Question

Calculate the molar solubility of lead(II) bromide (PbBr2). For lead(II) bromide, Ksp=4.67×10−6.

Answer 1

The molarities of the ions in a saturated solution satisfy the solubility equilibrium equation:
Ksp = [Pb²⁺]∙[Br⁻]²

Let x be the molar solubility of PbBr₂. When you dissolve x moles per liter of solution, you get a just saturated solution in which each salt molecule is dissolved and dissociated to one lead ion and two bromide ions. Assuming the solution already contains [Pb²⁺]₀ of lead ion and [Br⁻]₀ of bromide ions the ionic molarities in the saturated solution are:
[Pb²⁺] = [Pb²⁺]₀ + x
[Br⁻]² = [Br⁻]₀ + 2∙x

The solubility equilibrium can be rewritten as:
Ksp = ([Pb²⁺]₀ + x)∙([Br⁻]₀ + 2∙x)²

An equation which can be solved for each of three solutions.
[Pb²⁺]₀ + x)∙([Br⁻]₀

Neither lead nor bromide ions in the solution:
[Pb²⁺]₀ = [Br⁻]₀ = 0
=>
Ksp = x∙(2∙x)² = 4∙x³
=>
x = ∛( Ksp/4 )
= ∛( 4.67×10⁻⁶ / 4 )
= 0.0105(mol∙L⁻¹)