Kso= [Fe3+] [OH-]3 = 10-39
Solubilities of metal hydroxides:
If one leaves an orange solution of a ferric salt to stand, after a while it will clear, and an orange precipitate of Fe(OH)3(s) will form. The extent to which Fe3+ can exist in solution as a function of pH can be calculated from the solubility product, Kso. For Fe(OH)3(s) the expression for Kso is given by:
Kso = [Fe3+] [OH-]3 = 10-39 [2]
One thus finds that the maximum concentration of Fe3+ in solution is controlled by pH, as detailed on the next slide.
Note that we need [OH-] in expression 2, which is obtained from the pH from equation 3.
pKw = pH + pOH = 14 [3]
Thus, if the pH is 2, then pOH = 12, and so on. pOH is related to [OH-] in the same way as pH is related to [H+].
pH = -log [H+] [4]
pOH = -log [OH-] [5]
So, to calculate the maximum concentration of [ Fe3+ ] at pH 6.4, we use eqs. [3] to [5] to calculate that at pH 6.4, pOH = 7.6, so that [OH-] = 10-7.6 M. This is then used in equation [2] to calculate that [Fe3+] is given by:
Problem. What is the maximum [Fe3+] at pH 6.4?
From the previous page, at pH 6.4 we have [OH-] = 10-7.6 M. Thus, putting [OH-] = 10-7.6 M into equation 2, we get:
10-39 = [ Fe3+ ] x [ 10-7.6 ]3
[ Fe3+] = 10-39 / 10-22.8 = 10-16 M
Note that for a metal ion Mn+ of valence n that forms a solid hydroxide precipitate M(OH)n, the equation has the [OH-] raised to the power n. For example:
Pb2+ forms Pb(OH)2(s): Kso = 10-14.9 = [Pb2+] [OH-]2
Th4+ forms Th(OH)4(s): Kso = 10-50.7 = [Th4+] [OH-]4
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