Statement I: Aqueous solutions of ${NaCl}$ and ${CH}_3 {COONH}_4$ are neutral $\left(\mathrm{pH}=7.0\right.$ at $25^{\circ} \mathrm{C}).$ Statement II: Both the salts do not undergo hydrolysis.

If both statements are CORRECT, and Statement II is the CORRECT explanation of Statement I

If both statements are CORRECT, and Statement II is NOT the CORRECT explanation of Statement I

If Statement I is INCORRECT, but Statement II is CORRECT

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JEE CHEMISTRY MCQ: Equilibrium

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Analyzing +4 | -1

Statement I: Aqueous solutions of

{NaCl}

and

{CH}_3 {COONH}_4

are neutral

\left(\mathrm{pH}=7.0\right.

25^{\circ} \mathrm{C}).

Statement II: Both the salts do not undergo hydrolysis.

View answer and explanation

Correct option: C

If Statement I is CORRECT, but Statement II is INCORRECT

Option C is correct because while NaCl solutions are neutral due to not undergoing hydrolysis, CH3COONH4 solutions are also neutral because the hydrolysis of the cation and anion cancel each other out, but it DOES undergo hydrolysis.

More Questions from Equilibrium

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The indicator used in titrating oxalic acid with caustic soda solution is:

\mathrm{K}_1

and

\mathrm{K}_2

are equilibrium constant for reactions (i) and (ii)

\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\quad\ldots

(i)

\mathrm{NO}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{~N}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})\quad\ldots

(ii)

Then,

K{a_1},\,\,\,K{a_2}\,\,\,\mathrm{and}\,\,\,K{a_3}

are the respective ionisation constants for the following reactions

\begin{array}{l} {H_2}S\leftrightharpoons{H^ + } + H{S^ - }\\ H{S^ - }\leftrightharpoons{H^ + } + {S^{2 - }}\\ {H_2}S\leftrightharpoons2{H^ + } + {S^{2 - }} \end{array}

The correct relationship between

K{a_1},\,\,K{a_2}\,,K{a_3}

is The

\mathrm{pH}

0.1 \mathrm{~M}-\mathrm{N}_2 \mathrm{H}_4

solution is (For

\mathrm{N}_2 \mathrm{H}_4,

K_{{b}_1}=3.6 \times 10^{-6}, K_{{b}_2}=6.4 \times 10^{-12}, \log 2=0.3

\log 3=0.48

) Reaction

{\text{Ba}}{{\text{O}}_2}(s) \rightleftharpoons {\text{BaO}}(s) + {{\text{O}}_2}(g);\Delta H = - {\text{ve}}.

In equilibrium condition pressure of

\text{O}_2

depends on: A : At equilibrium concentration of the reactant and product does not change with time for a chemical reaction.
R: The rate of reaction is zero at equilibrium.