Consider the reaction below:Na2SO4(aq) + BaCl2(aq) → BaSO4(s) + 2NaCl(aq)Which is the net ionic equation for the reaction?

• A. Na^+(aq) + Cl^-(aq) → NaCl(aq)
• B. Na2SO4(aq) + BaCl2(aq) → BaSO4(s) + 2NaCl(aq)
• C. Ba^{2+}(aq) + 2Cl^-(aq) → BaCl2(aq)
• D. Ba^{2+}(aq) + SO_4^{2-}(aq) → BaSO_4(s)

Answer: (D)

Verified step by step guidance

1. Write the balanced molecular equation for the reaction: \(\mathrm{Na_2SO_4 (aq) + BaCl_2 (aq) \rightarrow BaSO_4 (s) + 2NaCl (aq)}\).
2. Write the full ionic equation by dissociating all strong electrolytes (soluble salts) into their ions: \$2\mathrm{Na^+ (aq)} + \mathrm{SO_4^{2-} (aq)} + \mathrm{Ba^{2+} (aq)} + 2\mathrm{Cl^- (aq)} \rightarrow \mathrm{BaSO_4 (s)} + 2\mathrm{Na^+ (aq)} + 2\mathrm{Cl^- (aq)}$.
3. Identify the spectator ions, which are ions that appear unchanged on both sides of the equation. Here, \(\mathrm{Na^+}\) and \(\mathrm{Cl^-}\) are spectator ions.
4. Remove the spectator ions from the full ionic equation to write the net ionic equation, which includes only the species that form the precipitate: \(\mathrm{Ba^{2+} (aq) + SO_4^{2-} (aq) \rightarrow BaSO_4 (s)}\).
5. Confirm that the net ionic equation shows the formation of the solid precipitate \(\mathrm{BaSO_4}\) from its ions in solution, representing the actual chemical change occurring.