A) \[C{{H}_{3}}COOH\] having \[N{{H}_{4}}OH\] \[({{10}^{-5}})\] and \[({{10}^{-5}})\] each
B) \[N{{H}_{4}}OH\] having \[4.0\] and \[3.0\] \[11.0\]each
C) \[10.0\] having \[\overset{O}{\mathop{\overset{||}{\mathop{C{{H}_{3}}-C-C{{H}_{2}}-C{{H}_{2}}-COOH}}\,}}\,\]\[\xrightarrow[(ii)\,\,{{H}_{2}}O+{{H}^{+}}]{(i)\,\,NaB{{H}_{4}}}\] and \[\beta \] each
D) \[\alpha \] having\[\alpha \] \[\beta \]and \[\beta \]each
Correct Answer: A
Solution :
Mill moles of \[C{{H}_{3}}Mgl+C{{H}_{3}}COC{{H}_{3}}\]millimol Added \[C{{H}_{3}}C{{H}_{2}}MgBr+C{{H}_{3}}CHO\] millimol \[C{{H}_{3}}C{{H}_{2}}MgBr+C{{H}_{3}}COC{{H}_{3}}\]gets completely neutralised that means solution turns acidic, so thats not the buffer solution.
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