A) \[100mL\] having \[0.15M\,N{{H}_{3}}\] and \[N{{H}_{4}}Cl\] each
B) \[100mL\] having \[0.2M\,N{{H}_{3}}\] and \[N{{H}_{4}}Cl\]each
C) \[100mL\] having \[0.2M\,N{{H}_{3}}\]and \[0.1M\,\,N{{H}_{4}}Cl\] each
D) \[100mL\] having \[0.05\text{ }M\text{ }N{{H}_{3}}\] and \[N{{H}_{4}}Cl\] each
Correct Answer: D
Solution :
Mill moles of \[N{{H}_{3}}=100\times 0.05=5m\] mol Added \[HCl\text{ }=\text{ }10\times 1.0\text{ }M=10m\]mol \[N{{H}_{3}}\] gets completely neutralised, solution turns acidic that is now not the buffer solution. In all other cases, m moles of \[N{{H}_{3}}\] are higher as compared to m moles of \[HCl\]. Thus, they remain as buffer.
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