A sample containing 0.0500 mol of Fe_2(SO_4)_3 is dissolved in enough water to m

Question

A sample containing 0.0500 mol of Fe_2(SO_4)_3 is dissolved in enough water to make 1.00 L of solution. This solution contains hydrated SO^2-_4 and Fe(H_2 O)^3+_6 ions. The latter behaves as an acid according to the equation: Fe (H_2 O)^3+_6 (aq) doubleheadarrow Fe (H_2 O)_5 OH^2+ (aq) + H^+ (aq) Calculate the expected osmotic pressure of this solution at 25 degree C if this acid dissociation is negligible. The actual osmotic pressure is of the solution is 6.80 atm at 25 degree C. Calculate the K_a for the acid dissociation reaction of Fe (H_2 O)^3+_6, assuming no ions cross the semipermeable membrane.

Explanation / Answer

Fe3+ +   6H2O « Fe(OH)63+ .........................(2)

Fe2(SO4)3 = 0.0500 moles

Volume of solution = 1.0L

No. of moles of (SO4 2- ) ions a sper equation (1) = 3 x 0.0500 moles = 0.15 mole

n (Fe(H2O)63+ ) as per equation (2)= 0.1 mole

Totalno. Of moles in solution : 0.15 + 0.1 = 0.25 moles

Osmotic Pressure OP = CRT

OP = 0.25 moles x 0.08206 atm. L./mol.K x 298K

Oamotic Pressure= 6.11 atm ................(3)

b) Fe(H2O)63+ (aq)       «   Fe(H2O)5OH2+ (aq)        + H+(aq)       

i = vint hoff’s factor

i = P/OP

i = 6.8 atm/6.11 atm = 1.11 .................(4)

a = dissociation degree

a= (i-1)/(n-1);

n = 2 ions

a = 0.11 – dissociation degree

Ka = a.C2

Ka = 6 x 10-4

  

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