બેઝિક બફર કેવી રીતે બનાવી શકાય ?

Question

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 $0.1 \mathrm{M} \mathrm{NaOH}$

NEET 2019, Diffcult

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Solution

c
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

milli mole of $\mathrm{NH}_{4} \mathrm{OH}=200 \times 0.1=20 \mathrm{mill}$ imole

$\mathrm{HCl}+\mathrm{NH}_{4} \mathrm{OH} \rightarrow \mathrm{NH}_{4} \mathrm{Cl}+\mathrm{H}_{2} \mathrm{O}^{-}$

$10 \quad \quad \quad 20\quad \quad \quad \quad -\quad \quad -$

$-\quad \quad \quad 10\quad \quad \quad \quad 10\quad \quad \quad -$

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