Question

Arrange $N{H_4}^ + ,{H_2}O,{H_3}{O^ + },{\text{HF and }}O{H^ - }$ in increasing order of acidic nature.
(a) $O{H^ - } < {H_2}O < N{H_4}^ + < {\text{HF}} < {H_3}{O^ + }$
(b) ${H_3}{O^ + } > {\text{HF}} > {H_2}O > N{H_4}^ + > O{H^ - }$
(c) $N{H_4}^ + < {\text{HF}} < {H_3}{O^ + } < {H_2}O < O{H^ - }$
(d) ${H_3}{O^ + } < N{H_4}^ + < {\text{HF}} < O{H^ - } < {H_2}O{\text{ }}$

Answer 1

Hint:Acidic nature tells about the acidic strength of an acid, and acidic strength for an acid ${\text{HA}}$ is the ability to dissociates into ${H^ + }$ cations, and ${A^ - }$ anions where ${A^ - }$ is called the conjugate base of the acid ${\text{HA}}$.

Complete step by step answer:
(1) Acidic nature or acidic strength of an acid is directly proportional to the stability of the conjugate base of that acid. An acid is more acidic in nature if it has a more stable conjugate base, and an acid is less acidic or weaker in nature if it has an unstable conjugate base.
(2) Conjugate acid of an acid ${\text{HA}}$ can be formed by removing ${H^ + }$ ion from the acid ${\text{HA}}$. In general, conjugate base one less hydrogen, and one more negative charge than the corresponding acid.
(3) For $N{H_4}^ + $, conjugate base will be $N{H_4}$, which will be formed by the following reaction:
$N{H_4}^ + \to N{H_4} + {H^ + }$, and the conjugate base $N{H_4}$ is more stable than $N{H_4}^ + $ as positive charge on electronegative Nitrogen is unstable, and $N{H_4}$ is a neutral molecule called ammonia.
For ${H_2}O$, conjugate base will be $O{H^ - }$, which will be formed by the following reaction:
${H_2}O \to O{H^ - } + {H^ + }$, but ${H_2}O$ is a neutral molecule, so, it will not lose hydrogen to become unstable.
For ${H_3}{O^ + }$, conjugate base will be ${H_2}O$, which will be formed by the following reaction:
${H_3}{O^ + } \to {H_2}O + {H^ + }$, we very well know that ${H_2}O$ is a stable molecule exist as water in nature, so, ${H_3}{O^ + }$ will readily lose hydrogen ion to form water, and act as a strong acid.
For ${\text{HF}}$, conjugate base will be ${F^ - }$, which will be formed by the following reaction:
${\text{HF}} \to {{\text{F}}^ - } + {H^ + }$, ${F^ - }$ is more stable than ${\text{HF}}$ because negative charge on an electronegative element Fluorine is more stable, so, ${\text{HF}}$ will readily lose hydrogen ion to act as a strong acid.
For $O{H^ - }$, conjugate base will be ${O^{2 - }}$, which will be formed by the following reaction:
$O{H^ - } \to {O^{2 - }} + {H^ + }$, and ${O^{2 - }}$ molecule is highly unstable as compared to $O{H^ - }$, so, it will act like a base, and gain hydrogen. Hence, it is the least acidic in nature.
Therefore, the final order will be $O{H^ - } < {H_2}O < N{H_4}^ + < {\text{HF}} < {H_3}{O^ + }$.
Hence option A is the correct answer.

Note:
Conjugate base is formed from an acid, and conjugate acid is formed from a base. Conjugate acid of a base can be formed by adding a ${H^ + }$ ion to a base. Conjugate acid has one more hydrogen, and one more positive charge then the corresponding base.