Oxalic acid can donate two protons to water in successive reactions: (1) H₂C₂O₄(aq) + H₂O(l) ⇌ H₃O⁺(aq) + HC₂O₄^-(aq)
(2) HC₂O₄^-(aq) + H₂O(l) ⇌ H₃O⁺(aq) + C₂O₄^2-(aq)
If K_c1 = 5.9 × 10^-2 and K_c2 = 6.4 × 10^-5 at 25°C,what is the value of K_c for reaction (3) ?
(3) H₂C₂O₄(aq) + 2 H₂O(l) ⇌ 2 H₃O⁺(aq) + C₂O₄^2-(aq)

Question

Quizplus · Accepted Answer

The overall equilibrium constant $K_c$ for reaction (3) is the product of the equilibrium constants for reactions (1) and (2), i.e., $K_{c1} \times K_{c2} = 5.9 \times 10^{-2} \times 6.4 \times 10^{-5} = 3.776 \times 10^{-6}$, which is approximately $3.8 \times 10^{-6}$.

Oxalic Acid Can Donate Two Protons to Water in Successive