Question

Calculate the molar solubility of Ag2 CrO4 in water. Use 1.10 x 10-12 as the solubility product constant of AgzCro4

7.42 x 10-7M
1.05 x 10-6M
6.50 x 105M
1.03 x 10-4M

Answer 1

6.50 x 105M

Step-by-step explanation:

Answer 2

• 1.03 × 10⁻⁴ M

Step-by-step explanation:

1) Write the chemical equation for the solubility equilibrium:

• Ag₂ CrO₄ ⇄ 2 Ag⁺ + CrO₄²⁻

2) Write the expression for the solubility product constant, Ksp:

• Ksp = [Ag⁺]² [CrO₄²⁻]

3) Relate the solubility with the concentrations:

• s = [Ag⁺] = [CrO₄²⁻]

4) Substitute in the Ksp equation:

• 1.10 × 10⁻¹² = s² × s = s³

5) Solve for s:

• 
  `s=\sqrt[3]{(1.10)(10^(-12))}=1.03` ×
  `10^(-4)`