Answer:
[tex]\rm Cu^{2+} + S^{2-} \to CuS[/tex].
Explanation:
Start by identifying the solubility of each species in the original chemical equation:
[tex]\rm CuBr_2\, (aq) + Na_2S\, (aq) \to CuS\, (s) + 2\, NaBr\, (aq)[/tex]. (Balanced.)
- [tex]\rm CuBr_2[/tex], [tex]\rm Na_2 S[/tex], and [tex]\rm NaBr[/tex] are all soluble in water.
- [tex]\rm CuS[/tex] is not soluble in water.
Rewrite species that exist as ions as the corresponding ions to obtain an ionic equation for this reaction. Those species include:
- Soluble ionic compounds ([tex]\rm CuBr_2[/tex], [tex]\rm Na_2 S[/tex], and [tex]\rm NaBr[/tex].)
- Strong acids and bases. (Not in this question.)
In particular:
- [tex]\rm CuBr_2[/tex] will be rewritten as [tex]\rm Cu^{2+} + 2\, Br^{-}[/tex].
- [tex]\rm Na_2 S[/tex] will be rewritten as [tex]\rm 2\, Na^{+} + S^{2-}[/tex].
- [tex]2\, \rm NaBr[/tex] will be rewritten as [tex]\rm 2\, Na^{+} + 2\, Br^{-}[/tex].
Species that do not ionize in water ([tex]\rm CuS[/tex] in this question) should not be rewritten. Hence, the ionic equation will be:
[tex]\rm Cu^{2+} + 2\, Br^{-} + 2\, Na^{+} + S^{2-} \to CuS + 2\, Na^{+} + 2\, Br^{-}[/tex].
Eliminate ions that appear on both sides of the equation to obtain a net ionic equation. In this ionic equation, those ions include:
- Two [tex]\rm Br^{-}[/tex] ions, and
- Two [tex]\rm Na^{+}[/tex] ions.
Hence, the net ionic equation will be:
[tex]\rm Cu^{2+} + S^{2-} \to CuS[/tex].
