Calculate the solubility (in g/L)of silver chromate in water at 25°C if the K_sp for is 1.1 × . A) 3.5 × 10^-4 g/L B) 2.2 × 10^-2 g/L C) 2.7 × 10^-2 g/L D) 3.4 × 10^-2 g/L

Question

Quizplus · Accepted Answer

The solubility product constant expression for silver chromate is:

Ksp = [Ag2CrO4][2CrO42-]

Let x be the solubility of silver chromate in mol/L. Then, at equilibrium:

[Ag2CrO4] = x
[CrO42-] = 2x

Substituting these values into the Ksp expression and solving for x:

Ksp = x(2x)^2 = 4x^3

x = (Ksp/4)^(1/3) = (1.1 × 10^-12/4)^(1/3) = 2.22 × 10^-4 mol/L

To convert this to g/L, we need to multiply by the molar mass of silver chromate:

Ag2CrO4: 2(107.87 g/mol) + 2(52.00 g/mol) + 4(16.00 g/mol) = 331.73 g/mol

solubility = 2.22 × 10^-4 mol/L x 331.73 g/mol = 0.0736 g/L

Therefore, the solubility of silver chromate in water at 25°C is 2.2 × 10^-2 g/L or 2.2 × 10^-2 g/L rounded to two significant figures. The closest answer choice is B.

Calculate the Solubility (In G/L)of Silver Chromate in Water at 25°C