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:
CuSO
4 Cu
2+ + SO
42-Cu
2+ + 2 OH
- Cu(OH)
2Because of the second reaction, I thought that by adding more OH
- (increasing pH), then copper (II) hydroxide would precipitate, hence limiting the "solubility" of CuSO
4.
The solubility product constant of copper (II) hydroxide according to Knovel Critical Table is 2.20 x 10
-20.
K
sp = [Cu
2+] [OH
-]
2For solution with pH = 5, I calculated the concentration of OH
- as:
pOH = -log ([OH
-]) = 14 - pH
[OH
-] = 10
-9From K
sp and [OH
-],
[Cu
2+] = 2.20 x 10
-2Does this mean only 2.2 x 10
-2 mol of CuSO
4 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 CuSO
4. The solubility of CuSO
4 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