NEET UG CHEMISTRY MCQ: Equilibrium

1 Hard

1 of 1

Analyzing +4 | -1

Given,

p K_b

values

\begin{array}{cccccccccccccccccccc} {{\text{Species}}}&{p{K_b}} \\ {N{H_3}}&{4.74} \\ {{A^ - }}&{4.00} \\ {{B^ - }}&{5.00} \\ {{C^ - }}&{6.00} \end{array}

queous solutions of

0.01 \mathrm{M}

salts of

\mathrm{HX}

(X=A, B, C)

and

\mathrm{NH}_4 \mathrm{OH}

have

\mathrm{pH}

in the order

View answer and explanation

Correct option: B

\mathrm{NH}_4 {C} < \mathrm{NH}_4 B < \mathrm{NH}_4 A

Option B (

\mathrm{NH}_4 {C} < \mathrm{NH}_4 B < \mathrm{NH}_4 A

) is correct because pH is inversely proportional to the strength of the conjugate acid, which is related to the

pK_b

values.

More Questions from Equilibrium

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}