What pH would yield the maximum separation of Mn 2+ from Sn 2+ in a solution tha

Question

What pH would yield the maximum separation of Mn2+ from Sn2+ in a solution that is 0.060 M in Mn2+, 0.060 M in Sn2+, and saturated in H2S? (Assume the green form of MnS.)

Metal Ion, Sulfide, Kspa, Ksp

What pH would yield the maximum separation of Mn2+ from Sn2+ in a solution that is 0.060 M in Mn2+, 0.060 M in Sn2+, and saturated in H2S? (Assume the green form of Mns.) Metal Ion Sulfide Sn2+ Mn MaS (pink form) 1 x 10 3 x 1010 3 x 10 l x 10-26 3 x 10- 3 x 10-14 SnS MnS (green form) Data are for 25 °C. See R. J. Meyers, J Chem. Ed, vol. 63, 1986, p. 687 PHmax separation

Explanation / Answer

MnS(s) + 2H+(aq) ------> Mn+2(aq) + H2S(g)

SnS(s) + 2H+(aq) --------> Sn+2(aq) + H2S(g)

Kspa = reaction coefficient Q when denominator H+ ion concentration is lowered Q value increases

It can be determined by using Ksp and Ka for acidic solutions

Kspa = Ksp/Kw × Ka1 = Ksp/ 10-21

When the solution is saturated with H2S(g) then the concentration is almost 0.1M H2S

Kspa of MnS (told to assume green) so = 3 x 107

3× 107 = [H2S] [Mn+2][H+]2

[Mn+2] = 0.06 M

[H2S] = 0.1 MM

3 x 107 = 0.06 x 0.1 /[H+]2

[H+]2 = 0.006/3 x 107 = 0.2 x 10-8

[H+] = 0.45 x 10-4

pH = -log [H+] = -log (0.45 x 10-4) = 4.35 ~ 4.4

b) Kspa of SnS = 1 x 10-5 = [Sn+2] [H2S] / [H+]2

[Sn+2] = 0.06 M

1 x 10-5 = 0.06 x 0.1 / [H+]2

[H+]2 = 0.006/1 x 10-5 = 6 x 102

[H+] = 24.5  

pH = -log [H+] = -log 24.5= 1.39 ~ 1.4

Above pH and H+ concentration show that the concentration of H+ in SnS i.e 24.5M is not possible. This infers that SnS can precipitate in all kind of solutions even in extreme acidic solution.

But below the pHmax of 4.4 MnS cannot be precipitated

So pHmax seperation is 4.4

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