1.
HCl(aq)+KOH(aq)→KCl(aq)+H2O(l)
Express your answer as a net ionic equation. Identify all of the phases in your answer.
2.
2HC2H3O2(aq)+Ca(OH)2(aq)→Ca(C2H3O2)2(aq)+2H2O(l)
Express your answer as a net ionic equation. Identify all of the phases in your answer.
you guys should just delete your website ya'll are useless af
To write the net ionic equation, first write the total (complete) ionic equation, remove ions that appear on both sides of the equation, the net ionic equation is what is left. Try that on this one. You should be left with H^+(aq) + OH^-(aq) ==> H2O(l)
I included the phases although your question didn't ask to do that.
For
2H^+(aq) + 2OH^-(aq)==> 2H2O(l)
1. HCl(aq) + KOH(aq) → KCl(aq) + H2O(l)
Net Ionic Equation: H+(aq) + OH-(aq) → H2O(l)
Phases:
HCl(aq) and KOH(aq) are both aqueous solutions,
KCl(aq) is also an aqueous solution,
H2O(l) is a liquid.
2. 2HC2H3O2(aq) + Ca(OH)2(aq) → Ca(C2H3O2)2(aq) + 2H2O(l)
Net Ionic Equation: 2H+(aq) + 2C2H3O2-(aq) + Ca2+(aq) + 2OH-(aq) → Ca(C2H3O2)2(aq) + 2H2O(l)
Phases:
HC2H3O2(aq) and Ca(OH)2(aq) are both aqueous solutions,
Ca(C2H3O2)2(aq) is also an aqueous solution,
H2O(l) is a liquid.
To write the net ionic equation, we first need to write the balanced molecular equation:
1. HCl(aq) + KOH(aq) → KCl(aq) + H2O(l)
Next, we need to break down each compound into its respective ions:
HCl(aq) → H+(aq) + Cl-(aq)
KOH(aq) → K+(aq) + OH-(aq)
KCl(aq) → K+(aq) + Cl-(aq)
H2O(l) remains as H2O(l)
Now, we can eliminate any spectator ions that appear on both sides of the equation. In this case, K+ and Cl- are spectator ions since they appear on both sides:
Net ionic equation: H+(aq) + OH-(aq) → H2O(l)
The phases for each species are:
H+(aq): aqueous (dissolved in water)
OH-(aq): aqueous (dissolved in water)
H2O(l): liquid