(N/A) The $s$-orbital is closer to the nucleus and more electronegative than the $p$-orbital. Therefore,the electronegativity of carbon increases as the percentage of $s$-character increases.
The order of electronegativity is: $sp \text{ } C > sp^{2} \text{ } C > sp^{3} \text{ } C$.
Due to this,the $sp$-hybridized carbon in ethyne $(HC \equiv CH)$ attracts the shared pair of electrons of the $C-H$ bond towards itself more strongly than $sp^{2}$ or $sp^{3}$ hybridized carbons.
This makes the hydrogen atom attached to the $sp$-hybridized carbon acidic in nature.
$50\% \text{ } s$-character in $sp$ carbon makes the $C-H$ bond polar,allowing the $H$ to be released as a proton $(H^{+})$ in the presence of strong bases.
Chemical reactions showing the acidic nature of ethyne:
$1$. Reaction with sodium metal:
$HC \equiv CH + Na \xrightarrow{475 \ K} HC \equiv C^{-}Na^{+} + \frac{1}{2} H_{2}$
$2$. Reaction with sodamide $(NaNH_{2})$:
$HC \equiv CH + NaNH_{2} \xrightarrow{liq. NH_{3}} HC \equiv CNa + NH_{3}$
$HC \equiv CH + 2NaNH_{2} \xrightarrow{liq. NH_{3}} Na^{+}C^{-} \equiv C^{-}Na^{+} + 2NH_{3}$
Thus,terminal alkynes exhibit weak acidic character.