Using chemical equations, show how the triprotic acid H3PO4 ionizes in water. Phases, such as (l) or (aq), are optional.
1. H3PO4 (aq) + H2O (l) ⇌ H2PO4⁻ (aq) + H3O⁺ (aq)
2. H2PO4⁻ (aq) + H2O (l) ⇌ HPO4²⁻ (aq) + H3O⁺ (aq)
3. HPO4²⁻ (aq) + H2O (l) ⇌ PO4³⁻ (aq) + H3O⁺ (aq)
Sure! The ionization of triprotic acid H3PO4 in water occurs in three steps. Here are the equations for each step:
Step 1:
H3PO4 + H2O ⇌ H3O+ + H2PO4-
Step 2:
H2PO4- + H2O ⇌ H3O+ + HPO42-
Step 3:
HPO42- + H2O ⇌ H3O+ + PO43-
This represents the successive ionization of phosphoric acid H3PO4, where it donates one hydrogen ion (H+) at each step, resulting in the formation of hydronium ions (H3O+) and consecutive polyatomic anions (H2PO4-, HPO42-, and PO43-).
To show how the triprotic acid H3PO4 ionizes in water, we can break down its ionization into three steps. Each step represents the acid donating one proton (H+) to water, resulting in the formation of its corresponding conjugate base and hydronium ion (H3O+).
Step 1:
H3PO4 (aq) + H2O (l) ⇌ H2PO4- (aq) + H3O+ (aq)
In this step, one hydrogen ion is transferred from H3PO4 to water, resulting in the formation of the dihydrogen phosphate ion (H2PO4-) and the hydronium ion (H3O+).
Step 2:
H2PO4- (aq) + H2O (l) ⇌ HPO42- (aq) + H3O+ (aq)
In this step, another hydrogen ion is donated from the dihydrogen phosphate ion (H2PO4-) to water, resulting in the formation of the hydrogen phosphate ion (HPO42-) and the hydronium ion (H3O+).
Step 3:
HPO42- (aq) + H2O (l) ⇌ PO43- (aq) + H3O+ (aq)
In the final step, the last hydrogen ion is donated from the hydrogen phosphate ion (HPO42-) to water, resulting in the formation of the phosphate ion (PO43-) and the hydronium ion (H3O+).
Overall, the ionization of the triprotic acid H3PO4 in water can be summarized as:
H3PO4 (aq) + 3 H2O (l) ⇌ H2PO4- (aq) + H3O+ (aq)
H2PO4- (aq) + 2 H2O (l) ⇌ HPO42- (aq) + H3O+ (aq)
HPO42- (aq) + H2O (l) ⇌ PO43- (aq) + H3O+ (aq)