I checked the pH and it is just under 4 (3.88 to be precise). It is kept in room temperature and I am wondering how can i dissolve this precipitate, which I am guessing is a Cu(OH)2.
p.s. (and if it helps) the solution was made with copper II sulfate pentahydrate.
any help would be deeply appreciated.
Thanking you in advance
May be water used for dissolving is not completely distilled , copper sulfate used by you should be having some ionic impurities or hydrolysis of copper sulfate takes place. I would suggest you to add one or two drops of concentrated sulfuric acid while preparing solution of the salt in distilled water to check the hydrolysis .
Dear Chrysanthou
Since your solution is acidic the formation of Cu(OH)2 is unlikely. First of all, check the quality of your distilled water. The EC of it should be lower than 1 microsimens/cm and its pH should be about 7. When the distilled water is not freshly produced some CO2 dissolve in it towards forming H2CO3 related carbonate ions which result in CuCO3 with very low solubility. To overcome this problem, add some HNO3 or H2SO4 to the distilled water to eliminate carbonates then add the CuSO4. The other point could be your chemical is not pure enough. The formation of Cu hydroxy-sulfates may be eliminated by acidification of the suspension.
Please look for any possible sources of alkalinization of the mentioned (acidic) pure CuSO4 solution, that being the case.
The solubility product constant of Cu(OH)2 is given by: Ksp = [Cu2+] [OH-]2 = 2.20•10-20 (units omitted). The CuSO4 salt can be taken as a strong electrolyte, thus (nearly) completely dissociated in aq. sol., so that CCuSO4 ≈ [Cu2+], where CCuSO4 is the nominal molar concentration of dissolved CuSO4 (formality, to be precise). Precipitation of Cu(OH)2 can be predicted for [OH-] > √(Ksp / |CCuSO4|) = 1.4832•10-10/ √|CCuSO4|. Recalling that pOH = - log10 (|[OH-]|) and pH = 14 - pOH, that corresponds to pH > 4 + log10 [1.4832 / √|CCuSO4|]; or for CCuSO4 = 0.1M, to pH > 4 + log10 [1.4832 / √0.1] = 4.67.
The pH of aq. sol. of CuSO4 can be predicted as follow:
a) Cu2+ contributes to pH by its hydrolysis:
i) Cu2+ + 2H2O ⇌ Cu(OH)+ + H3O+; Kh1 = 5.1·10-8 (*)
ii) Cu(OH)+ + 2H2O ⇌ Cu(OH)2 + H3O+; Kh2 = 7.9·10-14 (*)
b) SO42- contributes to pH accordingly the second ionization of H2SO4:
HSO4- + H2O ⇌ SO42- + H3O+; Ka2 = 1.2·10−2
c) Water contributes to pH by its autoionization (Kw = 1.0·10-14).
d) Charge balance writes:
2[Cu2+] + [Cu(OH)+] + [H3O+] = 2[SO42-] + [HSO4-] + [OH-]
or:
2{[Cu2+] - [SO42-]}·[H3O+] + {1- [SO42-]/Ka2}·[H3O+]2 = Kw - Kh1[Cu2+]
The following simplification can be accepted: CCuSO4 ≈ [Cu2+] ≈ [SO42-]
hence:
[H3O+] ≈ √[(Kw - Kh1·CCuSO4)/(1- CCuSO4/Ka2)]
or:
pH ≈ ½log10 [(1- CCuSO4/(1.2·10−2))/(1.0·10−14 - 5.1·10-8·CCuSO4)]
Last expression is not valid for (approx.) CCuSO4 ≤ 0.012 M. It is also not valid for (approx.) pH > 4 + log10 [1.4832 / √|CCuSO4|] (cf. my first post), since (apparently possible) precipitation of Cu(OH)2 at higher pH was not accounted for. It can be rather safely accepted for (approx.) 0.012 M < CCuSO4 < 0.145 M. In practice, it can be also used for higher CCuSO4 as long as no detectable Cu(OH)2 precipitation occurs while preparing the actual solution, what often happens to be the case. ― that solution can possibly be supersaturated in Cu(OH)2. The expression is approximate, chiefly because the hydrolytic constants for Cu2+ are not known with good accuracy. Units of constants were omitted.
For CCuSO4 = 0.1 M, we obtain: pH ≈ 4.58. By considering my earlier post, it can be confirmed that precipitation of Cu(OH)2 for that CuSO4 solution (0.1 M) should not occur (pH < 4.67).
The measured value reported in the enunciate seems not unreasonable.
(*) Proc. Indian Acad. Sci. (Chem. Sci.), 114(2), April 2002, 115–124
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