Question

Arrange $N{H}_4^{+},H_{2}O,H_3{O}^{+},HF$ and $O{H}^{-}$in increasing order of acidic nature

• A. $H_3{O}^{+}<N{H}_4^{+}<HF<O{H}^{-}<H_{2}O$
• B. $N{H}_4^{+}<HF<H_3{O}^{+}<H_{2}O<O{H}^{-}$
• C. $O{H}^{-}<H_{2}O<N{H}_4^{+}<HF<H_3{O}^{+}$
• D. $H_3{O}^{+}>HF>H_{2}O>N{H}_4^{+}>O{H}^{-}$

Answer 1

Answer: (C)

The order of acidic nature of $N{H}_4^{+},H_{2}O,H_3{O}^{+}$, $HF$ and $O{H}^{-}$is as
$O{H}^{-}<H_{2}O<N{H}_4^{+}<HF<H_3{O}^{+}$
$O{H}^{-}$is basic (ie, it tends to gain a proton). So, it is least acidic. $H_{2}O$ is a neutral species. HF is a weak acid because the bond dissociation energy of $HF$ is very high. $H_3{O}^{+}$is most acidic because it readily loose proton.
Presence of negative charge on the atom holding the electron pair increases the basicity. While presence of positive charge on the atom holding the electron pair decreases the basicity (or increases acidity).