What happens to ${\rm{pH}}$ of a solution when ${\rm{N}}{{\rm{H}}_4}{\rm{Cl}}$ crystal is added to a dilute solution of ${\rm{N}}{{\rm{H}}_4}{\rm{OH}}$?

NEET UG CHEMISTRY MCQ: Equilibrium

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Analyzing +4 | -1

What happens to

\mathrm{pH}

of a solution when

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

crystal is added to a dilute solution of

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

View answer and explanation

Correct option: A

Decreases

Option A (Decreases) is correct because adding

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

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

introduces a common ion (

\mathrm{N}{\mathrm{H}_4}^+

), suppressing the ionization of

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

and decreasing the pH.

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}