A buffer mixture of acetic acid and potassium acetate has ${\rm{pH}} = 5.24.$ The ratio of $\left[ {{\rm{C}}{{\rm{H}}_3}{\rm{CO}}{{\rm{O}}^ - }} \right]/\left[ {{\rm{C}}{{\rm{H}}_3}{\rm{COOH}}} \right]$ in this buffer is, $\left( {{\rm{p}}{K_a} = 4.74} \right):$

JEE CHEMISTRY MCQ: Equilibrium

1 Easy

1 of 1

Application +4 | -1

A buffer mixture of acetic acid and potassium acetate has

\mathrm{pH} = 5.24.

The ratio of

\left[ {\mathrm{C}{\mathrm{H}_3}\mathrm{CO}{\mathrm{O}^ - }} \right]/\left[ {\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}} \right]

in this buffer is,

\left( {\mathrm{p}{K_a} = 4.74} \right):

View answer and explanation

Correct option: A

3:1

Option A (3:1) is correct because it represents the ratio of [CH₃COO⁻]/[CH₃COOH] calculated using the Henderson-Hasselbalch equation.

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\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