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This question already has an answer here:

Why do Hydrogen Halides(with the exception of HF) form mostly strong acids? My guess is that the negatively charged Halide part of the molecule is attracted by the partially positive part of the water molecule. Thus it is separated more easily. Although because it has one more electron it is more stable once it has dissociated

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marked as duplicate by Buttonwood, andselisk, Mithoron, Melanie Shebel, airhuff May 4 at 0:20

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I would like to correct a few points firstly that the halide part is not negatively charged. Also the electronegativity of the halide is the dominating effect in this case.

Hydrofluoric Acid as we know is not a very strong acid as compared to other hydrogen halide acids, had it been the case of electronegativity this would not have been the case.

The dominating factor here is the sizes of the atoms forming the molecule. As the size of halide increases, the bond strength of the H-X bond decreases, making it easier to break the Hydrogen halide bond and act as a stronger acid.

So the acidic strength comes out to be:
HF < HCl < HBr < HI

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    $\begingroup$ Agree mostly but why "the halide part is not negatively charged"? What about $F^-$, $Cl^-$, $Br^-$ ,$I^-$ ...? $\endgroup$ – Buck Thorn May 3 at 14:59
  • $\begingroup$ The question stated that the negative charged fluoride existed before dissociation, which wont be the case $\endgroup$ – Sidharth Giri May 3 at 16:37
  • $\begingroup$ Ah, I interpreted that to mean the halogens are electronegative, but ok. $\endgroup$ – Buck Thorn May 3 at 16:57