HI(aq)+KOH(aq)→H2O(l)+KI(aq) express this as a complete ionic equation
H^+(aq) + I^-(aq) + K^+(aq) + OH^-(aq) ==> H2O(l) + K^+(aq) + (I^-)(aq)
The net ionic equation is
H^+(aq) + OH^-(aq) ==> H2O(l)
H^+(aq) + OH^-(aq) ==> H2O(l)
H+(aq)+I−(aq)+K+(aq)+OH−(aq)→H2O(l)+K+(aq)+I−(aq)
Well, let's add some humor to this reaction, shall we? Here's the complete ionic equation for this reaction:
Hilarious Ions (HI) + Kaleidoscope of Happiness (KOH) → Hilarious Water (H2O) + Kooky Iodine (KI)
I hope this equation brought a smile to your face!
To write the complete ionic equation, we need to split all the ions that are present in the reactants and products.
The given equation is: HI(aq) + KOH(aq) → H2O(l) + KI(aq).
First, let's start with the reactants:
HI(aq) dissociates into H+ and I- ions.
KOH(aq) dissociates into K+ and OH- ions.
So the ionic form of the reactants is:
H+(aq) + I-(aq) + K+(aq) + OH-(aq)
Now, let's move on to the products:
H2O(l) is a liquid and does not dissociate into ions.
KI(aq) dissociates into K+ and I- ions.
So the ionic form of the products is:
K+(aq) + I-(aq)
Now we can write the complete ionic equation by combining the ions that appear on both sides:
H+(aq) + I-(aq) + K+(aq) + OH-(aq) → H2O(l) + K+(aq) + I-(aq)
Note: In a complete ionic equation, we write all the ions that are present in the solution. However, it's important to note that spectator ions, which are ions that appear on both the reactant and product sides, can be omitted from the final equation. In this case, K+ and I- are spectator ions, so we can remove them to get a net ionic equation, which shows only the species that actually participate in the chemical reaction.