Explain: 'Hydrogen attached to a carbon having a triple bond is acidic in nature.' Also,write an appropriate reaction for it.

(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: $sp \ C > sp^{2} \ C > sp^{3} \ C$.
Due to this,the $sp$ hybridized carbon attracts the bonded electron pair more strongly towards itself. Consequently,the hydrogen atom attached directly to the $sp$ hybridized carbon becomes more acidic compared to the hydrogen atoms in alkanes and alkenes.
The acidity order is: $HC \equiv CH > H_{2}C=CH_{2} > CH_{3}-CH_{3}$.
Only the hydrogen atom attached to the triply bonded carbon in terminal alkynes (e.g.,$HC \equiv CH$,$CH_{3}C \equiv CH$) is acidic. In $R-C \equiv C-H$,only the terminal $H$ is acidic. In $R-C \equiv C-R$,no acidic $H$ is present.
Chemical reactions showing the acidic nature of terminal alkynes:
$1$. Reaction with sodium metal:
$HC \equiv CH + Na \xrightarrow{475 \ K} HC \equiv C^{-}Na^{+} + \frac{1}{2} H_{2}$
$2$. Reaction with sodamide $(NaNH_{2})$:
$HC \equiv CH + NaNH_{2} \xrightarrow{NH_{3}} HC \equiv CNa + NH_{3}$
$3$. Reaction with excess sodamide:
$HC \equiv CH + 2NaNH_{2} \xrightarrow{NH_{3}} Na^{+}C^{-} \equiv C^{-}Na^{+} + 2NH_{3}$