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

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

Answer: (A)

Verified step by step guidance

1. Write the balanced molecular equation for the reaction between beryllium chloride and sodium hydroxide: \(\mathrm{BeCl_2(aq) + 2NaOH(aq) \rightarrow 2NaCl(aq) + Be(OH)_2(s)}\).
2. Dissociate all strong electrolytes (soluble ionic compounds) into their ions to write the complete ionic equation: \(\mathrm{Be^{2+}(aq) + 2Cl^-(aq) + 2Na^+(aq) + 2OH^-(aq) \rightarrow 2Na^+(aq) + 2Cl^-(aq) + Be(OH)_2(s)}\).
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 complete ionic equation to write the net ionic equation: \(\mathrm{Be^{2+}(aq) + 2OH^-(aq) \rightarrow Be(OH)_2(s)}\).
5. Confirm that the net ionic equation shows the formation of the solid precipitate beryllium hydroxide from its aqueous ions, which is the key chemical change in this reaction.