Points A 100 M Aqueous Solution Of The Following Compounds
4 5 Points A 100 M Aqueous Solution Of The Following Compounds We
A 100 M aqueous solution of various compounds is considered to evaluate their physical and chemical properties based on various parameters, such as molality, colligative properties, pH, gas solubility, and chemical equilibrium. These parameters provide insights into the behavior of different compounds in aqueous environments, especially in relation to their solubility, effect on boiling and melting points, and reactivity.
The assessment involves understanding the impact of solute composition on melting points, analyzing colligative properties like boiling point elevation, calculating pH values in acid-base reactions, determining gas concentrations based on Henry’s law, and evaluating the equilibrium constant (Kc) for a chemical reaction. Each part requires applying principles from physical chemistry to interpret experimental data and theoretical concepts effectively.
Paper For Above instruction
Understanding the influence of solutes on the physical properties of solutions, along with the dynamics of gas solubility and chemical equilibria, showcases fundamental aspects of physical chemistry. These principles are pivotal in various scientific and industrial applications, including formulation of pharmaceuticals, improving chemical manufacturing processes, and environmental studies.
Impact of Solutes on Melting Point
The ideal scenario considers the colligative property known as freezing point depression, which depends directly on the number of particles dissociated in solution, rather than their identity. The compounds listed are ionic and their dissociation in water differs, affecting lowering of melting points.
- KClO₃ (potassium chlorate): Dissociates into K⁺ and ClO₃⁻ ions, producing 2 particles per formula unit.
- MgCl₂ (magnesium chloride): Dissociates into Mg²⁺ and 2 Cl⁻ ions, totaling 3 particles.
- NaF (sodium fluoride): Dissociates into Na⁺ and F⁻ ions, total 2 particles.
- NaCl (sodium chloride): Dissociates into Na⁺ and Cl⁻ ions, total 2 particles.
- MgSO₄ (magnesium sulfate): Dissociates into Mg²⁺ and SO₄²⁻ ions, resulting in 2 particles.
Given all else equal, the compound that produces the highest number of particles in solution will most effectively depress the melting point. MgCl₂ produces three particles per formula unit, which is higher than the others, thus leading to the greatest melting point depression.
Conclusion: MgCl₂ would produce the solution with the lowest melting point because of its higher dissociation into three ions per formula unit, aligning with the principles of colligative properties.
Effect of Solutes on Boiling Point – Determination of the Largest Change
The boiling point elevation depends on the molality of the solute and the ebullioscopic constant (Kb) of the solvent. The change in boiling point is calculated by ΔTb = i·Kb·m, where:
- i = van ’t Hoff factor (number of particles produced),
- Kb = ebullioscopic constant of the solvent,
- m = molality of the solute.
Given the Kb values:
- Acetic acid (3.07 m−1)
- Benzene (2.53 m−1)
- Nitrobenzene (5.24 m−1)
- Phenol (3.56 m−1)
- Water (0.512 m−1)
Assuming the same molality for each solution (1.0 M approximates to 1.0 mol/kg):
- The highest ΔTb occurs when the product i·Kb is maximized.
- Nitrobenzene has the highest Kb (5.24), which when combined with its typical dissociation factor (depending on its behavior as a weak electrolyte), would lead to the greatest boiling point elevation.
Conclusion: Nitrobenzene exhibits the largest change in boiling point when a 1.0 molal solute is added, because of its highest Kb value.
pH Calculation of the NaOH and H₂SO₄ Mixture
When 0.5 M NaOH is mixed with 0.5 M H₂SO₄ in equal volumes, their neutralization results in a solution whose pH depends on the relative strength and dissociation of the acids and bases involved.
H₂SO₄ completely dissociates in the first proton (pKa₁
- NaOH supplies OH⁻,
- H₂SO₄ supplies H⁺ ions.
Since both are at equal molarity (0.5 M) in equal volumes:
- The moles of NaOH = 0.5 mol per 100 mL,
- The moles of H₂SO₄ = 0.5 mol in 100 mL.
The stoichiometry of the neutralization:
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
- Moles of NaOH required to neutralize all H₂SO₄: 1 mol NaOH per 1 mol H₂SO₄.
- Given the concentrations, NaOH is in excess because each mole of H₂SO₄ requires 2 mol NaOH.
Calculating:
- Total moles H₂SO₄ = 0.5 mol,
- Total NaOH needed for complete neutralization = 1 mol NaOH,
- Available NaOH = 0.5 mol, insufficient to neutralize all H₂SO₄.
Thus, after some neutralization:
- Excess H₂SO₄ remains in solution,
- The pH is dominated by the remaining unreacted acid.
Calculating the remaining H₂SO₄:
- It would have reacted with 0.25 mol NaOH (due to the 2:1 ratio),
- Remaining H₂SO₄ = 0.25 mol,
- Total volume = 0.1 L + 0.1 L = 0.2 L.
Remaining H₂SO₄ concentration = 0.25 mol / 0.2 L = 1.25 M.
Given the strong acids and their partial dissociation, the pH can be approximated assuming strong acid behavior:
pH ≈ -log[H⁺], with [H⁺] ≈ 1.25 M, leading to:
pH ≈ -log(1.25) ≈ -0.10 (which is physically implausible; more precise calculations consider activity coefficients).
However, practically, the solution remains highly acidic with a pH close to 0 or slightly above, approximately 0.1 to 1.
Gas Solubility of Nitrogen and Oxygen in Water
Henry's law states that the amount of gas dissolved in a liquid is proportional to its partial pressure:
\[ C = \frac{P \times k_H}{RT} \]
At 25°C:
- \( k_{H,N_2} = 8.20 \times 10^{-7} \) molal/mmHg,
- \( k_{H,O_2} = 1.62 \times 10^{-6} \) molal/mmHg,
- Atmospheric pressure \( P_{total} = 760 \) mmHg,
- Air composition: 80% N₂, 20% O₂.
Partial pressures:
- \( P_{N_2} = 0.8 \times 760 = 608 \) mmHg,
- \( P_{O_2} = 0.2 \times 760 = 152 \) mmHg.
Calculating dissolved gases:
- N₂: \( C_{N_2} = 8.20 \times 10^{-7} \times 608 \approx 4.99 \times 10^{-4} \) mol/kg,
- O₂: \( C_{O_2} = 1.62 \times 10^{-6} \times 152 \approx 2.46 \times 10^{-4} \) mol/kg.
Masses:
- For 236 mL of water (~0.236 kg):
- N₂: \( 4.99 \times 10^{-4} \times 0.236 \approx 1.18 \times 10^{-4} \) mol,
- O₂: \( 2.46 \times 10^{-4} \times 0.236 \approx 5.80 \times 10^{-5} \) mol.
Molar masses:
- N₂: 28 g/mol,
- O₂: 32 g/mol.
Masses:
- N₂: \( 1.18 \times 10^{-4} \times 28 \approx 3.31 \times 10^{-3} \) g,
- O₂: \( 5.80 \times 10^{-5} \times 32 \approx 1.86 \times 10^{-3} \) g.
Result: About 3.31 mg of nitrogen and 1.86 mg of oxygen are dissolved in the water under these conditions.
Equilibrium Constant Calculation for SO₃ Reaction
The reaction involves the dissociation:
\[ 2SO_3(g) \rightleftharpoons 2SO_2(g) + O_2(g) \]
Initial concentration of SO₃:
\[ [SO_3]_{initial} = 0.128\, M \]
At equilibrium:
- \( [O_2]_{eq} = 0.0130\, M \),
- The change in SO₃: reacts to produce O₂; the extent of reaction (x):
From the stoichiometry:
- \( [O_2] = x = 0.0130\, M \),
- \( [SO_2] = 2x = 2 \times 0.0130 = 0.0260\, M \),
- \( [SO_3]_{eq} = 0.128 - 2x = 0.128 - 0.026 = 0.102\, M \).
The equilibrium expression:
\[ K_c = \frac{[SO_2]^2 [O_2]}{[SO_3]^2} \]
Plugging in values:
\[ K_c = \frac{(0.026)^2 \times 0.013}{(0.102)^2} \]
\[ K_c = \frac{0.000676 \times 0.013}{0.010404} \]
\[ K_c \approx \frac{8.79 \times 10^{-6}}{0.010404} \]
\[ K_c \approx 8.45 \times 10^{-4} \]
Final result: The equilibrium constant \( K_c \) for the reaction is approximately 8.45 × 10⁻⁴.
References
- Atkins, P., & de Paula, J. (2014). Physical Chemistry (10th ed.). Oxford University Press.
- Chang, R. (2010). Chemistry (10th ed.). McGraw-Hill Education.
- Laidler, K. J., Meiser, J. H., & Sanctuary, B. C. (1999). Physical Chemistry (4th ed.). Houghton Mifflin.
- Moran, M. G., & Shapiro, H. N. (2008). Principles of Measurement Systems (5th ed.). Wiley.
- Petrucci, R. H., Herring, F. G., Madura, J. D., & Bissonnette, C. (2017). General Chemistry: Principles & Modern Applications (11th ed.). Pearson.
- Schweser, E. (2010). Gas Laws and Henry's Law: Applications in Environmental Chemistry. Journal of Environmental Chemistry Reports, 2(3), 100-110.
- Steinfeld, J. I., Francisco, J. S., & Hase, W. L. (2009). Chemical Kinetics and Dynamics. Prentice Hall.
- Van Hook, W. A. (1974). Basic Concepts of Chemistry. Prentice Hall.
- Weast, R. C. (Ed.). (1984). CRC Handbook of Chemistry and Physics (64th ed.). CRC Press.
- Zumdahl, S. S., & Zumdahl, J. (2014). Chemistry: An Atoms First Approach. Cengage Learning.