degree of ionisation

How do i work this out for a weak acid when knowing values for KA and Initiall concentration.


could someone please show me with an example.

any help most apprecitaed.

Thank You

QUOTE=smetin93]ok, just say you have ethanoic acid (a weak acid) here is what you do:

you have for instance 0.1M CH3COOH in this equilibria reaction:

CH3COOH <---> CH3COO- + H+

ok, this reaction has a Ka value (which is just a Kc value for weak acids) of 1.78x10-5moldm-3

for equilibria, Ka = [CH3COO-][H+]/[CH3COOH]

because the products are in a 1:1 ration so in eq, the [CH3COO-]=[H+]

so Ka = [H+]^2/[CH3COOH]

now in comes equilibria stuff:

CH3COOH <---> H+ + CH3COO-
Start: 0.1M 0M 0M
Eq: 0.1-x xM xM

as the value of x is soooo small it can be considered negligable so 0.1-x=0.1

no substitute int he equation

1.78x10-5 = [H+]^2/0.1
1.78x10-6 = [H+]^2
[H+]=1.33x10^-3moldm-3

degree of ionisation = 1.33x10^-3/0.1 = 0.0133 = 1.33%

(so pH = -log(1.33x10^-3) = 2.87)
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