor (\( i \)) by the equation:
\[
\Delta T_f = i \cdot K_f \cdot m
\]
Where:
- \( \Delta T_f \) is the change in freezing point (0.193°C),
- \( K_f \) is the freezing point depression constant for water (which is approximately 1.86 °C kg/mol),
- \( m \) is the molality of the solution (0.1 molal).
Now we can rearrange this formula to find \( i \):
\[
i = \frac{\Delta T_f}{K_f \cdot m}
\]
Substituting the values:
\[
i = \frac{0.193°C}{1.86°C \cdot 0.1\, \text{molal}} = \frac{0.193}{0.186} \approx 1.037
\]
**Step 2: Determine the degree of ionization (\( \alpha \))**
The van't Hoff factor can also be expressed in terms of the degree of ionization (\( \alpha \)) for a weak acid, where the base form of the acid is HA and it partially ionizes into H⁺ and A⁻:
\[
i = 1 + \alpha
\]
From our \( i \) value,
\[
1.037 = 1 + \alpha \\
\alpha = 0.037
\]
This means that approximately 3.7% of the acid has ionized.
**Step 3: Relate the degree of ionization \( \alpha \) to \( K_a \)**
For a weak acid dissociation represented as:
\[
HA \rightleftharpoons H^+ + A^-
\]
The equilibrium concentration expressions relate to \( K_a \):
\[
K_a = \frac{[H^+][A^-]}{[HA]}
\]
In a 0.1 molal solution, if \( \alpha \) is the fraction that ionizes, then at equilibrium:
\[
[H^+] = [A^-] = 0.1 \cdot \alpha = 0.1 \cdot 0.037 = 0.0037\, \text{mol/kg} \quad \text{(both products)}
\]
\[
[HA] = 0.1 - 0.1 \cdot \alpha = 0.1 - 0.0037 = 0.0963\, \text{mol/kg}
\]
Now substituting these values into the \( K_a \) expression:
\[
K_a = \frac{(0.0037)(0.0037)}{0.0963} = \frac{0.00001369}{0.0963} \approx 0.0001427
\]
Thus, the \( K_a \) value for the weak acid is approximately:
\[
K_a \approx 1.43 \times 10^{-4}
\]
This value represents the strength of the weak acid in terms of its dissociation in water.