Which sequence of chemical equations correctly represents the stepwise ionization of the triprotic acid H3PO4 in water?

• A. H3PO4 + H2O → PO4^{3-} + 3H3O^+
• B. H3PO4 + H2O → H2PO4^- + H3O^+; H2PO4^- + H2O → HPO4^{2-} + H3O^+; HPO4^{2-} + H2O → PO4^{3-} + H3O^+
• C. H3PO4 + 3H2O → PO4^{3-} + 3H3O^+
• D. H3PO4 → H2PO4^- + H^+; H2PO4^- → HPO4^{2-} + H^+; HPO4^{2-} → PO4^{3-} + H^+

Answer: (B)

Verified step by step guidance

1. Understand that a triprotic acid like H3PO4 can lose three protons (H⁺) stepwise, each step representing one ionization equilibrium.
2. Write the first ionization step where H3PO4 donates one proton to water, forming the dihydrogen phosphate ion (H2PO4⁻) and hydronium ion (H3O⁺): \[H\3PO\4 + H\2O \rightarrow H\2PO\4^- + H\3O^+\]
3. Write the second ionization step where H2PO4⁻ donates a proton to water, forming the hydrogen phosphate ion (HPO4^{2-}) and another hydronium ion: \[H\2PO\4^- + H\2O \rightarrow HPO\4^{2-} + H\3O^+\]
4. Write the third ionization step where HPO4^{2-} donates the last proton to water, forming the phosphate ion (PO4^{3-}) and a hydronium ion: \[HPO\4^{2-} + H\2O \rightarrow PO\4^{3-} + H\3O^+\]
5. Note that each step involves water acting as a base accepting a proton, and the acid species losing one proton at a time, which is characteristic of stepwise ionization of polyprotic acids.