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A buffer solution is made by dissolving by $${ 10 }^{ -2 }$$ mole $${ CH }_{ 3 }COOH$$ and $${ 10 }^{ -2 }$$ mole of $${ CH }_{ 3 }{ COO }^{ \ominus  }{ Na }^{ \oplus  }$$ in 100 ml of water.
(i) What is the pH of solution ?
(ii) If this solution is poured into a container having 10,000 litre of $${ H }_{ 2 }O$$, then what is the new pH?


Solution

(i) $$[CH_3COOH]= \cfrac {10^{-2}}{100 \times 10^{-3}}=\cfrac {10^{-2}}{0.1}, [CH_3COONa]=\cfrac {10^{-2}}{0.1}$$
$$pH=pK_a+\log \cfrac {[CH_3COONa]}{[CH_3COOH]}=pK_a+\log \cfrac {10^{-2}/0.1}{10^{-2}/0.1}$$
$$pH=pK_a$$
(ii) Total volume= $$V= 10000+0.1= 10000.1$$
$$[CH_3COOH]=\cfrac {10^{-2}}{V}, [CH_3COONa]=\cfrac {10^{-2}}{V}$$
$$pH=pK_a+\log \cfrac {10^{-2}/V}{10^{-2}/V}=pK_a$$ .

Chemistry

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