Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: apathy on November 12, 2009, 12:28:27 AM
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Hiya!
I'm trying to calculate how much of copper (II) sulfate can be dissolved in solution with different pH. As far as I understand, if we decrease the pH of solution, more copper (II) sulfate can be dissolved. Following is my approach to this problem (which might be completely wrong, because I got really confused and starting to doubt my understanding of this whole equilibrium and solubility product concepts):
The reactions involved are:
CuSO4 ::equil:: Cu2+ + SO42-
Cu2+ + 2 OH- ::equil:: Cu(OH)2
Because of the second reaction, I thought that by adding more OH- (increasing pH), then copper (II) hydroxide would precipitate, hence limiting the "solubility" of CuSO4.
The solubility product constant of copper (II) hydroxide according to Knovel Critical Table is 2.20 x 10-20.
Ksp = [Cu2+] [OH-]2
For solution with pH = 5, I calculated the concentration of OH- as:
pOH = -log ([OH-]) = 14 - pH
[OH-] = 10-9
From Ksp and [OH-],
[Cu2+] = 2.20 x 10-2
Does this mean only 2.2 x 10-2 mol of CuSO4 can be dissolved in 1L of water if the pH is maintained at 5? Which would translate to approximately 3.51 g/L of anhydrous CuSO4. The solubility of CuSO4 given in the MSDS that I checked was 243 g/L at 0°C.
As all of you can probably tell at this point, I am completely clueless as to what I'm doing ???, so any helps would be greatly appreciated! Thanks in advance :)
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You discuss two different problems - solubility of copper sulfate which is temperature dependent (common ion - sulfate and ionic strenght also change solubility of this compound) and solubility of Cu(OH)2which starts to precipitate in sligthtly basic solution.
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Thanks for the reply.
Does it mean I am not supposed to relate them at all? Sorry, I guess I should've stated my objective first. I am trying to dissolve copper (II) sulfate in acidic or basic solution without getting copper (II) hydroxide to precipitate. My prof suggested that I should add sulfuric acid or sodium hydroxide to alter the pH of the solution.
I chose a pH of 5 in the calculation, because I've dissolved over 10g/L of CuSO4 without getting any precipitation at pH close to 5
Edit: added additional infos