The equivalent conductance of $\frac{M}{{32}}$ solution of a weak monobasic acid is $8.0\;{\rm{mhos}}\;{\rm{c}}{{\rm{m}}^2}$ and at infinite dilution is $400\;{\rm{mhos}}\;{\rm{c}}{{\rm{m}}^2}.$ The dissociation constant of this acid is:

NEET UG CHEMISTRY MCQ: Equilibrium

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Application +4 | -1

The equivalent conductance of

\frac{M}{{32}}

solution of a weak monobasic acid is

8.0\;\mathrm{mhos}\;\mathrm{c}{\mathrm{m}^2}

and at infinite dilution is

400\;\mathrm{mhos}\;\mathrm{c}{\mathrm{m}^2}.

The dissociation constant of this acid is:

View answer and explanation

Correct option: B

1.25 \times {10^{ - 5}}

Option B (

1.25 \times 10^{-5}

) is correct because it represents the calculated dissociation constant using the given equivalent conductance values.

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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}