Fe ^3+ ions are to be seperated from Hg^2+ ions from an aqueous solution in whic

Question

Fe ^3+ ions are to be seperated from Hg^2+ ions from an aqueous solution in which their concentrations are both 0.063m. The criterion for complete seperation is that the Fe^3+ ion concentration must be 10,000 times larger or 10,000 times smaller that the Hg^2+ ion concentration. Answer the questions below. (a) At what pH will Fe(OH)s precipitate from a 0.063M Fe3+ solution? Ksp= 2.8 × 10-39 Number (b) At what pH will Hg(OH)2 precipitate from a 0.063M Hg2+ solution? Ksp 3.1 x 102 Number (c) Which hydroxide precipitates at a lower O Fe(OH)a OH concentration? O Hg(OH)2 O Cannot be determined Number (d) In a solution buffered to a pH of 3.00, what is [Fe3 VHg2]? O Yes (e) Is the separation of Hg2+ from Fe3* 3.00 scooplete' in a solution buffered to a pH of O No O Cannot be determined

Explanation / Answer

A)   Fe(OH)3 <==> Fe3+ + 3 OH-
The solubility product expression of Fe(OH)3 is:
Ksp = [Fe3+][OH-]3
2.8x10-39 = 0.063 M [OH-]3

[OH-]3 = 4.44 x 10-38
[OH-]= 3.54 x 10-13M

[H+] = 10-14/pOH = 0.028 M

pH = -log[H+] = -log[0.028] = 1.55

At pH higher than 1.55, [OH-] will exceed the Ksp and hence, the Fe(OH)3 will precipitate.
The answer is : 1.55
B)   Hg(OH)2 <==> Hg2+ + 2OH-
The solubility product expression of Hg(OH)2 is:
Ksp = [Hg2+][OH-]2
3.1x10-26   = 0.063 M [OH-]2

[OH-]2 = 4.92 x 10-25
[OH-]= 7.014 x 10-13M

[H+] = 10-14/pOH = 0.014 M

pH = -log[H+] = -log[0.014] = 1.85

At pH higher than 1.85, [OH-] will exceed the Ksp and hence, the Hg(OH)2 will precipitate.
The answer is: 1.85

C)   Since, [OH-] concentration of Fe(OH)3 is lower than Hg(OH)2, Fe(OH)3 will precipitate at a lower concentration of OH-.
Hence, the answer is; Fe(OH)3.
D)   Since the solution is buffered the concentration of [OH-] does not change.
We have pH = 3 = -log[H+]
[H+] = 10-3M
[OH-] = 10-14/[H+] = 10-14/10-3 = 10-11M
Now construct ICE table for Fe(OH)3:
                   Fe(OH)3     <===>   Fe+3 +          3OH-
Initial               0             0       10-11M
Change           0           +x          no change
Equilibrium       0           x              10-11M

Ksp = [Fe3+][OH-]3

2.8x10-39 = [Fe3+][10-11]3
[Fe3+] = 2.8 x 10-6 M

Similarly we can calculate for Hg(OH)2:

construct ICE table for Hg(OH)2:
                    Hg(OH)2 <===> Hg+3 +       2OH-
Initial                0                     0             10-11M
Change           0                      +x       no change
Equilibrium       0                      x       10-11M

Ksp = [Hg2+][OH-]2

3.1x10-26 = [Hg2+][10-11]2
[Hg2+] = 3.1 x 10-4 M
Hence, [Fe3+]/[Hg2+] = 2.8 x 10-6 M / 3.1 x 10-4 M = 9.03 x 10-3

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