NEET UG CHEMISTRY MCQ: Equilibrium

1 Hard

1 of 1

Application +4 | -1

The

\mathrm{pH}

0.5 \mathrm{~L}

1.0 \mathrm{MNaCl}

after the electrolysis for

965 \mathrm{~s}

using

5.0 \mathrm{~A}

current (

100 \%

efficiency) is

View answer and explanation

Correct option: A

Option A (13) is correct because the pH is calculated based on the amount of NaOH produced during the electrolysis of NaCl.

More Questions from Equilibrium

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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}