Question

A lead pipe can get oxidized in the presence of water and oxygen to form a surface layer of lead hydroxide, Pb(OH)2. Write down the equilibrium chemical reaction that expresses the process for Pb(OH)2 dissolving in water to form Pb2+ and OH‐. Include physical states for all the reactants and products. Calculate the solubility of Pb2+ in water based on the solubility product, Ksp=1.43x10‐20

Answer 1

Solubility equation for Pb(OH)₂ (aq) in aqueous phase

Pb(OH)₂ (aq) Pb²⁺ (aq) + 2 HO^- (aq)

Let Solubility of Pb(OH)₂ is "S" mol/L

I.e. S mol/L of Pb(OH)₂ (aq) dissolve in water at equilibrium.

As 1 molecule of Pb(OH)₂ (aq) gives 1 mol Pb²⁺ (aq) ion & 2 mol HO^- (aq) ion we write,

[Pb²⁺] = "S" mol/L   and         [HO^-] = "2S" mol/L

The expression for solubility product of Pb(OH)₂ (aq) is

Ksp = [Pb²⁺][HO^-]².

Given: Ksp = 1.43x10^‐20 & [Pb²⁺] = "S" mol/L   and [HO^-] = "2S" mol/L

Placing these values we get,

1.43*10^‐20 = (S)(2S)².

4S³ = 1.43*10^‐20.

S³ = 14.3*10^‐21 / 4

S³ = 3.575*10^‐21.

On taking cube root of both sides.

S = 1.53*10^‐7. mol/L

Solubility of Pb2+ in water is 1.53*10^‐7. mol/L

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