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) + 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) h⁺(aq) + oh⁻(aq) → h₂o(l)

Explanation:

Step1: Identify Strong Electrolytes

HI is a strong acid (WA - strong acid), NaOH is a strong base (WB - strong base), and NaI is a soluble salt (ISS - ionic strong substance). So, they dissociate completely.
HI(aq) dissociates into \( \text{H}^+(aq) + \text{I}^-(aq) \), NaOH(aq) dissociates into \( \text{Na}^+(aq) + \text{OH}^-(aq) \), and NaI(aq) dissociates into \( \text{Na}^+(aq) + \text{I}^-(aq) \).

Step2: Write Total Ionic Equation

Substitute dissociated ions:
\( \text{H}^+(aq) + \text{I}^-(aq) + \text{Na}^+(aq) + \text{OH}^-(aq)
ightarrow \text{Na}^+(aq) + \text{I}^-(aq) + \text{H}_2\text{O}(l) \)

Step3: Cancel Spectator Ions

\( \text{Na}^+(aq) \) and \( \text{I}^-(aq) \) appear on both sides, so cancel them.

Step4: Net Ionic Equation

After canceling, we get: \( \text{H}^+(aq) + \text{OH}^-(aq)
ightarrow \text{H}_2\text{O}(l) \)

Answer:

\( \text{H}^+(aq) + \text{OH}^-(aq)
ightarrow \text{H}_2\text{O}(l) \) (corresponding to the option \( \text{H}^+(aq) + \text{OH}^-(aq)
ightarrow \text{H}_2\text{O}(l) \))