What is the net ionic equation for the reaction between BeCl_2(aq) and NaOH(aq)?

• A. Be^{2+}(aq) + 2Cl^-(aq) + 2Na^+(aq) + 2OH^-(aq) → Be(OH)_2(s) + 2Na^+(aq) + 2Cl^-(aq)
• B. BeCl_2(aq) + 2NaOH(aq) → 2NaCl(aq) + Be(OH)_2(s)
• C. Be^{2+}(aq) + 2OH^-(aq) → Be(OH)_2(s)
• D. Be(OH)_2(s) → Be^{2+}(aq) + 2OH^-(aq)

Answer: (C)

Verified step by step guidance

1. Identify the ions present in the reactants. From BeCl_2(aq), you have \(\text{Be}^{2+}(aq)\) and \(2\text{Cl}^-(aq)\), and from NaOH(aq), you have \(\text{Na}^+(aq)\) and \(\text{OH}^-(aq)\) ions.
2. Write the full ionic equation by separating all strong electrolytes into their ions: \(\text{Be}^{2+}(aq) + 2\text{Cl}^-(aq) + 2\text{Na}^+(aq) + 2\text{OH}^-(aq) \rightarrow \text{Be(OH)}_2(s) + 2\text{Na}^+(aq) + 2\text{Cl}^-(aq)\).
3. Identify the spectator ions, which are ions that appear unchanged on both sides of the equation. Here, \(\text{Na}^+(aq)\) and \(\text{Cl}^-(aq)\) are spectator ions.
4. Remove the spectator ions from the full ionic equation to write the net ionic equation. This leaves you with \(\text{Be}^{2+}(aq) + 2\text{OH}^-(aq) \rightarrow \text{Be(OH)}_2(s)\).
5. Confirm that the net ionic equation shows the formation of the solid precipitate \(\text{Be(OH)}_2(s)\) from the aqueous ions, which represents the actual chemical change occurring in the reaction.