Common Ion Effect

Homework:
p492 (9)
P492 (10,11,12)

Determine the molar solubility of Fe(OH)3 in a 0.300M Fe2(SO4)3 solution.

      Fe(OH)3(s)   <=>     Fe3+ (aq)      +       3 OH-(aq)

I       n/a                           0.600M*                    0
C     n/a                                 +x                       +3x
E     n/a                            0.600 + x                  +3x
                                  ~0.600 (small K)

initial Fe3+ = 0.600M - from the 0.300M Fe2(SO4)3,
where there are twice as many Fe as Fe2(SO4)3 units

Ksp = [Fe3+][OH-]3
2.6x10-39 = (0.600)(3x)3
2.6x10-39 = 0.600(27x3)
5.4x10-14 = x 
molar solubility of Fe(OH)3 in the 0.300M Fe2(SO4)3 solution
 is 5.4x10-14 mol/L