Which of the following is the correct net ionic equation for the reaction between ZnCl_2(aq) and NaOH(aq)?

• A. ZnCl_2(aq) + 2NaOH(aq) → Zn(OH)_2(s) + 2NaCl(aq)
• B. Zn(OH)_2(s) → Zn^{2+}(aq) + 2OH^-(aq)
• C. Zn^{2+}(aq) + 2OH^-(aq) → Zn(OH)_2(s)
• D. Zn^{2+}(aq) + 2Cl^-(aq) + 2Na^+(aq) + 2OH^-(aq) → Zn(OH)_2(s) + 2Na^+(aq) + 2Cl^-(aq)

Answer: (C)

Verified step by step guidance

1. Write the balanced molecular equation for the reaction: \(\mathrm{ZnCl_2(aq) + 2NaOH(aq) \rightarrow Zn(OH)_2(s) + 2NaCl(aq)}\).
2. Separate all strong electrolytes (soluble ionic compounds) into their ions to write the full ionic equation: \(\mathrm{Zn^{2+}(aq) + 2Cl^-(aq) + 2Na^+(aq) + 2OH^-(aq) \rightarrow Zn(OH)_2(s) + 2Na^+(aq) + 2Cl^-(aq)}\).
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: \(\mathrm{Zn^{2+}(aq) + 2OH^-(aq) \rightarrow Zn(OH)_2(s)}\).
5. Confirm that the net ionic equation shows the formation of the solid precipitate \(\mathrm{Zn(OH)_2}\) from its aqueous ions, which correctly represents the chemical change occurring in the reaction.