Question

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

Answer 1

Answer:

• 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