To determine the net ionic equation of the reaction given:
[tex]\[
Na^{+} + OH^{-} + H^{+} + Cl^{-} \rightarrow H_2O + Na^{+} + Cl^{-}
\][/tex]
we need to identify the ions that participate in the chemical change and exclude the spectator ions.
### Step-by-Step Solution:
1. Identify all species in the reaction:
- On the reactants side, we have [tex]\( Na^{+} \)[/tex], [tex]\( OH^{-} \)[/tex], [tex]\( H^{+} \)[/tex], and [tex]\( Cl^{-} \)[/tex].
- On the products side, we have [tex]\( H_2O \)[/tex], [tex]\( Na^{+} \)[/tex], and [tex]\( Cl^{-} \)[/tex].
2. Determine the spectators:
- Spectator ions are present on both sides of the equation without undergoing any change.
- Here, [tex]\( Na^{+} \)[/tex] and [tex]\( Cl^{-} \)[/tex] are spectator ions because they appear as both reactants and products in the same form.
3. Write the net ionic equation by removing spectator ions:
- Removing [tex]\( Na^{+} \)[/tex] and [tex]\( Cl^{-} \)[/tex], we are left with [tex]\( OH^{-} + H^{+} \rightarrow H_2O \)[/tex].
4. Now, check the options given:
A. [tex]\( NaOH (aq) + HCl (aq) \rightarrow H_2O (l) + NaCl (aq) \)[/tex]
- This represents a complete ionic equation and not the net ionic equation.
B. [tex]\( Na^{+} + OH^{-} + H^{+} + Cl^{-} \rightarrow H^{+} + OH^{-} + Na^{+} + Cl^{-} \)[/tex]
- This is equivalent to a rearrangement of the initial equation and does not represent the net ionic equation.
C. [tex]\( Na^{+} + Cl^{-} \rightarrow NaCl \)[/tex]
- This reaction does not represent the net ionic change because the combination of sodium and chloride ions are spectators in this context.
D. [tex]\( OH^{-} + H^{+} \rightarrow H_2O \)[/tex]
- This correctly represents the net ionic equation, involving only the ions that actively participate in the reaction to form water.
Therefore, the correct net ionic equation is:
[tex]\[ OH^{-} + H^{+} \rightarrow H_2O \][/tex]
and the correct answer is:
D. [tex]\( OH^{-} + H^{+} \rightarrow H_2O \)[/tex]
