JEE CHEMISTRY MCQ: Equilibrium

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Application +4 | -1

100 \mathrm{ml}

solution (I) of buffer containing

0.1(\mathrm{M}) \mathrm{HA}

and 0.2 (M)

\mathrm{A}^{-}

, is mixed with another solution (II) of

100 \mathrm{ml}

containing 0.2(M) HA and 0.3(M)

\mathrm{A^{-}}

.
After mixing what is the

\mathrm{pH}

of resulting solution?
Given

\mathrm{pK}_{\mathrm{a}}

\mathrm{HA}=5

View answer and explanation

Correct option: B

5+\log 5 / 3

Option B (

5+\log 5 / 3

) is correct because it is calculated by using the Henderson-Hasselbalch equation after determining the total moles of

A^-

and HA after mixing.

More Questions from Equilibrium

Practice All Chapter MCQs →

\mathrm{K}_1

\mathrm{K}_2

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

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

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

\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}

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

\mathrm{pH}

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

\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

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

\text{O}_2