Answer:
BaC₂O₄, then ZnC₂O₄, then Ag₂C₂O₄
Explanation:
1. Calculate the equilibrium concentrations of oxalate ion
Let [C₂O₄²⁻] = c
(a) Barium oxalate
BaC₂O₄ ⇌ Ba²⁺ + C₂O₄²⁻
E/mol·L⁻¹: 5.0 × 10⁻⁵ c
Ksp = [Ba²⁺][C₂O₄²⁻] = 5.0 × 10⁻⁵c = 1.5 × 10⁻⁸
c = (1.5 × 10⁻⁸)/(5.0 × 10⁻⁵) = 3.0 × 10⁻⁴ mol·L⁻¹
(b) Zinc oxalate
ZnC₂O₄ ⇌ Zn²⁺ + C₂O₄²⁻
E/mol·L⁻¹: 2.0 × 10⁻⁷ c
Ksp = [Zn²⁺][C₂O₄²⁻] = 2.0 × 10⁻⁷c = 1.35 × 10⁻⁹
c = (1.35 × 10⁻⁹)/(2.0 × 10⁻⁷) = 6.8 × 10⁻³ mol·L⁻¹
(c) Silver oxalate
Ag₂C₂O₄ ⇌ 2Ag⁺ + C₂O₄²⁻
E/mol·L⁻¹: 3.0 × 10⁻⁵ c
Ksp = [Ag⁺]²[C₂O₄²⁻] = (3.0× 10⁻⁵)²c = 9.0 × 10⁻¹⁰c = 1.1 × 10⁻¹¹
c = (1.1 × 10⁻¹¹)/(9.0 × 10⁻¹⁰) = 0.012 mol·L⁻¹
2. Decide the order of precipitation
BaC₂O₄ will precipitate when c > 3.0 × 10⁻⁴ mol·L⁻¹
ZnC₂O₄ will precipitate when c > 6.8 × 10⁻³ mol·L⁻¹
Ag₂C₂O₄ will precipitate when c > 0.028 mol·L⁻¹
This happens to be the order of increasing concentration of oxalate ion.
The order of precipitation is
BaC₂O₄, then ZnC₂O₄, then Ag₂C₂O₄
