Solubility Product Calculations (CIE A Level Chemistry)

Calculating the solubility product of a compound from its solubility

Calculate the solubility product of a saturated solution of lead(II) bromide, PbBr₂, with a solubility of 1.39 x 10^-3 mol dm^-3.

Answer

• Step 1: Write down the equilibrium equation:
  • PbBr₂ (s) ⇌ Pb²⁺ (aq) + 2Br^- (aq)
• Step 2: Write down the equilibrium expression:
  • K_sp = [Pb²⁺(aq)] [Br^- (aq)]²

• Step 3: Calculate the ion concentrations in the solution:
  • [PbBr₂(s)] = 1.39 x 10^-3 mol dm^-3
  • The ratio of PbBr₂ to Pb²⁺ is 1:1
    • [Pb²⁺(aq)] = [PbBr₂(s)] = 1.39 x 10^-3 mol dm^-3
  • The ratio of PbBr₂ to Br^- is 1:2
    • [Br^-(aq)] = 2 x [PbBr₂(s)] = 2 x 1.39 x 10^-3 mol dm^-3 = 2.78 x 10^-3 mol dm^-3

• Step 4: Substitute the values into the expression to find the solubility product:
  • K_sp = (1.39 x 10^-3) x (2.78 x 10^-3)²
  • K_sp = 1.07 x 10^-8

• Step 6:  Determine the correct units of K_sp:
  • K_sp = (mol dm^-3) x (mol dm^-3)²
  • K_sp = mol³ dm^-9
• Therefore, the solubility product is 1.07 x 10^-8 mol³ dm^-9

Calculating the solubility of a compound from its solubility product

Calculate the solubility of a saturated solution of copper(II) oxide, CuO, with a solubility product of 5.9 x 10^-36 mol² dm^-6.

Answer

• Step 1: Write down the equilibrium equation:
  • CuO (s) ⇌ Cu²⁺ (aq) + O^2- (aq)
• Step 2: Write down the equilibrium expression:
  • K_sp = [Cu²⁺ (aq)] [O^2- (aq)]

• Step 3: Simplify the equilibrium expression:
  • The ratio of Cu²⁺ to O^2- is 1:1
  • [Cu²⁺(aq)] = [O^2-(aq)] so the expression can be simplified to:
    • K_sp = [Cu²⁺ (aq)]²
• Step 4: Substitute the value of K_sp into the expression to find the concentration:
  • 5.9 x 10^-36 = [Cu²⁺ (aq)]²
  • [Cu²⁺ (aq)] = 
  • [Cu²⁺ (aq)] = 2.4 x 10^-18 mol dm^-3
• Since [CuO (s)] = [Cu²⁺ (aq)], the solubility of copper oxide is 2.4 x 10^-18 mol dm^-3