Question

4. A solution of CuSO4 will have a pH -* (1 Point) 7 =7 0 0​

Answer 1

Explanation: 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).