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Silver sulfate (Ag2SO4) is slightly soluble in water and it partly dissolves at the equilibrium according to the following balanced reaction (Ks = 1.4x10): Ag2SO4) = 2 Ag (aq) + S0 + (aq) i) Calculate the molar solubility (S) of Ag2SO4. [15 marks] ii) Calculate the solubility (S') of Ag2SO4 in grams per liter.

Respuesta :

Answer:

a) S = 0.0152 mol/L

b) S' = 4.734 g/L

Explanation:

  • Ag2SO4 ↔ 2Ag+  +  SO42-

           S                2S            S...............in the equilibrium

  • Ksp = 1.4 E-5 = [ Ag+ ]² * [ SO42-]

a) molar solubility:

⇒ Ksp = ( 2S) ² * S = 1.4 E-5

⇒ 4S² * S = 1.4 E-5

⇒ S = ∛ ( 1.4 E-5 / 4 )

⇒ S = 0.0152 mol/L

b) solubility ( S' ) in grams per liter:

∴ Mw Ag2SO4 = 311.799 g/mol

⇒ S' = 0.0152 mol/L * ( 311.799 g/mol )

⇒ S' = 4.734 g/L

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