Physical Chemistry
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.?
1 Answer
Naveen Kumar
CH3COOH..<...............>CH3COO-............+........H+
C.........................................0............................10-7..............(initially)
C-C*α.................................Cα............................Cα+10-7.............(at equilibrium)
Ka=C(α)2
α= 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*5L
and multiply the moles with its molecular formula then u would have the net amount of the acid.
C.........................................0............................10-7..............(initially)
C-C*α.................................Cα............................Cα+10-7.............(at equilibrium)
Ka=C(α)2
α= 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*5L
and multiply the moles with its molecular formula then u would have the net amount of the acid.
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