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In aqueous media, Pb2+ ions form a precipitate, PbO, which is an amphoteric oxide. In acidic medium,Analytical Chemistry Chemistry Question

Amphoteric lead oxide

In aqueous media, Pb2+ ions form a precipitate, PbO, which is an amphoteric oxide. In acidic medium, only Pb2+ species is present; with increasing pH, PbO and Pb(OH)3 – are formed in appreciable quantities. The important equilibria for lead species in aqueous medium are given below:
Reaction 1: PbO(s) + H2O(l) ⇌ Pb2+(aq) + 2 OH– (aq) Ks = 8.0 ⋅ 10–16
Reaction 2: PbO(s) + 2 H2O(l) ⇌ Pb(OH)3 –(aq) + H3O+(aq) Ka = 1.0 ⋅ 10–15

10.1.

The amphoteric PbO completely dissolves when pH is sufficiently low. When initial concentration of Pb2+ is 1.00 ⋅ 10–2 mol dm–3, what is the pH at which PbO starts to precipitate?

Model Answer

[Pb2+] = 1.00 ⋅ 10–2
Ks = [Pb2+] [OH–]2 = 8.0 ⋅ 10–16 ⇒ [OH–] = 2.8 ⋅ 10–7
Thus, [H3O+] = 3.5 ⋅ 10–8 and pH = 7.45

10.2.

Starting from the value in 10.1, when pH is increased to a certain value, precipitate is redissolved. At what pH value does the precipitate dissolve completely?

Model Answer

At relatively high pH, solubility is predominantly represented by Pb(OH)3– ion.
[Pb(OH)3–] = 1.00 ⋅ 10–2
Using Ka expression: [Pb(OH)3–][H3O+] = 1.0 ⋅ 10–15,
[H3O+] = 1.0 ⋅ 10–13, pH = 13.00
Using Ks, [Pb2+] = 8.00 ⋅ 10–14, insignificant when compared with 1.00 ⋅ 10–2 mol dm–3

10.3.

Write a general expression for solubility s (in mol dm–3) of PbO.

Model Answer

s = [Pb2+] + [Pb(OH)3–]

10.4.

Theoretically, the minimum solubility is achieved when pH is 9.40. Calculate the concentrations of all the species and their solubilities at this pH.

Model Answer

[H3O+] = 10^-9.40 ⇒ Using Kw = 1.00 ⋅ 10^-14 [OH-] = 2.5 ⋅ 10^-5
From Ks: [Pb2+] = 1.3 ⋅ 10^-6 From Ka: [Pb(OH3-] = 2.5 ⋅ 10^-6
Solubility: s = [Pb2+] + [Pb(OH3-] = 1.3 ⋅ 10^-6 + 2.5 ⋅ 10^-6 = 3.8 ⋅ 10^-6 mol dm–3

10.5.

Calculate the pH range where the solubility is 1.0 ⋅10–3 mol dm–3 or lower.

Model Answer

Pb2+ will be predominating in relatively low pH. [Pb2+] = 1.0 ⋅ 10^-3.
Ks = [Pb2+] [OH–]2 = 8.0 ⋅ 10–16

Thus, under acidic conditions its concentration is negligible and Pb2+ is the predominating species.
Pb(OH3- will be predominating in more basic pH.
[Pb(OH3-] = 1.00 ⋅ 10^-3 and from Ka: [H3O+] = 1.00 ⋅ 10–12
pH = 12.00 and [OH–] = 1.00 ⋅ 10–2
From Ks: [Pb2+] = 8.00 ⋅ 10–12 and therefore [Pb2+] << [Pb(OH3-]
The range of pH is 7.95 - 12.0.

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