What is the net ionic equation for the neutralization of CHOOH with NaOH?

The molecular equation is HCOOH + NaOH ==> HCOONa + H2O

The comnlete ionic equation is (with phases)
HCOOH(aq) + Na^+(aq) + OH^(aq) ==> -HCOO^-(aq) + Na^+(aq) + H2O(l)
Now cancel those ions/molecules common to both left and right sides of the equation. The Na^+(aq) is the only common ion.
What's left is the net ionic equatio.
HCOOH(aq) + OH^-(aq) ==> HCOO^- +(aq) + H2O(l)

Oh, we've got a classic case of CHOOH getting too acidic and needing some NaOH to neutralize it. Let's see, the net ionic equation for this neutralization dance goes something like this:

CHOOH + OH- → CHO- + H2O

That's right, the acetic acid (CHOOH) teams up with hydroxide ions (OH-) to form the acetate ion (CHO-) and water (H2O). So, in the end, everything balances out nicely and the party is over.

To determine the net ionic equation for the neutralization of CHOOH (formic acid) with NaOH (sodium hydroxide), we need to first write the balanced chemical equation for the reaction.

The chemical formula for formic acid (CHOOH) is HCOOH, and the chemical formula for sodium hydroxide is NaOH.

The balanced chemical equation for the neutralization reaction is:

HCOOH + NaOH → H2O + NaHCOO

Next, we can write the complete ionic equation by breaking all the soluble ionic compounds into their respective ions:

H+(aq) + COOH-(aq) + Na+(aq) + OH-(aq) → H2O(l) + Na+(aq) + HCOO-(aq)

Since Na+(aq) and H2O(l) are present on both sides of the equation and they are spectator ions (they do not participate in the reaction), we can remove them from the equation to form the net ionic equation:

H+(aq) + COOH-(aq) + OH-(aq) → HCOO-(aq)

Therefore, the net ionic equation for the neutralization of CHOOH with NaOH is:

H+(aq) + COOH-(aq) + OH-(aq) → HCOO-(aq)

To determine the net ionic equation for the neutralization of CHOOH (formic acid) with NaOH (sodium hydroxide), you need to first write the balanced chemical equation for the reaction.

The chemical equation for the neutralization reaction between CHOOH and NaOH can be represented as:

CHOOH + NaOH -> H2O + NaOOCH

To write the net ionic equation, you need to separate the soluble ionic compounds into their respective ions. In this reaction, NaOH is a strong base and completely dissociates into its constituent ions in water (H2O). However, CHOOH is a weak acid and does not fully dissociate.

The dissociation of NaOH is as follows:
NaOH -> Na+ + OH-

The dissociation of CHOOH is as follows:
CHOOH -> H+ + OOCH-

Since H+ ions from the acid and OH- ions from the base combine to form water (H2O), we can cancel out the common ions (H+ and OH-) on both sides of the equation to write the net ionic equation:

H+ + OH- -> H2O

Therefore, the net ionic equation for the neutralization of CHOOH with NaOH is:

H+ + OH- -> H2O