Let hydrolysis constant $\left({K}_{{h}}\right)$ of $\stackrel{+}{{N}} {H}_4$ be ${y}$. Then equilibrium constant $\left({K}_{{a}}\right)$ for conjugate acid of ${NH}_3$ at $25^{\circ} \mathrm{C}$ is :

NEET UG CHEMISTRY MCQ: Equilibrium

1 Hard

1 of 1

Understanding +4 | -1

Let hydrolysis constant

\left({K}_{{h}}\right)

\stackrel{+}{{N}} {H}_4

{y}

. Then equilibrium constant

\left({K}_{{a}}\right)

for conjugate acid of

{NH}_3

25^{\circ} \mathrm{C}

is :

View answer and explanation

Correct option: C

{y}

Option A (

\frac{1}{y} \times 10^{-14}

) is correct because the equilibrium constant (

{K_a}

) for the conjugate acid of

{NH}_3

is related to the hydrolysis constant (

{K}_{{h}}

) by the equation

{K_a} = \frac{{{K_w}}}{{{K_h}}}

Practice All Chapter MCQs →

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}