Answer:
H₄P₂O₇ + H₂O ⇄ H₃P₂O₇⁻ + H₃O⁺
Weak acid Base Strong conjugate base conjugate acid
H₃P₂O₇⁻ + H₂O ⇄ H₂P₂O₇⁻² + H₃O⁺
Weak acid Base Strong conjugate base conjugate acid
H₂P₂O₇⁻² + H₂O ⇄ HP₂O₇⁻³ + H₃O⁺
Weak acid Base Strong conjugate base conjugate acid
HP₂O₇⁻³ + H₂O ⇄ P₂O₇⁻⁴ + H₃O⁺
Weak acid Base Strong conjugate base conjugate acid
Explanation:
Pyrophosphoric acid → H₄P₂O₇
It's a poliprotic acid, which is able to release more than two protons. Pyrophosphoric acid is generated when we add 2 water to phosphorus (V) oxide .
P₂O₅ + 2H₂O → H₄P₂O₇
As it is able to release 4 protons, it will have 4 Ka for each equilibrium.
H₄P₂O₇ + H₂O ⇄ H₃P₂O₇⁻ + H₃O⁺
Weak acid Base Strong conjugate base conjugate acid
H₃P₂O₇⁻ + H₂O ⇄ H₂P₂O₇⁻² + H₃O⁺
Weak acid Base Strong conjugate base conjugate acid
H₂P₂O₇⁻² + H₂O ⇄ HP₂O₇⁻³ + H₃O⁺
Weak acid Base Strong conjugate base conjugate acid
HP₂O₇⁻³ + H₂O ⇄ P₂O₇⁻⁴ + H₃O⁺
Weak acid Base Strong conjugate base conjugate acid
As the water, is always receiving one H⁺ behaves as a base, so the H₃O⁺ will be the conjugate acid
H₄P₂O₇ is a weak acid, it donates the H⁺ to water, so the anions will be the strong conjugate base. They will react in water donating the protons until become P₂O₇⁻⁴
