Final answer:
The net ionic equation for the reaction between sulfuric acid (H2SO4) and sodium hydroxide (NaOH) is: 2H+ (aq) + 2OH- (aq) → 2H2O (l), which shows the H+ ions from the acid reacting with OH- ions from the base to form water.
Step-by-step explanation:
The balanced chemical equation for the reaction between sulfuric acid (H2SO4) and sodium hydroxide (NaOH) is already provided: H2SO4 (aq) + 2NaOH(aq) → Na2SO4 (aq) + 2H2O(l). To write the ionic equation for this reaction, first, we need to consider the dissociation of the strong acid and strong base in water. Sulfuric acid is a strong acid and ionizes to form 2H+ and SO42- ions. Sodium hydroxide is a strong base and ionizes to form Na+ and OH- ions.
The net ionic equation for the reaction is then:
2H+ (aq) + 2OH- (aq) → 2H2O (l)
This equation shows that H+ ions from the sulfuric acid react with OH- ions from the sodium hydroxide to form water. Since sodium sulfate (Na2SO4) is soluble, the Na+ and SO42- ions remain in solution and are not part of the net ionic equation.
In order to deduce the stoichiometry of this neutralization reaction, it's essential to understand that sulfuric acid is diprotic, meaning it can donate two H+ ions, which requires twice as many OH- ions from the treatment with sodium hydroxide to completely neutralize the acid, resulting in the formation of sodium sulfate and water.