Why is the solution of ammonium bifluoride more acidic than aqueous ammonium fluoride?

At low concentrations of aqueous $$\ce{NH4HF2}$$ $$(w = 1–2\,\%)$$ $$\mathrm{pH}\approx 3.$$

But aqueous $$\ce{NH4F}$$ $$(w = 20–40\,\%)$$ has $$\mathrm{pH}\approx 7.$$

Yet the equilibria of

\begin{align} \ce{HF &-> H+ + F-} &\quad K &= \pu{6.85E-6 M} \tag{1} \\ \ce{HF + F- &-> HF2-} &\quad K &= \pu{3.963 M^-1} \tag{2} \end{align}

I understand that $$\ce{NH4F}$$ comes from a weak acid and a weak base. So, the $$\mathrm{pH}$$ is around neutral. I don't understand how this equilibria makes $$\ce{NH4HF2}$$ more acidic. Does the second equilibria make the $$\ce{HF2-}$$ a stronger acid? I'm having a hard time understanding this.

• Does this answer your question? Dissociation equilibria of ammonium bifluoride in water Feb 23 '21 at 17:08
• You're asking about pretty much the same thing as earlier. Feb 23 '21 at 17:09
• HF is stronger acid than NH4+, is not it ? Feb 24 '21 at 6:43

This question and the one referenced by Mithoron address $$\ce{NH4HF2}$$ as an entity in itself which must be examined as a whole.
It is easier conceptually to rewrite the formula for the compound as $$\ce{NH4F·HF}$$. Consider adding $$\ce{NH4F}$$ to water: you get a $$\mathrm{pH}$$ near $$7.$$ In a separate container, add $$\ce{HF}$$ to water (approx. $$\pu{0.1 M});$$ you get a $$\mathrm{pH}\approx 1$$.
Combine the solutions and the $$\mathrm{pH}$$ will settle out somewhere between $$1$$ and $$7,$$ because of the common ion effect: $$\ce{F-}$$ from the $$\ce{NH4F}$$ inhibits the ionization of $$\ce{HF},$$ but the solution is still acidic. The approximate $$\mathrm{pH}$$ can be calculated from the ionization constant of $$\ce{HF}$$ and the concentrations of $$\ce{NH4F}$$ and $$\ce{HF}.$$
The concept of the $$\ce{HF2-}$$ moiety dominating the discussion for a solution is probably making the analysis murkier than it needs to be. It's just a hydrogen-bonded complex that happens to retain some semblance of existence in the solid. It is fairly stable: mp $$\pu{126 °C},$$ bp $$\pu{240 °C}$$ (with decomposition).
• "126$^0$ C" reads as "one hundred twenty-six to the power of zero coulombs". "NH$_4$F.HF" has a dot that doesn't make sense and the formula breaks across lines incorrectly. Please consider using proper formatting tools (mhchem, Unicode, plain MathJax) so that your notations make sense. Sep 27 '21 at 21:29