I am not an expert in chemistry, but these are the few facts I know from Standard XII chemistry books (these facts could be wrong or incomplete or modified by now).
Facts I know -
Ethane has 'sp$^3$' hybridisation, Ethene has 'sp$^2$', and Ethyne 'sp'.
The % of 's' orbital present follows in each of the hybridisation follows the order sp > sp$^2$ > sp$^3$.
Since 's' orbital has greater penetration (to the nucleus) depth, the H atom in Ethyne comes the closest to the Carbon atom nucleus than in Ethene and Ethane, hence C-H bond length of Ethyne is the lowest, followed by Ethene and then Ethane.
On the other hand, since 's' orbital has greater penetration depth, 'sp' hybridised orbitals has the most Electronegativity, followed by 'sp$^2$' and then 'sp$^3$'. Because of this, Ethyne is most acidic, since the carbanion (HC $\equiv$ C $^-$) is more stable than (H$_2$C = CH$^-$) and (H$_3$C $-$ CH$_2$$^-$). Ethyne can release $H^+$ more easily than Ethene and Ethane, hence Ethyne is most acidic, followed by Ethene and then Ethane.
Following a similar reasoning of the nature of 's' orbital, one finds that the C-H bond energy of Eythne is the highest, followed by Ethene and then Ethane.
How can Ethyne be the most acidic and yet have the shortest C-H bond length. Isn't there a general rule that "the shorter the bond length, the stronger the bond strength, and so the weaker the acid"? Shouldn't shorter X-H bond lengths (X is some atom) mean that X and H has greater bond strength and so it does not dissociate much into $X^-$ and $H^+$ ions, thereby not producing much $H^+$ ions?
Is there something wrong with the Facts that I stated or is the statement of my Question not true in general?