${\rm{N}}{{\rm{H}}_4}{\rm{Cl}}$ is acidic because

On hydrolysis ${\rm{N}}{{\rm{H}}_4}{\rm{Cl}}$ gives weak base ${\rm{N}}{{\rm{H}}_4}{\rm{OH}}$ and strong acid HCl

Nitrogen donates a pair of electrons

It is a salt of weak acid and strong base

On hydrolysis ${\rm{N}}{{\rm{H}}_4}{\rm{Cl}}$ gives strong base and weak acid

JEE CHEMISTRY MCQ: Equilibrium

1 Easy

1 of 1

Understanding +4 | -1

\mathrm{N}{\mathrm{H}_4}\mathrm{Cl}

is acidic because

View answer and explanation

Correct option: A

On hydrolysis

\mathrm{N}{\mathrm{H}_4}\mathrm{Cl}

gives weak base

\mathrm{N}{\mathrm{H}_4}\mathrm{OH}

and strong acid HCl

Option A is correct because ammonium chloride (

NH_4Cl

) is a salt of a weak base (

NH_4OH

) and a strong acid (HCl), making it acidic due to the hydrolysis of the ammonium ion.

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\mathrm{K}_1

\mathrm{K}_2

\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\quad\ldots

\mathrm{NO}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{~N}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})\quad\ldots

K{a_1},\,\,\,K{a_2}\,\,\,\mathrm{and}\,\,\,K{a_3}

\begin{array}{l} {H_2}S\leftrightharpoons{H^ + } + H{S^ - }\\ H{S^ - }\leftrightharpoons{H^ + } + {S^{2 - }}\\ {H_2}S\leftrightharpoons2{H^ + } + {S^{2 - }} \end{array}

K{a_1},\,\,K{a_2}\,,K{a_3}

\mathrm{pH}

0.1 \mathrm{~M}-\mathrm{N}_2 \mathrm{H}_4

\mathrm{N}_2 \mathrm{H}_4,

K_{{b}_1}=3.6 \times 10^{-6}, K_{{b}_2}=6.4 \times 10^{-12}, \log 2=0.3

\log 3=0.48

{\text{Ba}}{{\text{O}}_2}(s) \rightleftharpoons {\text{BaO}}(s) + {{\text{O}}_2}(g);\Delta H = - {\text{ve}}.

\text{O}_2