Oxalic acid (H2C2O4, Lewis structure shown) is a diprotic acid with the following Ka values
The Correct Answer and Explanation is:
The correct answer is A. A solution of NaHC₂O₄ is acidic because Ka(HC₂O₄⁻) > Kb(HC₂O₄⁻).
To understand this, we must analyze the nature of the salt NaHC₂O₄. Oxalic acid (H₂C₂O₄) is a diprotic acid, meaning it donates two protons sequentially. The first dissociation yields HC₂O₄⁻, and the second produces C₂O₄²⁻. When NaHC₂O₄ dissolves, it forms Na⁺ and HC₂O₄⁻ ions. Sodium ions do not affect the solution’s pH, so the key lies in evaluating the behavior of the HC₂O₄⁻ ion.
HC₂O₄⁻ is amphiprotic. It can either donate a proton (acting as an acid) or accept a proton (acting as a base). To determine whether it acts more strongly as an acid or a base, we compare its acid dissociation constant (Ka) to its base ionization constant (Kb).
Given:
- Ka for HC₂O₄⁻ = 1.5 × 10⁻⁴
- To find Kb for HC₂O₄⁻, we use the expression Kb = Kw / Ka, where Kw = 1.0 × 10⁻¹⁴
- Kb = (1.0 × 10⁻¹⁴) / (1.5 × 10⁻⁴) = 6.67 × 10⁻¹¹
Since Ka for HC₂O₄⁻ is several orders of magnitude larger than its Kb, the ion acts predominantly as an acid in solution. As a result, when NaHC₂O₄ dissolves in water, the HC₂O₄⁻ ions donate protons to water molecules, forming H₃O⁺ and lowering the pH below 7.
Therefore, the solution is acidic, and this confirms that option A is correct. The remaining choices either misrepresent the species involved or incorrectly compare the strength of acidic and basic behavior.
