Give a brief note on the acidic nature of alkynes.

(N/A) The $s$-orbital is closer to the nucleus than the $p$-orbital,so electronegativity increases as the $s$-character increases. The order of electronegativity is: $sp > sp^2 > sp^3$.
Due to this,the $sp$ hybridized carbon in an alkyne attracts the bonded electron pair most strongly towards itself. Consequently,the hydrogen atom attached directly to the $sp$ carbon is more acidic than the hydrogen atoms in alkanes or alkenes.
Acidic strength order: $\equiv C-H > =C-H > -C-H$.
Only the hydrogen atom attached to the triply bonded carbon of a terminal alkyne (like $HC \equiv CH$,$CH_3 C \equiv CH$,$CH_3 CH_2 C \equiv CH$) is acidic. In $R-C \equiv C-H$,only the terminal $H$ is acidic. In $R-C \equiv C-R$,there is no acidic hydrogen.
Chemical reactions showing the acidic nature of terminal alkynes:
Ethyne reacts with strong bases like sodium metal at high temperatures or sodamide $(NaNH_2)$ to form ethynide (acetylide) products.
$1$. Reaction with sodium metal:
$HC \equiv CH + Na \xrightarrow{475 \ K} HC \equiv C^- Na^+ + \frac{1}{2} H_2$ (Eq. $i$)
$2$. Reaction with sodamide:
$HC \equiv CH + NaNH_2 \xrightarrow{NH_3} HC \equiv CNa + NH_3$ (Eq. $ii$)
$HC \equiv CH + 2NaNH_2 \xrightarrow{NH_3} Na^+C^- \equiv C^-Na^+ + 2NH_3$ (Eq. $iii$)