# Homework Help: Complex solubility problem

1. Mar 10, 2008

### 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...

1. The problem statement, all variables and given/known data

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

2. Relevant equations

K = [Products] / [reactants]

3. 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.

2. Mar 10, 2008

### 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.

3. Mar 10, 2008

### 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 :).

Last edited: Mar 10, 2008
4. Mar 11, 2008

### 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.

5. Mar 12, 2008

### 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 ;).