Question

Formic acid (HCO_A) is a monoprotic acid that ionizes only partially in aqueous solutions. A 0.10-M formic acid solution is \(4.2 \%\) ionized. Assuming that the molarity and molality of the solution are the same, calculate the freezing point and the boiling point of 0.10 \(M\) formic acid.

Short answer

The freezing point of the 0.10 M formic acid solution is -0.372 °C, and the boiling point is 100.1024 °C.

Step-by-step solution

Calculate the molality of the solution

We are given that the molarity and molality are the same for this solution, so the molality is also 0.10 mol/kg.

Determine the molality of the ionized formic acid

Since the solution is 4.2% ionized, we can calculate the molality of the ionized formic acid with the following formula: Molality of ionized formic acid = (molality of formic acid) * (% ionization) = 0.10 mol/kg * 0.042 = 0.0042 mol/kg

Calculate the total molality of the solution

Since the solution is partially ionized, both the unionized formic acid and its ions contribute to the changes in freezing and boiling points. We can find the total molality by adding the molality of ionized formic acid to the molality of unionized formic acid: Molality of unionized formic acid = initial molality - molality of ionized formic acid = 0.10 mol/kg - 0.0042 mol/kg = 0.0958 mol/kg Total molality = molality of unionized formic acid + molality of ionized formic acid = 0.0958 mol/kg + 0.0042 mol/kg = 0.1 mol/kg

Calculate the freezing point depression and boiling point elevation

We will use the following formulas for calculating freezing point depression and boiling point elevation: ΔTf = Kf * m * i ΔTb = Kb * m * i Where ΔTf is the freezing point depression, ΔTb is the boiling point elevation, Kf is the freezing point depression constant, Kb is the boiling point elevation constant, m is the molality, and i is the van't Hoff factor (which is 2 for formic acid, as it forms 1 H+ ion and 1 HCO2- ion upon ionization) For water, Kf = 1.86 °C kg/mol and Kb = 0.512 °C kg/mol. We have the total molality and van't Hoff factor, so we can plug these values into the formulas: ΔTf = 1.86 °C kg/mol * 0.1 mol/kg * 2 ΔTf = 0.372 °C ΔTb = 0.512 °C kg/mol * 0.1 mol/kg * 2 ΔTb = 0.1024 °C

Calculate the freezing and boiling points of the solution

Now that we have the freezing point depression and boiling point elevation, we can find the freezing point and boiling point of the formic acid solution by subtracting ΔTf and adding ΔTb to the normal freezing and boiling points of water: Freezing point = normal freezing point of water - ΔTf = 0 °C - 0.372 °C = -0.372 °C Boiling point = normal boiling point of water + ΔTb = 100 °C + 0.1024 °C = 100.1024 °C Therefore, the freezing point of the 0.10 M formic acid solution is -0.372 °C, and the boiling point is 100.1024 °C.

Key concepts

Freezing Point Depression

Freezing point depression is a fascinating colligative property. It refers to the lowering of the freezing point of a solvent when a non-volatile solute is dissolved in it. This happens because the solute particles disrupt the solvent's ability to form a solid structure, requiring a lower temperature to achieve freezing.
In our exercise, formic acid in water demonstrates this phenomenon. Since formic acid ionizes in solution, it increases the number of particles, further lowering the freezing point aside from mere dissolution effects.
The formula to calculate freezing point depression is:

• ΔTf = Kf \( \cdot \) m \( \cdot \) i

Where:

• ΔTf is the change in freezing point.
• Kf is the freezing point depression constant specific to the solvent.
• m is the molality of the solution.
• i is the van’t Hoff factor, which represents the number of particles the solute forms in solution.

In this case, the freezing point was lowered by 0.372°C. Remember, this value needs to be subtracted from the normal freezing point of water, 0°C, giving us -0.372°C.

Boiling Point Elevation

Boiling point elevation is another intriguing colligative property. It occurs when the boiling point of a solvent is increased by the addition of a solute. Like freezing point depression, this is due to solute particles disrupting the solvent characteristics, requiring a higher temperature for vaporization.
For the formic acid solution in question, we calculate how much the boiling point is raised using the formula:

• ΔTb = Kb \( \cdot \) m \( \cdot \) i

Where:

• ΔTb is the change in boiling point.
• Kb is the boiling point elevation constant for the solvent.
• m is the solution's molality.
• i is the van’t Hoff factor.

In our example, the boiling point of the solution is elevated by 0.1024°C. This is added to water's normal boiling point of 100°C, resulting in a new boiling point of 100.1024°C. This adjustment happens because more energy is needed for solvent molecules to escape into the vapour phase when solute particles are present.

Ionization

Ionization is the process by which a molecule splits into ions, often when dissolved in a solvent like water. Some compounds, like formic acid, only ionize partially in solution, making them weak electrolytes.
In the exercise, we see that formic acid is 4.2% ionized. It means that for every mole of formic acid dissolved, only 4.2% converts into ions: H+ and HCO2-. This conversion to ions increases the number of solute particles, which is crucial for colligative properties like freezing point depression and boiling point elevation.
Measuring the extent of ionization helps determine the "effective" number of particles in solution. This is expressed as the van’t Hoff factor (i), crucial for determining the impact on colligative properties.

Molality

Molality is a measure of the concentration of a solution. It is defined as the number of moles of solute per kilogram of solvent, giving it the units of mol/kg. Unlike molarity, molality does not change with temperature, making it particularly useful in colligative properties calculations.
In the given exercise, the molality of the formic acid solution is 0.10 mol/kg. While molarity and molality are the same in this example, it is usually molality that better aids in determining the effects on boiling and freezing points.
This distinction is crucial because molality considers the mass of the solvent, not the volume of the solution, making it a stable measure when temperature fluctuations occur. For changes in boiling and freezing points, using molality instead of molarity ensures more accurate calculations.