Answer :
Okay, let's carefully determine the net ionic equation for the given reaction:
[tex]\[ 2 \text{KOH} (aq) + \text{H}_2\text{SO}_4(aq) \rightarrow 2 \text{H}_2\text{O} (\ell) + \text{K}_2\text{SO}_4(aq) \][/tex]
### Step-by-step solution:
1. Dissociate all strong electrolytes into their ions:
The reactants:
[tex]\[ 2 \text{KOH} (aq) \rightarrow 2 \text{K}^+ (aq) + 2 \text{OH}^- (aq) \][/tex]
[tex]\[ \text{H}_2\text{SO}_4(aq) \rightarrow 2 \text{H}^+ (aq) + \text{SO}_4^{2-} (aq) \][/tex]
The products:
[tex]\[ \text{K}_2\text{SO}_4(aq) \rightarrow 2 \text{K}^+ (aq) + \text{SO}_4^{2-} (aq) \][/tex]
[tex]\[ 2 \text{H}_2\text{O} (\ell) \text{ (is a liquid and does not dissociate into ions)} \][/tex]
2. Write the complete ionic equation including all ions:
[tex]\[ 2 \text{K}^+ (aq) + 2 \text{OH}^- (aq) + 2 \text{H}^+ (aq) + \text{SO}_4^{2-} (aq) \rightarrow 2 \text{H}_2\text{O} (\ell) + 2 \text{K}^+ (aq) + \text{SO}_4^{2-} (aq) \][/tex]
3. Identify the spectator ions:
Spectator ions are ions that appear on both sides of the equation and do not participate in the actual reaction. Here, the spectator ions are:
[tex]\[ 2 \text{K}^+ (aq) \text{ and } \text{SO}_4^{2-} (aq) \][/tex]
4. Remove the spectator ions to find the net ionic equation:
When we remove the spectator ions, we get:
[tex]\[ 2 \text{OH}^- (aq) + 2 \text{H}^+ (aq) \rightarrow 2 \text{H}_2\text{O} (\ell) \][/tex]
Simplify the coefficients:
[tex]\[ \text{OH}^- (aq) + \text{H}^+ (aq) \rightarrow \text{H}_2\text{O} (\ell) \][/tex]
Thus, the correct net ionic equation for the reaction is:
Choice D: [tex]\( \text{OH}^- (aq) + \text{H}^+ (aq) \rightarrow \text{H}_2\text{O} (\ell) \)[/tex]
[tex]\[ 2 \text{KOH} (aq) + \text{H}_2\text{SO}_4(aq) \rightarrow 2 \text{H}_2\text{O} (\ell) + \text{K}_2\text{SO}_4(aq) \][/tex]
### Step-by-step solution:
1. Dissociate all strong electrolytes into their ions:
The reactants:
[tex]\[ 2 \text{KOH} (aq) \rightarrow 2 \text{K}^+ (aq) + 2 \text{OH}^- (aq) \][/tex]
[tex]\[ \text{H}_2\text{SO}_4(aq) \rightarrow 2 \text{H}^+ (aq) + \text{SO}_4^{2-} (aq) \][/tex]
The products:
[tex]\[ \text{K}_2\text{SO}_4(aq) \rightarrow 2 \text{K}^+ (aq) + \text{SO}_4^{2-} (aq) \][/tex]
[tex]\[ 2 \text{H}_2\text{O} (\ell) \text{ (is a liquid and does not dissociate into ions)} \][/tex]
2. Write the complete ionic equation including all ions:
[tex]\[ 2 \text{K}^+ (aq) + 2 \text{OH}^- (aq) + 2 \text{H}^+ (aq) + \text{SO}_4^{2-} (aq) \rightarrow 2 \text{H}_2\text{O} (\ell) + 2 \text{K}^+ (aq) + \text{SO}_4^{2-} (aq) \][/tex]
3. Identify the spectator ions:
Spectator ions are ions that appear on both sides of the equation and do not participate in the actual reaction. Here, the spectator ions are:
[tex]\[ 2 \text{K}^+ (aq) \text{ and } \text{SO}_4^{2-} (aq) \][/tex]
4. Remove the spectator ions to find the net ionic equation:
When we remove the spectator ions, we get:
[tex]\[ 2 \text{OH}^- (aq) + 2 \text{H}^+ (aq) \rightarrow 2 \text{H}_2\text{O} (\ell) \][/tex]
Simplify the coefficients:
[tex]\[ \text{OH}^- (aq) + \text{H}^+ (aq) \rightarrow \text{H}_2\text{O} (\ell) \][/tex]
Thus, the correct net ionic equation for the reaction is:
Choice D: [tex]\( \text{OH}^- (aq) + \text{H}^+ (aq) \rightarrow \text{H}_2\text{O} (\ell) \)[/tex]
