The dissociation of water at $25^\circ C$ is $1.9 \times {10^{ - 7}}$ percent and the density of water is $1.0\,\,{\rm{g}}/{\rm{c}}{{\rm{m}}^3}.$ The ionisation constant of water is:

$3.42\,\, \times \,\,{10^{ - 6}}$

$3.42\,\, \times \,\,{10^{ - 8}}$

$1.00\,\, \times \,\,{10^{ - 14}}$

NEET UG CHEMISTRY MCQ: Equilibrium

1 Hard

1 of 1

Application +4 | -1

The dissociation of water at

25^\circ C

1.9 \times {10^{ - 7}}

percent and the density of water is

1.0\,\,\mathrm{g}/\mathrm{c}{\mathrm{m}^3}.

The ionisation constant of water is:

View answer and explanation

Correct option: D

2.00\,\, \times \,\,{10^{ - 16}}

Option D (

2.00\\ \times \\ {10^{ - 16}}

) is correct because the ionization constant is calculated from the degree of dissociation and the concentration of water.

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}