Answer:
a) S = 0.0152 mol/L
b) S' = 4.734 g/L
Explanation:
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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The answer is 2
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