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_4^2- and Fe(H_2O)_6^3+ ions. The latter behaves as an acid according to the equation: Fe(H_2O)_6^2+ (aq) reversible Fe(H_2O)_3OH^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_g for the acid dissociation reaction of Fe(H_2O)_6^3+, assuming no ions cross the semipermeable membrane.

Explanation / Answer

a)molarity=0.05 M

Osmotic pressure= c*RT*i

=0.05*(1/12)*(273+25)*5 (i=5 since Fe2(SO4)3 dissociates into 5 particles)

=1.24*5

=6.2 atm

b)let the dissociation be x.so,

i=1-(n-1)/(n*x) (where n is the number of particles after dissociation)

so i(net)=5+i

=6 - (n-1)/(n*x)

so,

6.8=(6.2/5)*(6 - (2-1)/(2*x))

or x=0.96875

so Ka=cx^2/(1-x)

=1.84*10^-3

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