The $pH$ values of $0.1\;M$ solution of ${\rm{HCl}},{\rm{C}}{{\rm{H}}_3}{\rm{COOH}},{\rm{N}}{{\rm{H}}_4}{\rm{Cl}}\;{\rm{ and }}\;{\rm{C}}{{\rm{H}}_3}{\rm{COONa}}$ will have the order

${\rm{HCl}} < C{{\rm{H}}_3}{\rm{COOH}} < N{{\rm{H}}_4}{\rm{Cl}} < C{{\rm{H}}_3}{\rm{COONa}}$

${\rm{C}}{{\rm{H}}_3}{\rm{COONa}} < {\rm{N}}{{\rm{H}}_4}{\rm{Cl}} < {\rm{C}}{{\rm{H}}_3}{\rm{COOH}} < HCl$

${\rm{N}}{{\rm{H}}_4}{\rm{Cl}} < {\rm{C}}{{\rm{H}}_3}{\rm{COONa}} < {\rm{C}}{{\rm{H}}_3}{\rm{COOH}} < HCl$

NEET UG CHEMISTRY MCQ: Equilibrium

1 Medium

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Analyzing +4 | -1

The

pH

values of

0.1\;M

solution of

\mathrm{HCl},\mathrm{C}{\mathrm{H}_3}\mathrm{COOH},\mathrm{N}{\mathrm{H}_4}\mathrm{Cl}\;\mathrm{ and }\;\mathrm{C}{\mathrm{H}_3}\mathrm{COONa}

will have the order

View answer and explanation

Correct option: A

\mathrm{HCl} < C{\mathrm{H}_3}\mathrm{COOH} < N{\mathrm{H}_4}\mathrm{Cl} < C{\mathrm{H}_3}\mathrm{COONa}

Option A (

\mathrm{HCl} < C{\mathrm{H}_3}\mathrm{COOH} < N{\mathrm{H}_4}\mathrm{Cl} < C{\mathrm{H}_3}\mathrm{COONa}

) is correct because it correctly orders the pH values from most acidic (lowest pH) to most basic (highest 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}