Degree of dissociation of ${\rm{N}}{{\rm{H}}_4}{\rm{OH}}$ in water is $1.8 \times {10^{ - 5}}$, then hydrolysis constant of ${\rm{N}}{{\rm{H}}_4}{\rm{Cl}}$ is

NEET UG CHEMISTRY MCQ: Equilibrium

1 Medium

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Application +4 | -1

Degree of dissociation of

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

in water is

1.8 \times {10^{ - 5}}

, then hydrolysis constant of

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

View answer and explanation

Correct option: D

5.55 \times {10^{ - 10}}

Option D is correct because the hydrolysis constant of NH4Cl can be calculated using the formula Kh = Kw/Kb.

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Mg{(HC{O_3})_2}

MgC{O_3}

Mg{(OH)_2}

MgC{O_3}

K_{\mathrm{P}}

\mathrm{CaCO}_3

4.5 \times 10^{-2}

\mathrm{C}(\mathrm{s})+\mathrm{CO}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g}),

K_{\mathrm{P}}

2.0 \mathrm{~atm}.

\mathrm{CaO}

\mathrm{CaCO}_3

+

\leftrightharpoons

\mathrm{PC}{\mathrm{l}_5}\leftrightharpoons\mathrm{PC}{\mathrm{l}_3} + \mathrm{C}{\mathrm{l}_2}

{K_{\mathrm{eq}}} = 0.41

\mathrm{p}{K_a}