Why does electronegativity create a stronger acid, but so does greater electron distribution?

I just learned that:

Acids get stronger with:

1. Increasing electronegativity of atom bonded to $\ce{H}$.
2. Size of atom bearing negative charge.
3. Hybridization of negative charge (more s character, $\ce{e-}$ closer to nucleus, more acidic).
4. Resonance stabilization (like size, greater distribution of electrons).

So basically, the more concentrated the negative charge (think $\ce{F-}$, $-1$ and very small), the tighter it's going to "hold on" to the proton. With resonance and large negative groups, the charge is less concentrated, so it's not going to "hold on" to the proton as well.

However, electronegativity is the tendency for an atom to "hog" electrons. Therefore, the more electronegative, the more electrons there are in a similarly sized space.

More concentrated electrons = more stable base = weaker acid.

But electronegativity = more concentrated electrons (even though size trumps it).
And electronegativity = stronger acid.

Feel free to edit if you see what I'm trying to say but can say it better!

• You're overlooking the decisive step: An acid is the stronger, the more energy it gains by becoming an anion. With that, you can rationalise. Chemistry is 95% pure logic.
– Karl
Sep 14, 2015 at 22:48
• Hey, more stable base means stronger acid! Sep 14, 2015 at 23:46
• Well, $\ce{HF}$ is a weak acid ... Sep 5, 2017 at 21:09

The main point about strong acids is ‘how stable is the conjugate base?’

The better stabilised the negative charge on a conjugate base of a neutral acid is, the happier the species will be to loose a proton and thus the more acidic the proton is. All of the points you noted just emphasise that.

Therefore, the only wrong assumption you are making is:

more stable base = weaker acid.

The correct way is:

more stable base = stronger acid.

Increasing EN means less acidic. Refer to the $$\mathrm{p}K_\mathrm{a}$$ values below.

$$\begin{array}{l|rrrr} \hline \ce{HX} & \ce{HF} & \ce{HCl} & \ce{HBr} & \ce{HI} \\ \hline \mathrm{p}K_\mathrm{a} & 3.2 & -7 & -9 & -10 \\ \hline \end{array}$$