Which one of the following equations is the net ionic equation for the reaction of aqueous sodium hydroxide (NaOH) with aqueous hydrochloric acid (HCl)? HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) H+(aq) + Cl‒(aq) + Na+(aq) + OH‒(aq) → Na+(aq) + Cl‒(aq) + H2O(l) H+(aq) + Cl‒(aq) + Na+(aq) + OH‒(aq) → NaCl(aq) + H2O(l) H+(aq) + OH‒(aq) → H2O(l)

H+(aq) +  OH-(aq) → H2O(l)

Explanation:

Step 1: Data given

aqueous sodium hydroxide = NaOH(aq)

aqueous hydrochloric acid =HCl(aq)

Step 2: The (un)balanced equation

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

For 1 mol HCl we need 1 mol NaOH to produce 1 mol NaCl and 1 mol H2O

This equation is already balanced.

Step 3: The net ionic equation

H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) → Na+(aq) + Cl-(aq) + H2O(l)

The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will , after canceling those spectator ions in both side, look like this:

H+(aq) +  OH-(aq) → H2O(l) (The last option is correct)

how about u try it first

the answer for sodium is 0.186

the answer for rubidium is 0.251