What is the net ionic equation for the reaction of formic acid, HCOOH, with NH3?

A. H+(aq) + NH3(aq) → NH4+(aq)
B. H+(aq) + OH–(aq) → H2O(l)
C. HCOOH(aq) + NH3(aq) → NH4+(aq) + HCOO–(aq)
D. HCOO–(aq) + H+(aq) + NH3(aq) → NH4HCOO(aq)
E. HCOOH(aq) + NH3(aq) → NH4HCOO(aq)

NH3(aq) + HCOOH(aq) ==> HCOONH4(aq) ==> HCOO^-(aq) + NH4^+(aq).

Doesn't that look like C (although I put an extra step in to help in understanding what was going on)?

To write the net ionic equation for the reaction of formic acid (HCOOH) with NH3, we need to first write the balanced chemical equation.

The balanced chemical equation for the reaction is:

HCOOH(aq) + NH3(aq) → NH4HCOO(aq)

Now, we can write the net ionic equation, which only includes the ions that participate in the reaction.

The net ionic equation for the reaction is:

H+(aq) + NH3(aq) → NH4+(aq)

Therefore, the correct answer is option A: H+(aq) + NH3(aq) → NH4+(aq).

To determine the net ionic equation for the reaction between formic acid (HCOOH) and ammonia (NH3), we need to identify the spectator ions (ions that do not participate in the reaction) and remove them from the overall ionic equation. This leaves us with the net ionic equation that represents the actual chemical change occurring.

First, let's write out the balanced overall ionic equation for the reaction:

HCOOH(aq) + NH3(aq) → NH4+(aq) + HCOO–(aq)

Now, let's break down the reactants and products into their respective ions:

HCOOH(aq): H+(aq) + HCOO–(aq)
NH3(aq): NH4+(aq) + OH–(aq)

Substituting these ion forms into the overall ionic equation, we get:

H+(aq) + HCOO–(aq) + NH4+(aq) + OH–(aq) → NH4+(aq) + HCOO–(aq)

Next, we can see that the HCOO–(aq) and the NH4+(aq) are the same on both sides of the equation. Therefore, they can be canceled out as spectator ions.

Simplifying the equation further, we are left with:

H+(aq) + OH–(aq) → H2O(l)

This is the net ionic equation for the reaction between formic acid and ammonia.

The correct answer is B. H+(aq) + OH–(aq) → H2O(l).

Net Ionic Equation: Do it yourself.