To determine the correct net ionic equation for the reaction between [tex]\( \text{Ca(OH)}_2 \)[/tex] and [tex]\( \text{H}_2\text{SO}_4 \)[/tex], we can follow these steps:
1. Write the balanced molecular equation for the reaction:
The balanced molecular equation for the reaction between calcium hydroxide and sulfuric acid is:
[tex]\[
\text{Ca(OH)}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{CaSO}_4 + 2\text{H}_2\text{O}
\][/tex]
2. Break down the strong electrolytes into their ions:
Both [tex]\(\text{Ca(OH)}_2\)[/tex] and [tex]\(\text{H}_2\text{SO}_4\)[/tex] are strong electrolytes, so they dissociate completely in water:
- [tex]\(\text{Ca(OH)}_2 \rightarrow \text{Ca}^{2+} + 2\text{OH}^-\)[/tex]
- [tex]\(\text{H}_2\text{SO}_4 \rightarrow 2\text{H}^+ + \text{SO}_4^{2-}\)[/tex]
3. Include the ions from the products:
[tex]\(\text{CaSO}_4\)[/tex] is a salt that can dissociate:
- [tex]\(\text{CaSO}_4 \rightarrow \text{Ca}^{2+} + \text{SO}_4^{2-}\)[/tex]
- However, the water ([tex]\( \text{H}_2\text{O} \)[/tex]) does not dissociate and remains as it is.
4. Write the full ionic equation by replacing each compound with its constituent ions:
[tex]\[
\text{Ca}^{2+} + 2\text{OH}^- + 2\text{H}^+ + \text{SO}_4^{2-} \rightarrow \text{Ca}^{2+} + \text{SO}_4^{2-} + 2\text{H}_2\text{O}
\][/tex]
5. Cancel the spectator ions:
Spectator ions are those ions that appear on both sides of the ionic equation. Here, [tex]\(\text{Ca}^{2+}\)[/tex] and [tex]\(\text{SO}_4^{2-}\)[/tex] are the spectator ions.
6. Write the net ionic equation by excluding the spectator ions:
Removing the spectator ions ([tex]\(\text{Ca}^{2+}\)[/tex] and [tex]\(\text{SO}_4^{2-}\)[/tex]), we get:
[tex]\[
2\text{H}^+ + 2\text{OH}^- \rightarrow 2\text{H}_2\text{O}
\][/tex]
Simplifying this equation to its simplest form gives:
[tex]\[
\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}
\][/tex]
Given the options, the correct net ionic equation is:
[tex]\[
\boxed{\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}}
\][/tex]
