Calculate the solubility of silver chromate, Ag₂CrO₄, in 0.005 M Na₂CrO₄. K_sp = 2.6 × 10^-12 A) 1.4 × 10^-4 M B) 3.4 × 10^-5 M C) 1.1 × 10^-5 M D) 1.6 × 10^-6 M E)

Question

Calculate the solubility of silver chromate, Ag₂CrO₄, in 0.005 M Na₂CrO₄. K_sp = 2.6 × 10^-12 A) 1.4 × 10^-4 M B) 3.4 × 10^-5 M C) 1.1 × 10^-5 M D) 1.6 × 10^-6 M E) < 1.0 × 10^-6 M

Quizplus · Accepted Answer

The solubility product expression for silver chromate is:

Ksp = [Ag2CrO4] = [Ag+]^2[CrO4 2-]

The balanced equation for the dissolution of Ag2CrO4 in water is:

Ag2CrO4(s) ⇌ 2Ag+(aq) + CrO4 2-(aq)

Since we are given an initial concentration of Na2CrO4, we can assume that there is excess CrO4 2- in solution, and that the dissolution reaction will proceed until equilibrium is reached. At equilibrium, we can assume that:

[Ag+] = 2x
[CrO4 2-] = (0.005 M) - 2x

Substituting these expressions into the Ksp expression, we get:

Ksp = (2x)^2[(0.005 M) - 2x] = 4x^2(0.005 - 2x)

Solving for x, we get:

4x^2(0.005 - 2x) = 2.6 × 10^-12
8x^3 - 0.02x^2 + 2.6 × 10^-12 = 0

Using a calculator or solver tool, we find that x = 1.05 × 10^-5.

Therefore, [Ag+] = 2x = 2.1 × 10^-5 M, and the solubility of Ag2CrO4 is:

[Ag2CrO4] = [Ag+]^2[CrO4 2-] = (2.1 × 10^-5)^2(0.005 - 2x) = 1.1 × 10^-5 M

So the correct answer is C.

Calculate the Solubility of Silver Chromate, Ag2CrO4, in 0

Calculate the Solubility of Silver Chromate, Ag₂CrO₄, in 0