What is the molar solubility of PbBr2 in three different solutions?

The Ksp of PbBr2 is 6.60× 10–6.

Lead bromide dissolves according to the reaction equation:
PbBr₂(s) ⇄ Pb²⁺(aq) + 2 Br⁻(aq)

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

1. Pure water
Neither lead nor bromide ions in the solution:
[Pb²⁺]₀ = [Br⁻]₀ = 0
=>
Ksp = x∙(2∙x)² = 4∙x³
=>
x = ∛( Ksp/4 )
= ∛( 6.60×10⁻⁶ / 4 )
= 1.18×10⁻² (mol∙L⁻¹)

2. 0.50M KBr solution
Each potassium bromide molecule gives one bromide ion to the solution. So the solution contains:
[Br⁻]₀ = 0.500
and no lead ions
[Pb²⁺]₀ = 0
=>
Ksp = x∙([Br⁻]₀ + 2∙x)²

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That leads to a cubic equation, which takes some math effort to solve. You can simplify, by a simple approximation. Due to small solubility x will small. Due to common ion effect it will be smaller than result from part 1. So let’s assume that x is much smaller than [Br⁻]₀ = 0.5 and approximate:
[Br⁻]₀ + 2∙x ≈ [Br⁻]₀
=>
Ksp = x∙[Br⁻]₀²
=>
x = Ksp / [Br⁻]₀²
= 6.60×10⁻⁶ / 0.500²
= 2.64×10⁻³ mol∙L⁻¹

As you can see x is about 19,000 times smaller than 0.5, i.e. the approximation was appropriate)

3. 0.50M Pb(NO₃)₂
Each lead nitrate molecule gives one lead ion to the solution. So the solution contains:
[Pb²⁺]₀ = 0.500
and no bromide ions
[Br⁻]₀ = 0
=>
Ksp = ([Pb²⁺]₀ + x)∙(2∙x)²

As in part 2 you can expect x to be small compared to [Pb²⁺]₀ and approximate:
[Pb²⁺]₀ + x ≈ [Pb²⁺]₀
=>
Ksp = [Pb²⁺]₀∙4∙x²
=>
x = √( Ksp /(4∙[Pb²⁺]₀)
= √( 6.60×10⁻⁶ / (4∙0.5) )
= 1.82×10⁻³ mol∙L⁻¹

Pbbr2

Slight correction the answer in number 2. It should be to the power of -5, not a -3.

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