What pH is needed to precipitate chromium from wastewater?

[Terp]

Hi everybody. I'm stuck half way through this problem. If anybody could help it's greatly appreciated. This may be a long post...

Homework Statement

Automobile parts are plated with chromium to resist corrosion and produce a shiny product. Wastewater from a chromium plating company contains dissolved chromium(III) at 18 mg/L. What pH should the wastewater be adjusted to in order to precipitate chromium as Cr(OH)3(s) so that the total aqueous chromium concentration is 0.8 mg/L? Consider the following equilibria:

Cr3+ + OH- <==> CrOH2+(aq) K1 = 10^10.0

Cr2+ + 2 OH- <==> Cr(OH)2+(aq) K2 = 10^18.3

Cr2+ + 3 OH- <==> Cr(OH)3 (aq) K3 = 10^24.0

Cr(OH)3 (s) <==> Cr3+ + 3 OH- Ks = 10^-30

Homework Equations

K = [Products] / [reactants]

The Attempt at a Solution

So I first started by writing all of the equilibrium equations:

K1 = [CrOH 2+]/[Cr3+][OH-]
K2 = [Cr(OH)2 +]/[Cr2+][OH-]^2
K3 = [Cr(OH)3]/[Cr2+][OH-]^3
Ks = [Cr3+][OH-]^3

I also calculated the initial concentration of Cr(III) in the waste water to be 3.46x10^-4 M. and I want the final concentration to be 1.5x10^-5 M

So I then setup a sum of the soluble metals equation:

Cr(III) = [Cr2+] + [Cr3+] + [CrOH 2+] + [Cr(OH)2 +] + [Cr(OH)3]

I then plugged in values from my equilibrium expression into this to obtain:

Cr(III) = [Cr2+] + Ks/[OH-]^3 + K1[Cr3+][OH-] + K2[Cr2+][OH]^2 + K3[Cr2+][OH-]^3

Now this is where I'm stuck. Is the Cr(III) on the left side of the equation a concentration or some other number? I'm not sure where to factor in the 1.5x10^-5 M ending concentration, nor do I know the [Cr2+] concentration. I'm also not positive if I plug in the initial or wanted concentration for [Cr3+] on the right side of the equation. Once I figure out how to get/use those two I think I just solve for [OH-] and take the -log - 14 to get the pH. Anybody have any clue?


BTW I'm sorry for posting here a lot recently, but this class has gotten tough. I'm trying to show as much work as I can figure out. Thanks a lot Smiley.

EDIT: Doing this the simple, but wrong (in this case) way, I got the pH to be 5.61. Just did 10^-30 = [Cr3+][OH-]^3, solve for OH, took the -log and subtracted it from 14.

 

[GCT]

The Chromium on the left is the total Chromium or the initial Cr(III) content, for this problem you need to set up the equation with the [ OH - + or - x ] in for the actual [OH-] and use the other Cr(III) value to account for the Cr(II) content by following the set of reaction equations.

 

[Terp]

Thanks for the response...give me a few minutes while I try to work this out :). When you say I need to set up the equation with [OH-] +- x, which equation do you mean? The final one? I don't quite follow you.

I'm working with the algebra here and I can't figure out a way to solve for [Cr2+] in terms of [OH] and a K value without adding more unknowns. When rearranging the second and third equilibrium equations it puts either [Cr(OH)3] or [Cr(OH)2 +] as a part of Cr 2+, both of which are unknown.

Again, thanks :).

 

[Kalirren]

Are you sure that you're supposed to be dealing with Cr(II) species at all in this problem? Because it looks to me like you may have copied some equations wrong:

Cr2+ + 2 OH- <==> [Cr(OH)2]+(aq) K2 = 10^18.3

Cr2+ + 3 OH- <==> Cr(OH)3 (aq) K3 = 10^24.0

Those equations aren't charge-balanced. It would make sense if all the Cr2+ were in fact Cr3+.

This also makes sense because you're absolutely correct that you need more information in order to completely determine the system (asuming you really -are- dealing with Cr2+). You have 7 species, (Cr3+, [Cr(OH)]2+, [Cr(OH)2]+, Cr(OH)3 (aq), Cr(OH)3 (s), OH-, Cr2+), and 6 constraints (the four equations, CrT = 18 mg/L, CrT(aq) = 0.8 mg/L), which gives you a degree of freedom in the end corresponding to the redox potential of the solution (and thus the Cr3+/Cr2+ balance). This degree of freedom would go away if you didn't have to deal with Cr2+ at all.

 

[Terp]

You're absolutely right. I didn't copy the equations wrong...my professor made a mistake in the homework problems.

That simplifies it a lot. Thanks ;).