Example Problem


Problem: What is the molar solubility of ferrous hydroxide in a buffer solution made of equal volumes of ammonia and ammonium chloride? Ksp of Fe(OH)2 = 7.9 E-15, Kb of NH3 = 1.8 E-5.


Solution: This problem looks scary, but some quick thinking is all you need to solve it. Let's see what's going on:


  1. Fe(OH)2 --> Fe2+ + 2OH¯
    Ksp = 7.9 E-15
  2. NH3 + H2O --> NH4+ + OH¯
    Kb = 1.8 E-5
  3. NH4+ + H2O --> NH3 + H3O+
    Ka = Kw/Kb = 5.6 E-9


But wait! Don't we have an equation for pOH when you know the concentrations of an acid and its conjugate base? Oh yeah! pOH = pKb -log( [B] / [HB] ). But wait again! [NH4+] = [NH3], which means that pOH is really just pKb and [OH¯] = Kb. So, we get the OH¯ concentration as 1.8 E-5. Now, using our handy Ksp formula for ferrous hydroxide, we can get the molar solubility of of Fe(OH)2.


Ksp = [Fe2+][OH¯]2
Ksp = 7.9 E-15 = X(2X)2
Molar Solubility = X = 1.3 E-5


You're almost done! Why don't you check out Multiple Solubility?


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