Final answer:
In the reaction between K2C2O4 and Pb(OH)2, PbC2O4 is formed as a precipitate. The complete ionic equation includes the dissociation of K2C2O4 into potassium (K+) and oxalate (C2O4^2-) ions, but sparingly soluble compounds like Pb(OH)2 may remain undissociated. The net ionic equation simply shows the formation of PbC2O4(s).
Step-by-step explanation:
The complete ionic and net ionic equations for the reaction between potassium oxalate (K2C2O4) and lead(II) hydroxide (Pb(OH)2) are derived from the principles that soluble ionic compounds dissociate into their constituent ions in aqueous solutions, and that insoluble compounds remain solid.
The complete ionic equation would dissociate all soluble reactants into their ions:
- K2C2O4(aq) → 2K+(aq) + C2O4^2-(aq)
- Pb(OH)2(aq) → Pb^2+(aq) + 2OH^-(aq)
Since Pb(OH)2 is only sparingly soluble in water and likely to exist as a precipitate, it should not be fully dissociated in the ionic equation. The precipitation reaction that occurs is represented by the net ionic equation:
- Pb^2+(aq) + C2O4^2-(aq) → PbC2O4(s)
In the case of the potassium ions (K+) and hydroxide ions (OH-) from Pb(OH)2, they are spectator ions and do not participate in the formation of the precipitate, so they are omitted from the net ionic equation.