NEET UG CHEMISTRY MCQ: Equilibrium

1 Hard

1 of 1

Analyzing +4 | -1

The

\mathrm{pH}

value of

0.1 \mathrm{~M}

solutions of

\mathrm{CH}_3 \mathrm{COONa} \text{(I)}, \mathrm{CH}_3 \mathrm{COOH} \text{(II)}, \mathrm{CH}_3 \mathrm{COONH}_4 \text{(III)}, \mathrm{NaOH} \text{(IV)}

and

{HCl}(\mathrm{V})

is in the order

View answer and explanation

Correct option: C

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<

III

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<

Option C (V

<

<

III

<

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IV) is correct because it accurately orders the pH of the given solutions from most acidic (HCl) to most basic (NaOH).

More Questions from Equilibrium

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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}