Question

A sample containing 0.0500 moles of Fe2(SO4)3 is dissolved in enough water to make 1.00 L of solution. This solution contains hydrated SO42 and Fe (H₂O)6 ³⁺ ions. The latter behaves as an acid: Fe (H2O)6 ³⁺ 3 4 Fe (OH) (H2O)5 ²⁺ 1 H⁺

a. Calculate the expected osmotic pressure of this solution at 25˚C if the above dissociation is negligible.

b. The actual osmotic pressure of the solution is

6.73 atm at 25˚C. Calculate Ka for the dissociation reaction of Fe (H₂O)₆ ^3+. (To do this calculation, you must assume that none of the ions goes through the

semipermeable membrane. Actually, this is not a good assumption for the tiny H⁺ ion.)

Short answer

a = Osmotic pressure = 0.611 am

b = Degree of dissociation = 0.0015

I = Van’t Hoff factor

M = concentration of solute in mol/L

R = gas constant

T = absolute temperature in k

Step-by-step solution

Step 1:

A number of moles of Fe₂(SO₄)₃ = 0.0500 moles and it is dissolved in 1 liter of solution. So, molarity M is 0.00500 mol/L

When Fe₂(SO₄)₃ dissociates it gives 2 moles of Fe⁺² and 3 moles of SO4-²

Fe₂(SO₄)₃ ----- 2 Fe⁺² + 3 SO4-²

Number of dissociated ions are 5. Therefore, Van’t Hoff factor i = 5

Gas constant = 0.08206 L.atm./kmol

Temperature = 25 ˚C T = (25 + 273) K = 298K

5 x 0.005 x 0.08206 x 298 = 0.611 atm

Step 2:

Actual osmotic pressure is 6.73 atm

i = M (theoretical)/M (observed)

= 0.611/6.73 = 0.09078

Degree of dissociation of Fe (H₂O)₆ ³⁺= M (theoretical) - M (observed)/ M (observed) (n-1)

0.611 – 6.73 / 6.73 x (7-1) = -6.119/ 40.38 = 0.0015