From the give options, which of the following compound will make best buffer solution?

H_{2} O, 1 M N a O H, 1 M H_{2} S O_{4}

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H_{2} O, 1 M C H_{3} C O O H, 1 M {N a}^{+} C H_{3} C O O^{-}

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H_{2} O, 1 M C H_{3} C O O H, 6 M {N a}^{+} C H_{3} C O O^{-}

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H_{2} O, 1 M C H_{3} C O O H, 1 M N a O H

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Solution

The correct option is A $H_{2} O, 1 M C H_{3} C O O H, 1 M {N a}^{+} C H_{3} C O O^{-}$
Since the acidic buffer is a mixture of weak acid plus its salt with a strong base. So $B$ is correct.
And there should be base as limiting reagent $({c o n c}^{n} o f b a s e < {c o n c}^{n} o f a c i d)$

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Similar questions

{S_{N}}^{2}

H_{2} O

^{-} O C H_{3},^{-} O H

C H_{3} C O O^{-}

C H_{3} C O O H + O H^{-} ⇋ C H_{3} C O O^{-} + H_{2} O + q_{1}

H^{+} + O H^{-} ⇋ H_{2} O + q_{2}

C H_{3} C O O H ⇋ C H_{3} C O O^{-} + H^{+}

C H_{3} C O O H + O H^{-} \to C H_{3} C O O^{-} + H_{2} O + q_{1}

H^{+} + O H^{-} \to H_{2} O + q_{2}

C H_{3} C O O H \to C H_{3} C O O^{-} + H^{+}

C H_{3} C O O H + H_{2} O ⇌ C H_{3} C O O^{-} + H_{3} O^{+}; K_{1} = 1.8 \times 10^{- 5}

2 H_{2} O ⇌ H_{3} O^{+} + O H^{-}, K_{2} = 1 \times 10^{14}

C H_{3} C O O H + O H^{-} ⇌ C H_{3} C O O^{-} + H_{2} O

x \times 10^{9}

H_{2} O, C l^{-}

C H_{3} C O O^{-}, O H^{-}

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