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NEET · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)
Which will make basic buffer?
- A \(50 \mathrm{mL}\) of \(0.1 \mathrm{M} \mathrm{NaOH}+25 \mathrm{mL}\) of \(0.1 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\)
- B \(100 \mathrm{mL}\) of \(0.1 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}+100 \mathrm{mL}\) of \(0.1 \mathrm{M} \mathrm{NaOH}\)
- C \(100 \mathrm{mL}\) of \(0.1 \mathrm{M} \mathrm{HCl}+200 \mathrm{mL}\) of \(0.1 \mathrm{M} \mathrm{NH}_{4} \mathrm{OH}\)
- D \(100 \mathrm{mL}\) of \(0.1 \mathrm{M} \mathrm{HCl}+100 \mathrm{mL}\) of \(0.1 \mathrm{M} \mathrm{NaOH}\)
Answer & Solution
Correct Answer
(C) \(100 \mathrm{mL}\) of \(0.1 \mathrm{M} \mathrm{HCl}+200 \mathrm{mL}\) of \(0.1 \mathrm{M} \mathrm{NH}_{4} \mathrm{OH}\)
Step-by-step Solution
Detailed explanation
Basic buffer is mixture of weak base and salt of weak base with strong acid milli mole of \(\mathrm{HCl}=100 \times 0.1=10\) milli mole
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