Flag Physical Chemistry> ionic equilibrium...
question mark

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

Rishav raj , 12 Years ago
Grade
anser 1 Answers
Naveen Kumar

Last Activity: 10 Years ago

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.

Provide a better Answer & Earn Cool Goodies

Enter text here...
star
LIVE ONLINE CLASSES

Prepraring for the competition made easy just by live online class.

tv

Full Live Access

material

Study Material

removal

Live Doubts Solving

assignment

Daily Class Assignments


Ask a Doubt

Get your questions answered by the expert for free

Enter text here...