The $\left[ {{{\rm{H}}^ + }} \right]$ ion concentration in $0.001\;M$ acetic acid is $1.34 \times {10^{ - 4}}\;{\rm{g}}\;{\rm{ ion}}\;/{\rm{litre}}{\rm{.}}$ The $\left[ {{{\rm{H}}^ + }} \right]$ ion concentration of $0.164\;{\rm{g}}\;{\rm{ of}}\;{\rm{ C}}{{\rm{H}}_3}{\rm{COONa}}$ is added to a litre of $0.001\;M$ ${\rm{C}}{{\rm{H}}_3}{\rm{COOH}}$ will be:

NEET UG CHEMISTRY MCQ: Equilibrium

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The

\left[ {{\mathrm{H}^ + }} \right]

ion concentration in

0.001\;M

acetic acid is

1.34 \times {10^{ - 4}}\;\mathrm{g}\;\mathrm{ ion}\;/\mathrm{litre}\mathrm{.}

The

\left[ {{\mathrm{H}^ + }} \right]

ion concentration of

0.164\;\mathrm{g}\;\mathrm{ of}\;\mathrm{ C}{\mathrm{H}_3}\mathrm{COONa}

is added to a litre of

0.001\;M

\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}

will be:

View answer and explanation

Correct option: A

9 \times {10^{ - 6}}

Option A (

9 \times {10^{ - 6}}

) is correct because it calculates the hydrogen ion concentration after adding sodium acetate, considering the common ion effect.

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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}