Question about equilibrium

[cc609]

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Hello! I had a question about when hydrozoic acid (HN3) is put into a solution and figuring out pH.

I would assume that:

HN3 + H2O --> H3O+ + 3N-

but the answer was same reactants but with products ---> H3O + + N3-
...aka the difference of 3N- vs N3-

I know nitrogen is a diatomic like H2, Cl2, Br2...so I don't know about N3 etc but technically if there was a Cl2, it would go to 2 Cl- as well so when do you know which is which? If anything, I thought if you balance it , it is still technically 3 Nitrogen atoms, but for finding pH with given Ka

Ka= [H3O+][N3-]/[HN3]

when given Ka, you'd set H3O+ and N3- concentrations as X and do X^2/[HN3] correct? but if it was 3 N-, I would have done [x][3x^3] instead of X^2 so I am so confused.

If someone could explain this to me I would appreciate it so much!

 

[FeralisExtremum]

Think about in practical terms. When you deprotonate something, does its structure drastically change? Not really. Generally it's the same thing it was before, minus an H. If HN3 acts as an acid, it will lose its H and become N3. The individual atoms of the molecule don't break apart.

You may be confusing this with what you see when ionic compounds are dissolved in aqueous solutions, for example:
BiI3 (s) ↔ Bi 3+ (aq) + 3 I- (aq)
You may ask based on the above, why didn't this become I3 in the products, as N3 did? Since it's ionic, we're dealing with positive and negative charged atoms that are individually dissociating (as this is a metal atom bonded to a nonmetal atom). The structure is not the same as a covalent, molecular compound.

Also remember that a structure can have covalent and ionic components to it. For example, Mg(NO3)2 would break apart into Mg 2+ and 2 (NO3)-. Notice the ionic components come apart, the molecular ones don't.

If this is overwhelming, a rough shortcut is that if the question is dealing with pKa's, pH, solutions, etc. then it's most likely a Bronsted-Lowry acid/base equilibrium question. That means one reactant is going to act as a Bronsted-Lowry acid and donate a proton, so the product will be its conjugate base - just remove an H and account for the change in charge. The other reactant will act as a Bronsted-Lowry base and accept a proton, so the product will be its conjugate acid - add an H and account for the change in charge. That will give you both your products.