Question

Calculate the approximate freezing point of a solution of \(162 \mathrm{~g}\) of \(\mathrm{HBr}\) in \(500 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}\), assuming that the acid is \(90 \%\) Ionized.

Short answer

The approximate freezing point of the solution is given by \(T_{f} \approx 273.15\mathrm{~K} - \Delta T_{f}\), where \(\Delta T_{f} = 1.86\mathrm{~K\cdot kg/mol} \cdot \frac{162\mathrm{~g}}{80.912\mathrm{~g/mol} \cdot 0.5\mathrm{~kg}} \cdot (1 + (0.9 - 1)(2))\). Calculate the freezing point depression and subtract it from \(273.15\mathrm{~K}\) to find the approximate freezing point of the solution.

Step-by-step solution

Calculate the moles of HBr

First, we need to convert the mass of \(\mathrm{HBr}\) to moles. The molar mass of \(\mathrm{HBr}\) is: \(M_{\mathrm{HBr}} = M_\mathrm{H} + M_\mathrm{Br} = 1.00784 \mathrm{~g/mol} + 79.904\mathrm{~g/mol} = 80.912\mathrm{~g/mol}\) Now, we can calculate the moles of \(\mathrm{HBr}\): \(n_\mathrm{HBr} = \frac{162\mathrm{~g}}{80.912\mathrm{~g/mol}}\)

Calculate the molality of the solution

Now, we need to calculate the molality of the solution. Recall that the molality is defined as the moles of solute per kilogram of solvent. Converting grams of water to kilograms, we have: \(m = \frac{n_\mathrm{HBr}}{0.5\mathrm{~kg}}\)

Determine the van 't Hoff factor

As we are given that the acid is \(90 \%\) ionized, we need to calculate the van 't Hoff factor (\(i\)). For a weak acid, \(i\) can be calculated as: \(i = 1 + (\alpha - 1)n_{i}\) where \(\alpha\) is the degree of ionization and \(n_{i}\) is the number of ions produced per formula unit. In this case, \(\mathrm{HBr}\) dissociates into two ions, \(\mathrm{H}^{+}\) and \(\mathrm{Br}^{-}\), so: \(i = 1 + (0.9 - 1)(2)\)

Calculate the freezing point depression

Now that we have all the values needed, we can calculate the freezing point depression using the formula: \(\Delta T_{f} = K_{f} \cdot m \cdot i\) For water, the molal freezing point depression constant \(K_f\) is \(1.86\mathrm{~K\cdot kg/mol}\). Plugging in the values calculated earlier: \(\Delta T_{f} = 1.86\mathrm{~K\cdot kg/mol} \cdot \frac{162\mathrm{~g}}{80.912\mathrm{~g/mol} \cdot 0.5\mathrm{~kg}} \cdot (1 + (0.9 - 1)(2))\)

Calculate the approximate freezing point

Finally, we need to subtract the freezing point depression from the normal freezing point of water, which is \(0^\circ\mathrm{C}\) or \(273.15\mathrm{~K}\): \(T_{f} \approx 273.15\mathrm{~K} - \Delta T_{f}\) Calculate the final answer and express it in either Celsius or Kelvin, depending on the desired unit.

Key concepts

Molality

Molality is a measure of the concentration of a solution, specifically the amount of solute per kilogram of solvent. It differs from molarity, which is solute per liter of solvent. Molality is useful in colligative properties because it is temperature-independent, making calculations more consistent.

To find molality, calculate the moles of the solute, and then divide by the mass of the solvent in kilograms. In the given exercise, we convert 162 g of HBr into moles, using its molar mass of approximately 80.912 g/mol.

Once we have the moles, we divide by the solvent's mass (0.5 kg of water) to find molality. This step enables us to proceed with calculating colligative properties such as freezing point depression.

Van 't Hoff Factor

The Van 't Hoff factor (\(i\)) is essential in colligative properties. It accounts for the number of particles a compound forms in solution. This factor modifies the concentration term to simulate the real effects on properties like boiling point elevation and freezing point depression.

For ionic compounds, this factor depends on the degree of ionization. In our exercise, HBr ionizes 90% in water, forming H⁺ and Br⁻ ions, which gives a total of 2 ions per molecule when completely ionized.

Mathematically, the Van 't Hoff factor can be determined using the formula: \[i = 1 + (\alpha - 1)n_{i}\] where \(\alpha\) is the degree of ionization, and\(n_{i}\) is the number of ions per molecule. For HBr, this results in a Van 't Hoff factor less than 2, lowered due to incomplete ionization.

Ionization

Ionization involves the process by which a molecule forms ions, mostly occurring when dissolved in water. In our example, HBr, a strong acid, ionizes into hydrogen (H⁺) and bromide (Br⁻) ions, with a 90% ionization rate.

This percentage signifies how much of the original HBr becomes ions in the solution. Knowing the degree of ionization helps adjust the expected impact of the solution on boiling or freezing points since it predicts the actual count of dissolved particles.

Understanding ionization is crucial for calculating the Van 't Hoff factor accurately, as it impacts how much the freezing point of a solution is lowered in colligative property calculations.

Colligative Properties

Colligative properties depend solely on the number of solute particles in a solution, not their identity. These properties include boiling point elevation, lowering of vapor pressure, osmotic pressure, and crucially for our exercise, freezing point depression.

The freezing point depression occurs because solute particles interfere with the formation of the solid phase, requiring more energy (or lower temperature) to freeze. In the exercise, this effect was demonstrated using the formula: \[\Delta T_{f} = K_{f} \cdot m \cdot i\]where \(\Delta T_{f}\) is the freezing point depression, \(K_{f}\) is the cryoscopic constant of the solvent, \(m\) is the molality, and \(i\) is the Van 't Hoff factor.

Understanding colligative properties helps explain why adding substances like salt to water decreases its freezing point, a principle that extends to solutions like the one described in the exercise.