Write a net ionic equation for the reaction of the aqueous species ammonia and hydrofluoric acid.

I don't think mine is coming out right- I don't know how NH3 reacts with other species, because I end up getting hydrogen as both an anion and a cation in the same species.

NH3(aq) is NH3 + HOH ==> NH4^+ + OH^-
HF ==> H^+ + F^-
NH4F is soluble in water.

So if NH4F is soluble in water, would the net ionic equation be
H+ + OH- --> H2O?

Or would the net ionic equation deal with the NH4F?

I didn't indicate above that HF is a weak acid but it is.
I have never seen an equation like this before. Check my thinking.
NH3(aq) is NH3 + HOH ==> NH4^+ + OH^-
and if we were trying to be exact we would write NH4^+(aq) + OH^-(aq).
NH3 + HOH + HF ==> NH4^+ + OH^- + H^+ + F^-
NH3 + HOH + HF ==> NH4^+ + HOH + F^-
but HOH cancels on each side. Since the problem specifically states aqueous ammonis, then I think the following is reasonable.
NH3(aq) + HF ==> NH4^+(aq) + F^-(aq)
The problem is that I don't know how your prof counts an aqueous solution of NH3. It is ammonia gas dissolved in water and it is a weak base which dissociates slightly into NH4^+ and OH^- but there is a good deal of NH3 remaining undissociated. In addition, do we show the HF dissolving in the solution, too, and writing it as HF(aq). I don't know. But I think what I have written above is reasonable. The prof may simply be looking for
NH3(aq) + HF ==> NH4^+(aq) + F^-(aq) with the possible addition of HF(aq) replacing HF.

I am a little uncomfortable with the answer I wrote above BECAUSE, most of the NH3 stays in the form of NH3 and most of the HF stays in the form of HF; therefore, the molecular species far out number the ionic ones. In such cases we usually defer to the molecular form. However, writing "no reaction" just doesn't seem appropriate. I hope I haven't completely confused you.

I think the best answer is NH3(aq) + HF(aq) --> NH4^+(aq) + F^-(aq). This equation shows the reactants and products in their aqueous form, which is what the question asked for.

To write the net ionic equation for the reaction between ammonia (NH3) and hydrofluoric acid (HF), we need to understand the dissociation of these compounds in water.

First, let's consider the dissociation of ammonia in water:
NH3 + HOH (water) → NH4+ (ammonium ion) + OH- (hydroxide ion)

Next, let's consider the dissociation of hydrofluoric acid in water:
HF + HOH (water) → H3O+ (hydronium ion) + F- (fluoride ion)

Now that we understand the dissociation of NH3 and HF in water, we can write the net ionic equation:

NH3(aq) + HF(aq) → NH4+(aq) + F-(aq)

In this net ionic equation, we only include the species that undergo a change or reaction. The NH4+ and F- ions are the result of the reaction between NH3 and HF.

It is important to note that NH3 is a weak base and does not completely dissociate in water. Similarly, HF is a weak acid and does not completely dissociate in water. Therefore, there may still be some NH3 and HF molecules present in the solution.

If you are writing this equation for a specific purpose, such as a chemistry class or a lab report, it is always best to consult your professor or teacher for any specific requirements they may have regarding the representation of the reaction.

Based on the information provided, the net ionic equation for the reaction between aqueous ammonia (NH3(aq)) and hydrofluoric acid (HF) can be written as:

NH3(aq) + HF(aq) -> NH4+(aq) + F-(aq)

This equation represents the formation of ammonium ion (NH4+) and fluoride ion (F-) in the solution. The hydroxide ion (OH-) is not involved in the reaction between ammonia and hydrofluoric acid.