For an aqueous solution of HF, determine the van’t Hoff factor assuming… A solution is made by dissolving 0.0300 mol of HF in 1.00 kg of water. The solution was found to freeze at -0.0644 degree C. Calculate the value of i and estimate the percent ionization of HF in this solution.
The Correct Answer and Explanation is :
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Step 1: Freezing Point Depression Equation
To calculate the van’t Hoff factor (i) and percent ionization of HF, we will use the freezing point depression formula:
[
\Delta T_f = i \cdot K_f \cdot m
]
Where:
- (\Delta T_f) is the freezing point depression, which is the difference between the freezing point of pure water (0°C) and the freezing point of the solution.
- (i) is the van’t Hoff factor, which represents the number of particles the solute dissociates into.
- (K_f) is the cryoscopic constant (freezing point depression constant) for water, which is 1.86°C·kg/mol.
- (m) is the molality of the solution, defined as moles of solute per kilogram of solvent.
Step 2: Calculate Freezing Point Depression
The freezing point of pure water is 0°C. The solution freezes at -0.0644°C, so the freezing point depression is:
[
\Delta T_f = 0°C – (-0.0644°C) = 0.0644°C
]
Step 3: Determine Molality
The molality (m) is given by:
[
m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.0300 \text{ mol}}{1.00 \text{ kg}} = 0.0300 \text{ mol/kg}
]
Step 4: Use the Freezing Point Depression Equation
Now, we can plug values into the equation:
[
0.0644°C = i \cdot 1.86 \cdot 0.0300
]
Solve for (i):
[
i = \frac{0.0644}{1.86 \cdot 0.0300} = \frac{0.0644}{0.0558} = 1.15
]
Step 5: Estimate Percent Ionization
HF is a weak acid and only partially dissociates in water. The van’t Hoff factor (i) corresponds to the number of ions in solution. For HF, the dissociation is:
[
\text{HF} \rightleftharpoons \text{H}^+ + \text{F}^-
]
For 100% ionization, (i = 2) (one HF molecule dissociates into two ions). The observed value of (i = 1.15) suggests partial dissociation. The percent ionization can be calculated by comparing the observed (i) to the expected value for full dissociation:
[
\text{Percent Ionization} = \left( \frac{i – 1}{1} \right) \times 100 = (1.15 – 1) \times 100 = 15\%
]
Conclusion
- The van’t Hoff factor (i) is 1.15.
- The percent ionization of HF in the solution is approximately 15%.