Final answer:
To write the net ionic equation for the reaction between H₃PO₄ and NaOH, we consider the dissociation of these compounds in solution and eliminate the spectator ions. The net ionic equation is H₃PO₄ (aq) + OH⁻ (aq) -> H₂O (l) + H₂PO₄⁻ (aq).
Step-by-step explanation:
The question involves writing the net ionic equation for the reaction between H₃PO₄ (phosphoric acid) and NaOH (sodium hydroxide). In the reaction, H₃PO₄ can dissociate into H₂PO₄⁻ and H⁺⁻ (first ionization), which can further dissociate into HPO₄²⁻ and H⁺⁻ (second ionization), and eventually into PO₄³⁻ and H⁺⁻ (third ionization). As NaOH is a strong base, it fully dissociates into Na⁺ and OH⁻ ions in solution. The complete ionic equation would show all the ions present, and the net ionic equation would focus on the ions that actually participate in the reaction, eliminating the spectator ions.
For the reaction between H₃PO₄ and NaOH, assuming we are dealing with the initial ionization of H₃PO₄, the equation would be balanced as follows:
- H₃PO₄ (aq) + NaOH (aq) -> Na₂HPO₄ (aq) + H₂O (l)
- H₃PO₄ (aq) would dissociate to form H²⁻ (aq) + H₂PO₄⁻ (aq)
- NaOH (aq) would dissociate to form Na⁺ (aq) + OH⁻ (aq)
- Since Na⁺ (aq) is a spectator ion, it is removed from the net ionic equation.
The net ionic equation is therefore:
H₃PO₄ (aq) + OH⁻ (aq) -> H₂O (l) + H₂PO₄⁻ (aq)