Problem 144
Question
What would be the \(\mathrm{pH}\) of a \(0.1 \mathrm{molal}\) aqueous solution of a monoprotic acid 'HA', that freezes at \(-0.2046^{\circ} \mathrm{C} ?\) \(\left[\mathrm{K}_{\mathrm{f}}\left(\mathrm{H}_{2} \mathrm{O}\right)=1.86^{\circ} \mathrm{mol}^{-1} \mathrm{~kg} ;\right.\) assuming molality \(=\) molarity]
Step-by-Step Solution
Verified Answer
The pH of the solution is approximately 1.26.
1Step 1: Determine the Depression in Freezing Point
Find the depression in freezing point (ΔTₓ), which is the difference between the normal freezing point of water and the given freezing point of the solution. Normal freezing point of water is 0°C and the given freezing point is -0.2046°C.\[ΔT_x = 0 - (-0.2046) = 0.2046 °C\]
2Step 2: Apply Freezing Point Depression Formula
Use the equation for freezing point depression: \[ΔT_x = i imes K_f imes m\]where \(i\) is the van't Hoff factor, \(K_f\) is the cryoscopic constant, and \(m\) is the molality. For a monoprotic acid like HA, \(i=2\) (since it breaks down into two ions, \(H^+\) and \(A^-\)). Substitute the values to find \(m\):\[0.2046 = 2 \times 1.86 \times m\]Solve for \(m\):\[m = \frac{0.2046}{2 \times 1.86} \approx 0.055 \, \text{molal}\]
3Step 3: Calculate Ionization and Hydronium Ion Concentration
Since the initial concentration of \(HA\) is 0.1 molal and hydrolysis results in an additional concentration of hydronium ions, we know the effective concentration of ions. Since 0.055 molal ions are formed in solution (from freezing point depression), subtract from the initial concentration:\[[H^+] = 0.055\]The degree of dissociation, \(\alpha\), can be determined as fraction of original acid that dissociates:\[\alpha = \frac{0.055}{0.1} = 0.55\]
4Step 4: Calculate pH
Now that we have the concentration of hydronium ions, calculate the pH:\[pH = -\log[H^+]\]\[pH = -\log(0.055) \approx 1.26\]
5Step 5: Conclusion
Based on the calculations, the pH of the 0.1 molal solution of monoprotic acid \(HA\) which freezes at -0.2046°C is approximately 1.26.
Key Concepts
Freezing Point DepressionMonoprotic AcidVan't Hoff FactorDegree of Dissociation
Freezing Point Depression
Freezing point depression is a colligative property, which means it depends on the number of solute particles in a solution, rather than their identity. When a solute is added to a solvent, it interferes with the solvent's freezing process, causing the freezing point to decrease.
In this problem, the solution of monoprotic acid HA in water leads to a measured freezing point of -0.2046°C. Knowing the normal freezing point of pure water is 0°C, the depression in freezing point, ΔTₓ, can be calculated as 0.2046°C.
The degree of decrease in temperature is directly proportional to the molal concentration of the solute, given by the formula \( ΔT_x = i \times K_f \times m \), where \(i\) is the van't Hoff factor, \(K_f\) is the cryoscopic constant, and \(m\) is the molality of the solution.
In this problem, the solution of monoprotic acid HA in water leads to a measured freezing point of -0.2046°C. Knowing the normal freezing point of pure water is 0°C, the depression in freezing point, ΔTₓ, can be calculated as 0.2046°C.
The degree of decrease in temperature is directly proportional to the molal concentration of the solute, given by the formula \( ΔT_x = i \times K_f \times m \), where \(i\) is the van't Hoff factor, \(K_f\) is the cryoscopic constant, and \(m\) is the molality of the solution.
Monoprotic Acid
A monoprotic acid is an acid that can donate only one proton (or hydrogen ion, \(H^+\)) per molecule in an aqueous solution. Simple examples include hydrochloric acid (HCl) and acetic acid (CH₃COOH).
In the context of this exercise, the monoprotic acid HA dissociates in water to produce one hydronium ion \(H^+\) and one conjugate base ion \(A^-\). Therefore, when calculating the van't Hoff factor \(i\), it is assumed to fully dissociate into these two particles in dilute conditions.
Understanding how these acids dissociate is crucial for calculating the pH and other properties of the solution, as the dissociation increases the number of particles in the solution, affecting colligative properties like freezing point depression.
In the context of this exercise, the monoprotic acid HA dissociates in water to produce one hydronium ion \(H^+\) and one conjugate base ion \(A^-\). Therefore, when calculating the van't Hoff factor \(i\), it is assumed to fully dissociate into these two particles in dilute conditions.
Understanding how these acids dissociate is crucial for calculating the pH and other properties of the solution, as the dissociation increases the number of particles in the solution, affecting colligative properties like freezing point depression.
Van't Hoff Factor
The van't Hoff factor, represented by \(i\), is a measure of the effect of solute molecules on the freezing point depression and boiling point elevation. It tells us how many particles a solute dissociates into in a solution.
For the monoprotic acid HA in this exercise, the assumed van't Hoff factor is 2, indicating it fully dissociates into one \(H^+\) ion and one \(A^-\) ion. This assumption is vital for simplifying calculations in solutions where full dissociation can be expected.
Knowing \(i\) helps in the application of the formula \(ΔT_x = i \times K_f \times m\) to solve for the molality \(m\) of ions in the solution after dissociation, which is crucial for determining the freezing point depression and further calculating the degree of dissociation and pH.
For the monoprotic acid HA in this exercise, the assumed van't Hoff factor is 2, indicating it fully dissociates into one \(H^+\) ion and one \(A^-\) ion. This assumption is vital for simplifying calculations in solutions where full dissociation can be expected.
Knowing \(i\) helps in the application of the formula \(ΔT_x = i \times K_f \times m\) to solve for the molality \(m\) of ions in the solution after dissociation, which is crucial for determining the freezing point depression and further calculating the degree of dissociation and pH.
Degree of Dissociation
Degree of dissociation (\(\alpha\)) refers to the fraction of the original solute molecules that dissociate into ions in solution. It's a crucial concept for assessing the extent of dissociation and subsequent calculations of the solution's properties.
In this problem, the initial concentration of the monoprotic acid, HA, is 0.1 molal. After applying the freezing point depression calculations, we find that it dissociates to form 0.055 molal ions (\([H^+]\) ions).
The degree of dissociation \(\alpha\) is calculated as \(\alpha = \frac{0.055}{0.1} = 0.55\). This value shows that 55% of the acid molecules are dissociated in the solution. Understanding \(\alpha\) is essential for predicting the concentration of ions and, in this case, aiding further in the determination of the pH.
In this problem, the initial concentration of the monoprotic acid, HA, is 0.1 molal. After applying the freezing point depression calculations, we find that it dissociates to form 0.055 molal ions (\([H^+]\) ions).
The degree of dissociation \(\alpha\) is calculated as \(\alpha = \frac{0.055}{0.1} = 0.55\). This value shows that 55% of the acid molecules are dissociated in the solution. Understanding \(\alpha\) is essential for predicting the concentration of ions and, in this case, aiding further in the determination of the pH.
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