A sample containing 0.0500mol of Fe2(SO4)3 is dissolved in enough water to make 1.00L of solution. This solution contains hydrated SO4 2- and Fe(H2O)6 3+ ions. The latter behaves as an acid:

Fe(H2O)6 3+ --> Fe(H2O)5OH 2+ + H+

a) Calculate the expected osmotic pressure of this solution at 25�0†2C if the above dissociation is neglibile.
b) The actual osmotic pressure of the solution is 6.73 atm at 25�0†2C. Calculate Ka for the dissociation reaction of Fe(H2O)6 3+.

Thanks in advance!

Formulas to consider:
Osmotic pressure = MRT
R = 0.08206 --> constant
T = 298 K

a) To calculate the expected osmotic pressure of the solution, we need to determine the concentration of the solute particles in the solution.

1. Calculate the number of moles of solute particles:
- Fe2(SO4)3 dissociates into 2 Fe^3+ ions and 3 SO4^2- ions.
- 0.0500 mol of Fe2(SO4)3 will produce (2 mol Fe^3+/mol Fe2(SO4)3) × 0.0500 mol = 0.100 mol Fe^3+ ions.
- 0.0500 mol of Fe2(SO4)3 will produce (3 mol SO4^2-/mol Fe2(SO4)3) × 0.0500 mol = 0.150 mol SO4^2- ions.

2. Calculate the total number of moles of solute particles:
- The total number of moles of solute particles is the sum of the moles of Fe^3+ and SO4^2- ions.
Total moles = 0.100 mol + 0.150 mol = 0.250 mol

3. Calculate the concentration of the solute particles:
- The concentration of the solute particles is the total number of moles divided by the volume of the solution: 0.250 mol / 1.00 L = 0.250 M

4. Calculate the osmotic pressure using the formula:
Osmotic pressure = MRT
- M is the molarity of the solute particles (0.250 M)
- R is the ideal gas constant (0.08206 L·atm/(mol·K))
- T is the temperature in Kelvin (298 K)

Osmotic pressure = (0.250 M) × (0.08206 L·atm/(mol·K)) × (298 K) = 6.08 atm

Therefore, the expected osmotic pressure of the solution at 25°C is 6.08 atm if the dissociation of Fe(H2O)6 3+ is negligible.

b) To calculate Ka for the dissociation reaction of Fe(H2O)6 3+, we can use the equation for osmotic pressure and rearrange it to solve for Ka:

Osmotic pressure = MRT = (n/V)RT = CRT

Where C is the concentration of the solute particles in terms of molarity.

1. Rearrange the equation to solve for C:
C = (Osmotic pressure) / (RT)

2. Calculate the concentration C:
C = 6.73 atm / (0.08206 L·atm/(mol·K) × 298 K) = 0.269 M

3. Since Fe(H2O)6 3+ dissociates according to the given equation:
Fe(H2O)6 3+ → Fe(H2O)5OH 2+ + H+

The concentration of Fe(H2O)6 3+ ions is the same as the concentration of OH^- ions and is represented by x.

4. Based on the stoichiometry of the equation, the concentration of H+ ions is also equal to x.

5. The concentration of Fe(H2O)5OH 2+ ions can be considered negligible.

6. Now, we can write the equilibrium expression and substitute the values:
Ka = [H+][OH^-] / [Fe(H2O)6 3+]
= x × x / (0.269 - x)

7. Since the reaction is an acid dissociation reaction, the concentration of H+ ions is equal to the concentration of OH^- ions. So:
Ka = x^2 / (0.269 - x)

8. The actual osmotic pressure is given as 6.73 atm, which corresponds to the concentration of the solute particles (Fe(H2O)6 3+ and OH^- ions) as 0.269 M.

9. Substitute the osmotic pressure (C) and solve for x:
6.73 atm = x^2 / (0.269 - x)

10. Solve the equation for x using numerical methods or trial and error. By solving this equation, we find that x ≈ 0.103.

11. Substitute the value of x into the equilibrium expression to calculate Ka:
Ka = (0.103)^2 / (0.269 - 0.103) = 0.676

Therefore, Ka for the dissociation of Fe(H2O)6 3+ is approximately 0.676.