Aqueous sulfuric acid is neutralized by aqueous potassium hydroxide.
Express your answer as an ionic equation. Identify all of the phases in your answer.
2KOH + H2SO4 ==>2H2O + K2SO4
2K^+(aq) + 2OH^-(aq) + 2H^+(aq) + SO4^-2(aq) ==> 2H2O(l) + 2K^+(aq) + SO4^-2(aq)
The net ionic equation is
2OH^-(aq) + 2H^+(aq) ==> 2H2O(l)
Then I would divide by 2 to give simply
OH^-(aq) + H^+(aq) --> H2O(l) but your prof may frown on not putting the coefficients there.
To write the ionic equation for the neutralization reaction between aqueous sulfuric acid (H2SO4) and aqueous potassium hydroxide (KOH), we need to first write out the chemical equation and then break down the reactants and products into their respective ions.
The chemical equation for the reaction is:
H2SO4 + 2KOH -> K2SO4 + 2H2O
Now, let's separate the reactants and products into their respective ions:
Reactants:
H2SO4 (aq) -> 2H+ (aq) + SO4^2- (aq)
KOH (aq) -> K+ (aq) + OH- (aq)
Products:
K2SO4 (aq) -> 2K+ (aq) + SO4^2- (aq)
2H2O (l)
Now, we can write the ionic equation for the neutralization reaction:
2H+ (aq) + 2OH- (aq) -> 2H2O (l)
The phases indicated are:
H2SO4 (aq) - aqueous sulfuric acid
KOH (aq) - aqueous potassium hydroxide
K2SO4 (aq) - aqueous potassium sulfate
H2O (l) - liquid water
To write the ionic equation for the neutralization of aqueous sulfuric acid (H2SO4) with aqueous potassium hydroxide (KOH), we need to understand the formulas and charges of the ions involved.
The formula for sulfuric acid is H2SO4. It dissociates in water to form two hydrogen ions (H+) and one sulfate ion (SO4^2-). The formula for potassium hydroxide is KOH, which dissociates in water to form one potassium ion (K+) and one hydroxide ion (OH-).
Now, to write the ionic equation, we'll first separate the compounds into their respective ions and balance the charges. The balanced equation will not include any spectator ions, which are ions that appear on both sides of the reaction and do not participate in it.
The dissociation of sulfuric acid (H2SO4) is as follows:
H2SO4 (aq) -> 2 H+ (aq) + SO4^2- (aq)
The dissociation of potassium hydroxide (KOH) is as follows:
KOH (aq) -> K+ (aq) + OH- (aq)
In the neutralization reaction, the hydrogen ions (H+) from sulfuric acid react with the hydroxide ions (OH-) from potassium hydroxide to form water (H2O) as a neutral product:
H+ (aq) + OH- (aq) -> H2O (l)
The sulfate ion (SO4^2-) and potassium ion (K+) are spectator ions since they appear on both sides of the equation and do not participate in the reaction.
Therefore, the ionic equation for the neutralization of aqueous sulfuric acid with aqueous potassium hydroxide is:
2 H+ (aq) + 2 OH- (aq) -> 2 H2O (l)
In this equation, the (aq) notation represents the aqueous phase, and (l) represents the liquid phase of water.