A sample containing 0.0500 mol of Fe2(SO4)3 is dissolved in enough water to make
ID: 1070780 • Letter: A
Question
A sample containing 0.0500 mol of Fe2(SO4)3 is dissolved in enough water to make 1.00 L of solution. This solution contains hydrated SO42- and Fe(H2O)63+ ions. The latter behaves as an acid according to the equation: Fe(H2O)63+ (aq) Fe(H2O)5OH2+ (aq) + H+ (aq)
a. Calculate the expected osmotic pressure of this solution at 25° C if this acid dissociation is negligible.
b. The actual osmotic pressure is of the solution is 6.80 atm at 25° C. Calculate the Ka for the acid dissociation reaction of Fe(H2O)63+, assuming no ions cross the semipermeable membrane.
Thank You!
Explanation / Answer
Answer (a)
(1) Fe2(SO4)3 <----> 2Fe3++ 3SO42-
(2) Fe3+ + 6H 2O <----> Fe(OH)63+
n (SO42-) = 0.15 mole
n (Fe(H2O)63+) = 0.1 mole
Total: 0.15 + 0.1 = 0.25 mole
P0= CRT = 0.25 * 1000 * 8.314 * 298 = 0.62 MPa = 6.11 atm
Answer (b)
Fe(H2O)63+ <----> Fe(H2O)5OH2+ + H+
i = P / P0 = 6.8 /6.11 = 1.11
= (i - 1) / (n - 1) , n = 2 ions
= 0.11 – dissociation degree
Ka = 2C = 6 x 10-4
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