Question

what is the balanced net ionic equation for the reaction below? hi(aq) + naoh(aq) → nai(aq) + h₂o(l) be watchful for iss, wa, and wb! h⁺(aq) + oh⁻(aq) → h₂o(l) h⁺(aq) + i⁻(aq) + oh⁻(aq) + na⁺ → h₂o(l) + nai(aq) hi(aq) + naoh(aq) → h₂o(l) +nai(aq) hi(aq) + oh⁻(aq) + na⁺ → h₂o(l) + i⁻(aq)

Explanation:

1. First, identify the strong electrolytes (ISS - Ionic Substances in Solution, WA - Weak Acid, WB - Weak Base). HI is a strong acid (dissociates completely into \(H^+(aq)\) and \(I^-(aq)\)), NaOH is a strong base (dissociates completely into \(Na^+(aq)\) and \(OH^-(aq)\)), and NaI is a strong electrolyte (dissociates into \(Na^+(aq)\) and \(I^-(aq)\)).
2. The total ionic equation would be \(H^+(aq)+I^-(aq)+Na^+(aq)+OH^-(aq)

ightarrow Na^+(aq)+I^-(aq)+H_2O(l)\).

1. To get the net ionic equation, cancel the spectator ions (\(Na^+(aq)\) and \(I^-(aq)\)) from both sides. This leaves \(H^+(aq)+OH^-(aq)

ightarrow H_2O(l)\).

Answer:

A. \(H^+(aq) + OH^-(aq)
ightarrow H_2O(l)\)