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.