Question

# 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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