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