In 5L sol. of ascetic acid , having alpha=1%, Ka=1.8*10^-5. The amount of acetic acid present in sol.?

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Naveen Kumar · Answer

CH3COOH..<...............>CH3COO - ............+........H+ C.........................................0............................10 -7 ..............(initially) C-C*&alpha;.................................C&alpha;............................C&alpha;+10 -7 .............(at equilibrium) Ka=C(&alpha;) 2 &alpha;= 0.01(given 1%) So we can get C from the above formula and after that (C-C&alpha;) would be the net concentration of acetic acid. Now we can also calculate the mole of acid left in 5L=C*5L and multiply the moles with its molecular formula then u would have the net amount of the acid.

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In 5L sol. of ascetic acid , having alpha=1%, Ka=1.8*10^-5. The amount of acetic acid present in sol.?

In 5L sol. of ascetic acid , having alpha=1%, Ka=1.8*10^-5. The amount of acetic acid present in sol.?

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