The ksp of pbbr2 is 6.60*10-6. What is the molar solubility of pbbr2 in pure water?

The molar solubility is 0.0118 mole / liter = 1.18 * 10⁻² M

Explanation:

1) Ksp stands for product solubility constant.

2) The product solubility constant is the equilibrium constant used for the aqueous ionization (dissolution) of low soluble ionic compounds.

3) The ionization or equilibrium reaction for the PbBr₂ is:

PbBr₂ (s) ⇄ Pb²⁺ (aq) + 2 Br (aq) ⁻

4) Sotichiometry:

PbBr₂ (s) ⇄ Pb²⁺ (aq) + 2 Br (aq) ⁻

Ao - x moles x 2x

5) Calculations:

Per definition of Ksp:

Ksp = [Pb²⁺] [Br (aq) ⁻]² = (x) (2x) ² = 4x³

Then, you have to solve: 6.60*10⁻⁶ = 4x³

x³ = 6.60*10⁻⁶ / 4

x = 0.0118 mole / liter = 1.18 * 10⁻² M