Answer :
When each of the given ionic compounds is dissolved in water, they dissociate into their respective ions. Let’s break down each compound and determine the ions that are present in an aqueous solution.
1. Sodium iodide ([tex]\( \text{NaI} \)[/tex]):
- Sodium iodide dissociates completely in water.
- The dissociation can be represented as:
[tex]\[ \text{NaI} \rightarrow \text{Na}^+ + \text{I}^- \][/tex]
- Hence, the ions present in an aqueous solution of sodium iodide are [tex]\( \text{Na}^+ \)[/tex] and [tex]\( \text{I}^- \)[/tex].
2. Potassium sulfate ([tex]\( \text{K}_2\text{SO}_4 \)[/tex]):
- Potassium sulfate dissociates into two potassium ions and one sulfate ion in water.
- The dissociation can be represented as:
[tex]\[ \text{K}_2\text{SO}_4 \rightarrow 2\text{K}^+ + \text{SO}_4^{2-} \][/tex]
- Thus, the ions present in an aqueous solution of potassium sulfate are [tex]\( 2\text{K}^+ \)[/tex] and [tex]\( \text{SO}_4^{2-} \)[/tex].
3. Sodium cyanide ([tex]\( \text{NaCN} \)[/tex]):
- Sodium cyanide dissociates completely in water.
- The dissociation can be represented as:
[tex]\[ \text{NaCN} \rightarrow \text{Na}^+ + \text{CN}^- \][/tex]
- Therefore, the ions present in an aqueous solution of sodium cyanide are [tex]\( \text{Na}^+ \)[/tex] and [tex]\( \text{CN}^- \)[/tex].
4. Barium hydroxide ([tex]\( \text{Ba}(\text{OH})_2 \)[/tex]):
- Barium hydroxide dissociates into one barium ion and two hydroxide ions in water.
- The dissociation can be represented as:
[tex]\[ \text{Ba}(\text{OH})_2 \rightarrow \text{Ba}^{2+} + 2\text{OH}^- \][/tex]
- Hence, the ions present in an aqueous solution of barium hydroxide are [tex]\( \text{Ba}^{2+} \)[/tex] and [tex]\( 2\text{OH}^- \)[/tex].
5. Ammonium sulfate ([tex]\( (\text{NH}_4)_2\text{SO}_4 \)[/tex]):
- Ammonium sulfate dissociates into two ammonium ions and one sulfate ion in water.
- The dissociation can be represented as:
[tex]\[ (\text{NH}_4)_2\text{SO}_4 \rightarrow 2\text{NH}_4^+ + \text{SO}_4^{2-} \][/tex]
- Therefore, the ions present in an aqueous solution of ammonium sulfate are [tex]\( 2\text{NH}_4^+ \)[/tex] and [tex]\( \text{SO}_4^{2-} \)[/tex].
Summarizing, the ions present in an aqueous solution for each compound are:
- [tex]\( \text{NaI} \)[/tex]: [tex]\( \text{Na}^+ \)[/tex] and [tex]\( \text{I}^- \)[/tex]
- [tex]\( \text{K}_2\text{SO}_4 \)[/tex]: [tex]\( 2\text{K}^+ \)[/tex] and [tex]\( \text{SO}_4^{2-} \)[/tex]
- [tex]\( \text{NaCN} \)[/tex]: [tex]\( \text{Na}^+ \)[/tex] and [tex]\( \text{CN}^- \)[/tex]
- [tex]\( \text{Ba}(\text{OH})_2 \)[/tex]: [tex]\( \text{Ba}^{2+} \)[/tex] and [tex]\( 2\text{OH}^- \)[/tex]
- [tex]\( (\text{NH}_4)_2\text{SO}_4 \)[/tex]: [tex]\( 2\text{NH}_4^+ \)[/tex] and [tex]\( \text{SO}_4^{2-} \)[/tex]
1. Sodium iodide ([tex]\( \text{NaI} \)[/tex]):
- Sodium iodide dissociates completely in water.
- The dissociation can be represented as:
[tex]\[ \text{NaI} \rightarrow \text{Na}^+ + \text{I}^- \][/tex]
- Hence, the ions present in an aqueous solution of sodium iodide are [tex]\( \text{Na}^+ \)[/tex] and [tex]\( \text{I}^- \)[/tex].
2. Potassium sulfate ([tex]\( \text{K}_2\text{SO}_4 \)[/tex]):
- Potassium sulfate dissociates into two potassium ions and one sulfate ion in water.
- The dissociation can be represented as:
[tex]\[ \text{K}_2\text{SO}_4 \rightarrow 2\text{K}^+ + \text{SO}_4^{2-} \][/tex]
- Thus, the ions present in an aqueous solution of potassium sulfate are [tex]\( 2\text{K}^+ \)[/tex] and [tex]\( \text{SO}_4^{2-} \)[/tex].
3. Sodium cyanide ([tex]\( \text{NaCN} \)[/tex]):
- Sodium cyanide dissociates completely in water.
- The dissociation can be represented as:
[tex]\[ \text{NaCN} \rightarrow \text{Na}^+ + \text{CN}^- \][/tex]
- Therefore, the ions present in an aqueous solution of sodium cyanide are [tex]\( \text{Na}^+ \)[/tex] and [tex]\( \text{CN}^- \)[/tex].
4. Barium hydroxide ([tex]\( \text{Ba}(\text{OH})_2 \)[/tex]):
- Barium hydroxide dissociates into one barium ion and two hydroxide ions in water.
- The dissociation can be represented as:
[tex]\[ \text{Ba}(\text{OH})_2 \rightarrow \text{Ba}^{2+} + 2\text{OH}^- \][/tex]
- Hence, the ions present in an aqueous solution of barium hydroxide are [tex]\( \text{Ba}^{2+} \)[/tex] and [tex]\( 2\text{OH}^- \)[/tex].
5. Ammonium sulfate ([tex]\( (\text{NH}_4)_2\text{SO}_4 \)[/tex]):
- Ammonium sulfate dissociates into two ammonium ions and one sulfate ion in water.
- The dissociation can be represented as:
[tex]\[ (\text{NH}_4)_2\text{SO}_4 \rightarrow 2\text{NH}_4^+ + \text{SO}_4^{2-} \][/tex]
- Therefore, the ions present in an aqueous solution of ammonium sulfate are [tex]\( 2\text{NH}_4^+ \)[/tex] and [tex]\( \text{SO}_4^{2-} \)[/tex].
Summarizing, the ions present in an aqueous solution for each compound are:
- [tex]\( \text{NaI} \)[/tex]: [tex]\( \text{Na}^+ \)[/tex] and [tex]\( \text{I}^- \)[/tex]
- [tex]\( \text{K}_2\text{SO}_4 \)[/tex]: [tex]\( 2\text{K}^+ \)[/tex] and [tex]\( \text{SO}_4^{2-} \)[/tex]
- [tex]\( \text{NaCN} \)[/tex]: [tex]\( \text{Na}^+ \)[/tex] and [tex]\( \text{CN}^- \)[/tex]
- [tex]\( \text{Ba}(\text{OH})_2 \)[/tex]: [tex]\( \text{Ba}^{2+} \)[/tex] and [tex]\( 2\text{OH}^- \)[/tex]
- [tex]\( (\text{NH}_4)_2\text{SO}_4 \)[/tex]: [tex]\( 2\text{NH}_4^+ \)[/tex] and [tex]\( \text{SO}_4^{2-} \)[/tex]
