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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

Respuesta :

Edufirst

Answer:

  • 1.03 × 10⁻⁴ M

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:

  • [tex]s=\sqrt[3]{(1.10)(10^{-12})}=1.03[/tex] × [tex]10^{-4}[/tex]
jrgould72

Answer:

6.50 x 105M

Explanation:

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