# Why is hydrobromic acid a stronger acid than hypobromous acid?

If my understanding is correct, bromous acid ($\ce{HBrO2}$) is a stronger acid than hypobromous acid ($\ce{HBrO}$) because the additional electronegative oxygen atom draws the electron away from the hydrogen atom, making dissociation for $\ce{H+}$ easier.

Then why is hydrobromic acid stronger than hypobromous acid, despite the latter having one more electronegative atom to pull the electron? Are other factors involved here? And how do we decide which factor weighs more?

$\mathrm{p}K\mathrm{a}$ data:

The structure of $\ce{HBr}$ and $\ce{HBrO}$ is different, and also the place from where the proton leaves the molecule is also diferent.So, we have to take account of this and also the stability of the conjugate base formed after deprotonation.
In $\ce{HBrO, }$ the proton is attached to the $\ce{O}$ atom of the molecule.The molecule can lose proton as, $$\ce{H -O -Br <=> H^+ + (O -Br)^-}$$ The negative charge here mainly resides on the Oxygen, which makes $3$ lone pairs on $\ce{O}$ atom, adjacent to that, there is also electron rich $\ce{Br}$ atom, which also has $3$ lone pairs on it, so there are inter electronic repulsions. For the conjugate base to get stabilisation, the weak $\ce{-I}$ effect of $\ce{Br}$ atom is not at all sufficient. Instead, if the lone pairs can somehow delocalise that will contribute better to stability of the conjugate base. But the lone pair resides on the $\ce{2p}$ orbital of Oxygen, which can't at all effectively overlap with filled$\ce{4p}$ orbital of $\ce{Br}$, and $\ce{4d}$ orbitals are hugely high in energy. So, due to high electronic repulsion, no chances of delocalisation, and very weak $-I$ effect of $\ce{Br}$ can't make the conjugate base stable at all. So, $\ce{HOBr}$ has high $\ce{pK_a}$.
But if you come to $\ce{HBr}$, the bond between $\ce{H}$ and $\ce{Br}$ is very weak. More over if you dissolve it in a sloution, the weak bond is easily broken, and the individual cations and anions can get extensively solvated which makes the whole process of dissociation thermodynamically very much favourable.That's why, the $\ce{pK_a}$ of $\ce{HBr}$ is very low which means it can dissociate very easily in a solvent.