To determine which reaction will produce a solution that does not conduct electricity, we need to examine the nature of the substances involved and their behavior in water.
1. [tex]$CH_3OH(l) \rightarrow CH_3OH(aq)$[/tex]:
- Methanol ([tex]$CH_3OH$[/tex]) is a neutral organic molecule. When it dissolves in water, it does not dissociate into ions. Since electrical conductivity in a solution arises from the presence of ions, methanol in water will not conduct electricity.
2. [tex]$CaBr_2(s) \rightarrow Ca^{2+}(aq) + 2Br^{-}(aq)$[/tex]:
- Calcium bromide ([tex]$CaBr_2$[/tex]) is an ionic compound. When it dissolves in water, it dissociates into calcium ions ([tex]$Ca^{2+}$[/tex]) and bromide ions ([tex]$Br^{-}$[/tex]). These ions are free to move in solution, enabling it to conduct electricity.
3. [tex]$HF(g) + H_2O(l) \longleftrightarrow H_3O^{+}(aq) + F^{-}(aq)$[/tex]:
- Hydrofluoric acid ([tex]$HF$[/tex]) is a weak acid. When it dissolves in water, it partially ionizes to produce hydronium ions ([tex]$H_3O^+$[/tex]) and fluoride ions ([tex]$F^-$[/tex]). Even though the ionization is not complete, the ions present in the solution will still conduct electricity.
4. [tex]$NH_3(g) + H_2O(l) \longleftrightarrow NH_4^{+}(aq) + OH^{-}(aq)$[/tex]:
- Ammonia ([tex]$NH_3$[/tex]) is a weak base. When it dissolves in water, it reacts with water to produce ammonium ions ([tex]$NH_4^+$[/tex]) and hydroxide ions ([tex]$OH^-$[/tex]). These ions are also capable of conducting electricity.
From the above analysis, we can see that the reaction involving methanol ([tex]$CH_3OH$[/tex]) is the one that does not produce any ions in solution. Therefore, this reaction will produce a solution that does not conduct electricity.
Answer:
The correct reaction that produces a solution that does not conduct electricity is:
[tex]\[ \boxed{CH_3OH(l) \rightarrow CH_3OH(aq)} \][/tex]
