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

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

Grade:12

1 Answers

Vikas TU
14149 Points
4 years ago
Der student 
CH3COOH and multiply the moles with its molecular formula then u would have the net amount of the acid.
Now we can also calculate the mole of acid left in 5L=C*5L
So we can get C from the above formula and after that (C-Cα) would be the net concentration of acetic acid.
α= 0.01(given 1%)
Ka=C(α)2
C-C*α......Cα......Cα+10-7.  .(at equilibrium)
C...........0.......10-7.     (initially)
CH3COO- + H+

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