Two buffers, $\mathrm{X}$ and $\mathrm{Y}$ of $\mathrm{pH}, 4.0$ and $6.0,$ respectively, are prepared from acid $\mathrm{HA}$ and the salt $\mathrm{NaA}.$ Both the buffers are $0.50 \mathrm{~M}$ in $\text{HA}.$ What would be the $\mathrm{pH}$ of the solution obtained by mixing equal volumes of the two buffers? The value of $K_{{a}}$ for $\mathrm{HA}=1.0 \times 10^{-5}(\log 5.05=0.7).$

JEE CHEMISTRY MCQ: Equilibrium

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Analyzing +4 | -1

Two buffers,

\mathrm{X}

and

\mathrm{Y}

\mathrm{pH}, 4.0

and

6.0,

respectively, are prepared from acid

\mathrm{HA}

and the salt

\mathrm{NaA}.

Both the buffers are

0.50 \mathrm{~M}

\text{HA}.

What would be the

\mathrm{pH}

of the solution obtained by mixing equal volumes of the two buffers? The value of

K_{{a}}

for

\mathrm{HA}=1.0 \times 10^{-5}(\log 5.05=0.7).

View answer and explanation

Correct option: D

5.7

Option D (5.7) is correct because mixing the two buffers results in a pH that can be calculated using the Henderson-Hasselbalch equation after considering the new concentrations of HA and A-.

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.