Answer:
A. K₂C₂O₄ Potassium oxalate
B. CuC₂O₄ Copper oxalate
C. Bi₂(C₂O₄)₃ Bismuth (III) oxalate
D. Pb(C₂O₄)₂ Lead (IV) oxalate
E. (NH₄)₂C₂O₄ Ammonium oxalate
F. HC₂O₄⁻ Acid oxalate
Explanation:
C₂O₄⁻² → oxalate anion
This is the conjugate base from the H₂C₂O₄ which is the oxalic acid. A weak dyprotic acid that can release 2 protons.
A. 2K⁺ + C₂O₄⁻² → K₂C₂O₄ Potassium oxalate
It can be formed by the neutralization of the acid with the base
H₂C₂O₄ + 2KOH → K₂C₂O₄ + 2H₂O
B. Cu²⁺ + C₂O₄⁻² ⇄ CuC₂O₄ ↓
This is a precipitate.
C. 2Bi³⁺ + 3C₂O₄⁻² ⇄ Bi₂(C₂O₄)₃ ↓
This is a precipitate.
D. Pb⁴⁺ + 2C₂O₄⁻² ⇄ Pb(C₂O₄)₂ ↓
This is a precipitate.
E. 2NH₄⁺ + C₂O₄⁻² ⇄ (NH₄)₂C₂O₄ ↓
This is a precipitate.
F. This is the conjugate strong base, for the weak acid
H⁺ + C₂O₄⁻² ⇄ HC₂O₄⁻
HC₂O₄⁻ + H₂O ⇄ C₂O₄⁻² + H₃O⁺ Ka
HC₂O₄⁻ + H₂O ⇄ H₂C₂O₄ + OH⁻ Kb
HC₂O₄⁻ is an amphoteric compound
A. K₂C₂O₄ →Potassium oxalate
B. CuC₂O₄ →Copper oxalate
C. Bi₂(C₂O₄)₃ →Bismuth (III) oxalate
D. Pb(C₂O₄)₂ →Lead (IV) oxalate
E. (NH₄)₂C₂O₄ →Ammonium oxalate
F. HC₂O₄⁻ →Acid oxalate
Given:
It is the conjugate base from the H₂C₂O₄ which is the oxalic acid. A weak diprotic acid that can release 2 protons.
The reactions involved in this are:
A. 2K⁺ + C₂O₄⁻² → K₂C₂O₄
B. Cu²⁺ + C₂O₄⁻² ⇄ CuC₂O₄ ↓
C. 2Bi³⁺ + 3C₂O₄⁻² ⇄ Bi₂(C₂O₄)₃ ↓
D. Pb⁴⁺ + 2C₂O₄⁻² ⇄ Pb(C₂O₄)₂ ↓
E. 2NH₄⁺ + C₂O₄⁻² ⇄ (NH₄)₂C₂O₄ ↓
↓- This symbol denotes the precipitate formation.
F. H⁺ + C₂O₄⁻² ⇄ HC₂O₄⁻
HC₂O₄⁻ is an amphoteric compound.
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