The solubility of ${\rm{Ca}}{{\rm{F}}_2}$ in pure water is $2.3 \times {10^{ - 6}}\;{\rm{mol }}\;{\rm{d}}{{\rm{m}}^{ - 3}}$. Its solubility product will be

NEET UG CHEMISTRY MCQ: Equilibrium

1 Hard

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Application +4 | -1

The solubility of

\mathrm{Ca}{\mathrm{F}_2}

in pure water is

2.3 \times {10^{ - 6}}\;\mathrm{mol }\;\mathrm{d}{\mathrm{m}^{ - 3}}

. Its solubility product will be

View answer and explanation

Correct option: B

48.66 \times {10^{ - 18}}

Option D (

48.66 \times {10^{ - 18}}

) is correct because it is the solubility product calculated from the given solubility of

CaF_2

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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}