Answer:
- There are 2 ionization constants associated with oxalic acid (H₂C₂O₄).
Step-by-step explanation:
The ionization of an acid is the process through which the acid loses its hydrogen ions, H⁺.
Monoprotic acids ionize one hydogen ion in one step. For instance, nitric acid, HNO₃.
Diprotic acids ionize into two hydrogen ions, each in one step, i.e. a total fo two steps. This is the case of oxalic acid (H₂C₂O₄). Other diprotic acids are sulfuric acid (H₂SO₄) and carbonic acid (H₂CO₃), among many others.
Triprotic acids ionizes into three hydrogen ions, each into one step, for a total of three steps. For instance, phosphoric acid (H₃PO₃).
Each ionizatiion step has its own ionization constant. So, since oxalic acid has two acid hydrogens, there are two ionization steps and two ionization constants.
The steps and expressions fot the ionization constants for oxalic acid are given by these equations:
- H₂C₂O₄ (aq) + H₂O (l) ⇄ H₃O⁺(aq) + HC₂O₄⁻(aq)
Ka₁ = [H₃O⁺ (aq)] [HC₂O₄⁻ (aq)] / [H₂C₂O₄(aq)]
- HC₂O₄⁻(aq) + H₂O (l) ⇄ H₃O⁺(aq) + C₂O₄²⁻(aq)
Ka₂ = [H₃O⁺ (aq)] [C₂O₄²⁻ (aq)] / [HC₂O₄⁻ (aq)]
You can find the values of the two ionization constants (Ka₁ and Ka₂) at different temperatures in the literature (tables).