JEE CHEMISTRY MCQ: Equilibrium

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Analyzing +4 | -1

50 \mathrm{~ml}, 0.1 \mathrm{~M} \mathrm{~NaHCO_3}

aqueous solution is mixed with

50 \mathrm{~ml}, 0.8 \mathrm{~M} \mathrm{~Na}_2 \mathrm{CO}_3

aqueous solution at

25^{\circ} \mathrm{C}

. The

\mathrm{pH}

of resulting solution exists between colour transition range of an indicator which is:

\left(\mathrm{K}_{\mathrm{a}_1}=4 \times 10^{-7}, \mathrm{K}_{\mathrm{a}_2}=2 \times 10^{-11}\right.

for

\left.\mathrm{H}_2 \mathrm{CO}_3\right)

View answer and explanation

Correct option: D

Malachite green (11.4 to 13)

Option D is correct because the pH of the resulting buffer solution is calculated to be 11.6, which falls within the color transition range of Malachite green (11.4 to 13).

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.