Calculate ionic product of ${H}_2 {O}$ at $50^{\circ} \mathrm{C}$.

NEET UG CHEMISTRY MCQ: Equilibrium

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Calculate ionic product of

{H}_2 {O}

50^{\circ} \mathrm{C}

View answer and explanation

Correct option: A

4\times 10^{-14}

Option A (

4\times 10^{-14}

) is correct because the ionic product of water (

K_w

) increases with temperature, and at 50°C, it's approximately

4 \times 10^{-14}

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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}