Common-Ion Effect (College Board AP® Chemistry): Revision Note

Calculations using the K_sp values and the concentration of the common ion

Predict whether a precipitate of CaSO₄ will form if a solution of 1.0 x 10^-3 M CaSO₄ is mixed with an equal volume of 1.0 x 10^-3 M Na₂SO₄.

K_sp CaSO₄ = 2.0 x 10^-5 

Answer:

Step 1: Write down the equilibrium equation and expression:

CaSO₄ (s) ⇌ Ca²⁺ (aq) + SO₄^2- (aq)

K_sp = [Ca²⁺ (aq)] [SO₄^2- (aq)]

Step 2: Determine the ion concentrations in the solution:

• There are equal volumes of each solution

• This means that the total solution was diluted by a factor of 2

• The new concentration of the Ca²⁺ ion is halved:

[Ca²⁺] =  $\frac{1.0\times {10}^{-3}}{2}$

[Ca²⁺] = 5.0 x 10^-4 M

• The sulfate ion concentration remains the same, because both solutions contain 1.0 x 10^-3 M sulfate: [SO₄^2-] = (1.0 x 10^-3 + 1.0 x 10^-3) ÷ 2 = 1.0 x 10^-3 M

Step 3: Substitute the values into the expression:

Q = [Ca²⁺ (aq)] x [SO₄^2- (aq)]

Q = (5.0 x 10^-4) x (1.0 x 10^-3)

Q = 5.0 x 10^-7

Step 4: Determine if a precipitate will form:

• As Q (5.0 x 10^-7) is smaller than the K_sp value (2.0 x 10^-5), the CaSO₄ precipitate will not be formed

The K_sp for lead chloride, PbCl₂ is 1.7 x 10^-5 at 25 °C.

Calculate the molar solubility of PbCl₂ at 25 °C which is in a solution of 0.10 M NaCl.

Answer:

Step 1: Write down the equilibrium equation and expression:

PbCl₂ (s) $\rightleftharpoons$ Pb²⁺ (aq) + 2Cl^– (aq)

K_sp = [Pb²⁺ (aq)][Cl^– (aq)]²

Step 2: Set up an equilibrium table:

Step 3: Place values into expression:

K_sp = [Pb²⁺ (aq)][Cl^– (aq)]² 

1.7 x 10^-5 = (x)(2x + 0.10)² 

• The value for x is very small, so we can approximate and use 0.10 M for 2x + 0.10

1.7 x 10^-5 = (x)(0.10)²

x = 1.7 x 10^-3 M

• Check: 2x = 3.4 x 10^-3 M, which is 3.4% of 0.10 M, so the approximation is valid

Additional Note

• Compare this with the molar solubility of PbCl₂ in pure water:

K_sp = [Pb²⁺ (aq)][Cl^– (aq)]²

1.7 x 10^-5 = (x)(2x)² 

1.7 x 10^-5 = 4x³

x = $\sqrt[3]{\frac{1.7\times {10}^{-5}}{4}}$

x = 0.016 M

• The addition of a common ion (Cl^–) decreased the solubility of PbCl₂ by roughly a factor of 10

• Adding a common ion changes Q, not K_sp; only a change in temperature changes K_sp

• Naming “the common-ion effect” is not a justification. Explain that the added ion makes Q greater than K_sp, so the reverse reaction (precipitation) occurs until Q = K_sp, leaving less salt dissolved