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Thermodynamics and pH

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Please see the attached file for the fully formatted problems.

1) A quantity of 7.480g of an organic compound is dissolved in water to make 300.0 mL of solution. The solution has an osmotic pressure of 1.43 atm at 27 degrees Celsius. The analysis of this compound shows that it contains 41.8 percent C, 4.7 percent H, 37.3 percent O and 16.3 percent N. Calculate the molecular formula of the compound.

2) The Ka for benzoic acid is 6.5 x 10^-5. Calculate the pH of a 0.10 M benzoic acid solution.

3) If 20.0 mL of 0.10 M Ba(NO3)2 are added to 50.0 mL of 0.10 M Na2CO3, will BaCO3 precipitate?

4) Compare the molar solubility of Mg(OH)2 in water and in a solution buffered at a pH of 9.0.

5) Calculate the pH of the 0.20M NH3/0.20M NH4Cl buffer. What is the pH of the buffer after the addition of 10.0 mL of 0.10M HCl to 65.0 mL of the buffer?

6) Given the following absolute entropies, determine So for the reaction

SO3(g) + H2O(l) H2SO4(l)
So (J/K mol)
SO3 256.2
H2O 69.9
H2SO4 156.9

7) Given the following free energies of formation, calculate Go for the reaction

3NO2(g) + H2O(l) 2HNO3(l) + NO(g)
G (kJ/mol)
H2O(l) -237.2
HNO3(l) -79.9
NO(g) 86.7
NO2(g) 51.8

8) Nitrosyl chloride (NOCl) decomposes at elevated temperatures according to the equation

2NOCL(g) 2NO(g) + Cl2(g)

Use the following information to calculate KP for this reaction at 227oC:

Ho = 81.2 kJ
So = 128 J/K

9) Given the following free energies of formation, calculate KP for the reaction below at 298 K.

SO2(g) + NO2(g) SO3(g) + NO(g)
G (kJ/mol)
SO2(g) -300.4 kJ/mol
SO3(g) -370.4 kJ/mol
NO(g) 86.7 kJ/mol
NO2(g) 51.8 kJ/mol

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Solution Summary

This solution explains:

1) How to calculate the molecular formula of a compound based on percent composition.
2) How to calculate the pH of a weak acid.
3) How to calculate molar solubility.
4) How to calculate the pH of a buffer solution.
5) How to use Hess's law to calculate enthalpy and free energy of reactions.
6) How to calculate the equilibrium constant of a reaction using Gibbs free energy.

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