To determine the complete ionic equation for the chemical reaction [tex]\(NaOH (aq) + HCl (aq) \rightarrow H_2O (l) + NaCl (aq)\)[/tex], let's break down the reaction into its constituent ionic forms.
1. Dissociation of Reactants:
- [tex]\(NaOH (aq)\)[/tex] dissociates into [tex]\(Na^+ (aq)\)[/tex] and [tex]\(OH^- (aq)\)[/tex]:
[tex]\[ NaOH (aq) \rightarrow Na^+ (aq) + OH^- (aq) \][/tex]
- [tex]\(HCl (aq)\)[/tex] dissociates into [tex]\(H^+ (aq)\)[/tex] and [tex]\(Cl^- (aq)\)[/tex]:
[tex]\[ HCl (aq) \rightarrow H^+ (aq) + Cl^- (aq) \][/tex]
2. Combination of Ions to Form Products:
- [tex]\(H_2O (l)\)[/tex] is a non-dissociated liquid:
[tex]\[ H_2O (l) \text{ remains as } H_2O (l) \][/tex]
- [tex]\(NaCl (aq)\)[/tex] dissociates into [tex]\(Na^+ (aq)\)[/tex] and [tex]\(Cl^- (aq)\)[/tex]:
[tex]\[ NaCl (aq) \rightarrow Na^+ (aq) + Cl^- (aq) \][/tex]
Given the complete dissociation and recombination, we can write the complete ionic equation as follows:
[tex]\[ Na^+ (aq) + OH^- (aq) + H^+ (aq) + Cl^- (aq) \rightarrow H_2O (l) + Na^+ (aq) + Cl^- (aq) \][/tex]
After writing the complete ionic equation, we observe that [tex]\(Na^+ (aq)\)[/tex] and [tex]\(Cl^- (aq)\)[/tex] appear on both sides of the equation. These are spectator ions and do not change during the course of the reaction. Thus, if we removed the spectator ions, we would get the net ionic equation.
However, for the complete ionic equation before simplification, the correct form is:
[tex]\[ Na^+ (aq) + OH^- (aq) + H^+ (aq) + Cl^- (aq) \rightarrow H_2O (l) + Na^+ (aq) + Cl^- (aq) \][/tex]
Therefore, the correct answer choice that represents the complete ionic equation is:
D. [tex]\(Na^+ (aq) + OH^- (aq) + H^+ (aq) + Cl^- (aq) \rightarrow H_2O (l) + Na^+ (aq) + Cl^- (aq)\)[/tex]
