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Formic acid \(\left(\mathrm{HCO}_{2} \mathrm{H}\right)\) 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

Expert verified
The freezing point of the 0.10 M formic acid solution is -0.194 掳C, and the boiling point is 100.053 掳C.

Step by step solution

01

Calculate the van't Hoff factor

Given that a 0.10 M formic acid solution is 4.2% ionized, we can determine the van't Hoff factor. The van't Hoff factor represents the number of particles produced after ionization. For formic acid, the ionization reaction is: HCO2H (aq) \(\leftrightarrows\) HCO2鈦 (aq) + H鈦 (aq) The van't Hoff factor, i, can be calculated using the following formula: i = 1 + degree of ionization In this case, the degree of ionization is 4.2%, which is equal to 0.042, so, i = 1 + 0.042 = 1.042
02

Calculate the freezing point

Use the freezing point depression formula to determine the new freezing point: 螖Tf = i 脳 Kf 脳 m where 螖Tf is the freezing point depression, i is the van't Hoff factor, Kf is the freezing point depression constant for water (1.86 掳C kg/mol), and m is the molality (equal to molarity, as given in the problem, 0.10 M). 螖Tf = 1.042 脳 1.86 掳C kg/mol 脳 0.10 mol/kg = 0.194 掳C The freezing point of the formic acid solution is the difference between the freezing point of pure water (0 掳C) and the freezing point depression: Tf = 0 掳C - 0.194 掳C = -0.194 掳C
03

Calculate the boiling point

Use the boiling point elevation formula to determine the new boiling point: 螖Tb = i 脳 Kb 脳 m where 螖Tb is the boiling point elevation, i is the van't Hoff factor, Kb is the boiling point elevation constant for water (0.512 掳C kg/mol), and m is the molality (equal to molarity, as given in the problem, 0.10 M). 螖Tb = 1.042 脳 0.512 掳C kg/mol 脳 0.10 mol/kg = 0.053 掳C The boiling point of the formic acid solution is the sum of the boiling point of pure water (100 掳C) and the boiling point elevation: Tb = 100 掳C + 0.053 掳C = 100.053 掳C In conclusion, the freezing point of the 0.10 M formic acid solution is -0.194 掳C, and the boiling point is 100.053 掳C.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Freezing Point Depression
Freezing point depression is a colligative property that occurs when a solute is dissolved in a solvent, causing the freezing point of the solution to be lower than that of the pure solvent. This phenomenon is important in understanding how solutes influence physical properties of solutions. It arises because the presence of solute particles interferes with the formation of a solid lattice of solvent molecules, which is necessary for freezing.

The freezing point depression is calculated using the formula:
  • \( \Delta T_f = i \times K_f \times m \)
Where:
  • \(\Delta T_f\) is the change in the freezing point.
  • \(i\) is the van't Hoff factor, which indicates the number of particles in solution.
  • \(K_f\) is the freezing point depression constant of the solvent (for water, it is 1.86 掳C kg/mol).
  • \(m\) represents the molality of the solution.
In the case of formic acid, this calculation reveals how slightly acidic conditions can adjust the freezing point by a small margin, demonstrating this property in action.
Boiling Point Elevation
Boiling point elevation is another essential colligative property that describes how a solution's boiling point increases when a solute is added to a solvent. Similar to freezing point depression, this effect happens because the solute particles disrupt the solvent's molecular interactions, requiring more energy (heat) to reach the boiling point.

The relationship can be calculated as:
  • \( \Delta T_b = i \times K_b \times m \)
Where:
  • \(\Delta T_b\) is the change in boiling point.
  • \(i\) denotes the van't Hoff factor.
  • \(K_b\) is the boiling point elevation constant (for water, it is 0.512 掳C kg/mol).
  • \(m\) indicates the molality of the solution.
For formic acid, the calculation reflects how even a modest level of ionization impacts its boiling point. This increase is often smaller than the freezing point depression, but it is crucial for processes like distillation.
Van't Hoff Factor
The van't Hoff factor, often denoted as \(i\), is a measure of the number of particles a compound forms in solution, and it plays a critical role in colligative properties calculations. For electrolytes like formic acid, which partially dissociates into ions in solution, determining the van't Hoff factor is essential for accurately predicting changes in boiling and freezing points.

Calculating \(i\) involves using the degree of ionization of the solute:
  • \( i = 1 + \text{degree of ionization} \)
In the case of formic acid, the degree of ionization is 4.2%, or 0.042 as a decimal. Therefore,
  • \( i = 1 + 0.042 = 1.042 \)
This factor helps determine how significant the effect of the solute will be on the solution's colligative properties, demonstrating that even a slight ionization can alter the physical characteristics of a solution significantly.
Ionization of Monoprotic Acids
Ionization refers to the process where an acid releases protons when dissolved in water, forming ions in solution. Monoprotic acids like formic acid (HCO鈧侶) are characterized by their ability to donate only one proton per molecule in the ionization process. This partial ionization significantly influences the acid's behavior in solutions.

For formic acid in particular, the ionization can be represented by the chemical equation:
  • \( \text{HCO}_2\text{H} (aq) \leftrightarrows \text{HCO}_2^- (aq) + \text{H}^+ (aq) \)
In solutions, this equilibrium means that not all acid molecules dissociate completely, which is quantified by the degree of ionization. For a 0.10 M solution, a 4.2% ionization results in a van't Hoff factor indicating the acidic environment.

This insight helps to understand how weakly ionizing acids alter solution properties, especially in calculations involving colligative properties, without fully dissociating as strong acids would.

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Most popular questions from this chapter

Calculate the molarity and mole fraction of acetone in a 1.00 -m solution of acetone \(\left(\mathrm{CH}_{3} \mathrm{COCH}_{3}\right)\) in ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) . (Density of acetone \(=0.788 \mathrm{g} / \mathrm{cm}^{3} ;\) density of ethanol \(=\) 0.789 \(\mathrm{g} / \mathrm{cm}^{3} .\) ) Assume that the volumes of acetone and ethanol add.

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