Solubility Product Calculations (Cambridge International A-Level Chemistry)

Solubility Product Calculations

• Calculations involving the solubility product(K_sp) may include::

  • Calculating the solubility product of a compound from its solubility
  • Calculating the solubility of a compound from the solubility product

Worked Example 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) ⇌ Pb**²⁺(aq) + 2Br^-(aq)

• **Step 2: **Write down the equilibrium expression:

K_sp** = [Pb**²⁺(aq)] [Br^-(aq)]²

• **Step 3: **Calculate the ion concentrations in the solution:

  • [PbBr₂(s)] = 1.39 x 10^-3mol dm^-3
  • The ratio of PbBr₂ to Pb²⁺ is 1:1

[Pb²⁺(aq)] = [PbBr₂(s)] = 1.39 x 10^-3mol dm^-3

• The ratio of PbBr₂ to Br^- is 1:2

[Br^-(aq)] = 2 x [PbBr₂(s)] = 2 x 1.39 x 10^-3mol dm^-3 = 2.78 x 10^-3mol 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** = mol ³ dm^-9

• Therefore, the solubility product is 1.07 x 10^-8 mol³ dm^-9

Worked Example 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** = [Cu**²⁺(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** = [Cu**²⁺(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)] =$\sqrt{5.9\times10^{-36}}$

[Cu²⁺(aq)] = 2.4 x 10^-18mol dm^-3

• Since [CuO (s)] = [Cu²⁺ (aq)], the solubility of copper oxide is 2.4 x 10^-18 mol dm^-3

Exam Hint Remember that the solubility product is only applicable to very slightly soluble salts and cannot be used for soluble salts such as:

• Group 1 element salts
• All nitrate salts
• All ammonium salts
• Many sulfate salts
• Many halide salts (except for lead(II) halides and silver halides)