Solubility of Cr hydroxide in buffered solution

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The solubility of Cr(OH)3 in a buffered solution at pH 8.673 requires calculating the pOH to find the hydroxide ion concentration. The Ksp value of Cr(OH)3 is 6.0256 E -31, which can be used in conjunction with the hydroxide concentration to determine the chromium ion concentration. By substituting the calculated [OH-] into the Ksp expression, the concentration of Cr+3 can be derived. Finally, this concentration can be converted into grams of Cr(OH)3 per liter. Understanding the relationship between pH, pOH, and Ksp is crucial for solving the solubility problem.
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Homework Statement


What is the solubility (gram/liter) of Cr(OH)3 (103.02 g/mol) in a solution buffered at pH = 8.673? The pKsp of Cr(OH)3 is 30.220.


Homework Equations


Ksp = [Cr][OH]^3
Cr(OH)3 -> Cr + 3OH


The Attempt at a Solution


I have already solved that Ksp = 6.0256 E -31, but I'm not entirely sure what to do with the buffer pH. I tried using it to find the pOH and subsequently [OH], but for some reason I'm not getting the right answer. Do I plug [OH] into the Ksp equation after finding out the concentration using the pH or is there a step I'm missing?
 
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Sounds like you are approaching it correctly. Use the concentration of [OH-] you obtain from the pOH in the Ksp expression and determine [Cr+3]. From that revelation back out how many grams of Cr(OH)3 will produce that concentration of [Cr+3].
 
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