using chemical equations, show how the triprotic acid H3PO4 ionizes. Phases are optional. Ka2
The correct answer and explanation is:
The ionization of phosphoric acid (H₃PO₄), a triprotic acid, occurs in three steps, each corresponding to the loss of a proton (H⁺) from the molecule. The ionization can be represented by the following chemical equations:
First ionization:
H3PO4(aq)⇌H+(aq)+H2PO4−(aq)H_3PO_4 (aq) \rightleftharpoons H^+ (aq) + H_2PO_4^- (aq)
Second ionization:
H2PO4−(aq)⇌H+(aq)+HPO42−(aq)H_2PO_4^- (aq) \rightleftharpoons H^+ (aq) + HPO_4^{2-} (aq)
Third ionization:
HPO42−(aq)⇌H+(aq)+PO43−(aq)HPO_4^{2-} (aq) \rightleftharpoons H^+ (aq) + PO_4^{3-} (aq)
Each ionization step is characterized by its corresponding acid dissociation constant (Ka). The Ka values for each step decrease with each successive ionization, meaning that the first ionization is the strongest, and the third ionization is the weakest.
The first dissociation (Ka₁) represents the loss of a proton from H₃PO₄ to form H₂PO₄⁻. This reaction is relatively strong, with Ka₁ being about 7.5×10−37.5 \times 10^{-3}. The second dissociation (Ka₂) involves the loss of a proton from H₂PO₄⁻ to form HPO₄²⁻. The value of Ka₂ is smaller, approximately 6.2×10−86.2 \times 10^{-8}, showing that this ionization is weaker than the first. The third dissociation (Ka₃) is even weaker, with Ka₃ being around 4.8×10−134.8 \times 10^{-13}, indicating that HPO₄²⁻ loses its proton to form PO₄³⁻ only under very specific conditions.
The successive decrease in Ka values shows that each proton is more difficult to lose than the previous one. The Ka values also reflect the increasing stability of the ions formed in each step. The process of ionization in H₃PO₄ is essential in understanding its role as a buffer in biological systems, where it helps maintain pH balance by accepting or donating protons depending on the conditions.