Solubility-Product Constant (College Board AP® Chemistry): Revision Note

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 molar solubility of 1.39 x 10^-3 M.

Answer:

Step 1: Write down the equilibrium equation and expression:

PbBr₂ (s) ⇌ Pb²⁺ (aq) + 2Br^- (aq)

K_sp = [Pb²⁺(aq)] [Br^- (aq)]²

Step 2: Set up an equilibrium table:

Step 3: Calculate the ion concentrations in the solution:

Molar solubility, s = 1.39 x 10^-3 M

• The ratio of PbBr₂ to Pb²⁺ is 1:1

[Pb²⁺(aq)] = s

[Pb²⁺(aq)] = 1.39 x 10^-3 M

• The ratio of PbBr₂ to Br^- is 1:2

[Br^-(aq)] = 2s

[Br^-(aq)] = 2 x 1.39 x 10^-3 M

[Br^-(aq)] = 2.78 x 10^-3 M

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

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

K_sp and molar solubility are different quantities

• K_sp is the equilibrium constant

• The molar solubility, s, is calculated from K_sp

Each ion's concentration is its coefficient × s, and that term is then raised to the coefficient: for PbBr₂, K_sp = (s)(2s)² = 4s³

Calculating the solubility of a compound from its solubility product

Calculate the molar solubility of copper(II) hydroxide, Cu(OH)₂, with a solubility product of 2.2 x 10^-20.

Answer:

Step 1: Write down the equilibrium equation and expression:

Cu(OH)₂ (s) ⇌ Cu²⁺ (aq) + 2OH^- (aq)

K_sp = [Cu²⁺ (aq)] [OH^- (aq)]²

Step 2: Set up an equilibrium table:

Step 3: Simplify the equilibrium expression:

• The ratio of Cu²⁺ to OH^- is 1:2

  • [Cu²⁺(aq)] = x

  • [OH^-(aq)] = 2x

• So, the expression can be simplified to:

K_sp = (x)(2x)² = 4x³

Step 4: Substitute the value of K_sp into the expression to find the concentration:

2.2 x 10^-20 = 4x³

x = $\sqrt[3]{\frac{2.2\times {10}^{-20}}{4}}$​​

x = 1.8 x 10^-7 M

• The molar solubility of copper(II) hydroxide is 1.8 x 10^-7 M

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

• All sodium and potassium salts

• All nitrate salts

• All ammonium salts

These are the only solubility rules you need to recall; any other solubility information will be given in the question. A salt with K_sp > 1 is soluble

A 1.70 x 10^-3 M solution of calcium nitrate is mixed with an equal volume of 1.50 × 10^-3 M potassium sulfate.

Predict whether calcium sulfate (K_sp = 2.00 x 10^-5) will precipitate.

Answer:

Step 1: Write the equation:

Ca(NO₃)₂ + K₂SO₄ → CaSO₄ + 2KNO₃

Step 2: Determine the concentration of the calcium and sulfate ions 

• You must halve the concentrations of the solutions as you are doubling the volume

  • The question says you are adding equal volumes of both solutions

[Ca²⁺ (aq)] = $\frac{1.70\times {10}^{-3}}{2}$ = 8.50 x 10^-4 M

[SO₄^2- (aq)] = $\frac{1.50\times {10}^{-3}}{2}$ = 7.50 x 10^-4 M

Step 3: Calculate Q:

Q = [Ca²⁺ (aq)] x [SO₄^2- (aq)]

Q = 8.50 x 10^-4 x 7.50 x 10^-4

Q = 6.38 x 10^-7 

Step 4: Compare to K_sp:

• Therefore, there will be no precipitation as Q is less than K_sp (of calcium sulfate)