Question

Silver chromate is sparingly soluble in aqueous solutions. The Ksp of Ag₂CrO₄ is 1.12× 10⁻¹². What is the solubility (in mol/L) of silver chromate in 1.50 M potassium chromate aq solution? In 1.50 M silver nitrate aq solution? In pure water?

Answer 1

The solubility of silver chromate (Ag₂CrO₄) can be determined using its solubility product constant (Ksp) and the concentrations of the ions in the solution.

The dissolution of silver chromate in water can be represented by the following equation:

Ag₂CrO₄(s) ⇌ 2Ag⁺(aq) + CrO₄²⁻(aq)

From this, we can write the expression for the solubility product constant as:

Ksp = [Ag⁺]²[CrO₄²⁻]

Given that the Ksp of Ag₂CrO₄ is 1.12×10⁻¹², we can solve for the solubility of Ag₂CrO₄ in different solutions.

In pure water: Since Ag₂CrO₄ dissociates into 2 moles of Ag⁺ for every mole of Ag₂CrO₄, if we let the solubility of Ag₂CrO₄ be S, then the concentration of Ag⁺ will be 2S and the concentration of CrO₄²⁻ will be S. Substituting these into the Ksp expression gives:

Ksp = (2S)² * S = 4S³

Solving this for S (the solubility of Ag₂CrO₄ in mol/L) gives:

S = (Ksp / 4)^(1/3)

In 1.50 M potassium chromate (K₂CrO₄) solution: The presence of CrO₄²⁻ ions from K₂CrO₄ will decrease the solubility of Ag₂CrO₄ due to the common ion effect. In this case, the concentration of CrO₄²⁻ is no longer S, but is instead 1.50 M. The Ksp expression becomes:

Ksp = (2S)² * 1.50

Solving this for S gives:

S = sqrt(Ksp / (4 * 1.50))

In 1.50 M silver nitrate (AgNO₃) solution: Similarly, the presence of Ag⁺ ions from AgNO₃ will decrease the solubility of Ag₂CrO₄. In this case, the concentration of Ag⁺ is no longer 2S, but is instead 1.50 M. The Ksp expression becomes:

Ksp = 1.50² * S

Solving this for S gives:

S = Ksp / (1.50²)

By substituting the given value of Ksp into these equations, you can calculate the solubility of Ag₂CrO₄ in each of these solutions.

Answer 2

The solubility of silver chromate in pure water, potassium chromate solution, and silver nitrate solution is determined by the Ksp value and common ion effects, taking into account the reaction stoichiometry and solubility equilibrium expressions.

Step-by-step explanation:

To determine the solubility of silver chromate in different solutions using its Ksp, we must take into account common ion effects and the solubility expression based on the dissolution reaction:

Ag2CrO4(s) ⇌ 2Ag+(aq) + CrO42-(aq),

with the Ksp expression being Ksp = [Ag+]2 [CrO42-]. Given the Ksp value of 1.12 × 10-12, we use the molar solubility S to represent the solubility of silver chromate.

In pure water, the ionic product would be Ksp = (2S)2 × S = 4S3 = 1.12 × 10-12, solving this gives the solubility in pure water.

In 1.50 M potassium chromate solution, the concentration of CrO42- is high and makes Ag+ solubility decrease significantly due to common ion effect. For silver nitrate solution, a similar process applies where high concentration of Ag+ lowers the CrO42- solubility.