$2.5\;mL\;of\;\frac{2}{5}\;M$ weak monoacidic base ${K_b} = 1 \times {10^{ - 12}}\;\;{\rm{at}}\;25^\circ C$ is titrated with $\frac{2}{{15}}\;M\;{\rm{HCl}}$ in water at $25^\circ C$ The concentration of ${{\rm{H}}^ + }$ at equivalence point is ${K_w} = 1 \times {10^{ - 14}}\;{\rm{at}}\;25^\circ C$

NEET UG CHEMISTRY MCQ: Equilibrium

1 Hard

1 of 1

Application +4 | -1

2.5\;mL\;of\;\frac{2}{5}\;M

weak monoacidic base

{K_b} = 1 \times {10^{ - 12}}\;\;\mathrm{at}\;25^\circ C

is titrated with

\frac{2}{{15}}\;M\;\mathrm{HCl}

in water at

25^\circ C

The concentration of

{\mathrm{H}^ + }

at equivalence point is

{K_w} = 1 \times {10^{ - 14}}\;\mathrm{at}\;25^\circ C

View answer and explanation

Correct option: D

2.7 \times {10^{ - 2}}\;M

Option D (

2.7 \times {10^{ - 2}}\;M

) is correct because it is the hydrogen ion concentration calculated at the equivalence point after the weak base is neutralized by the strong acid.

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}