butanoic acid ionizes as below:
CH3CH2CH2COOH-----> CH3CH2CH2COO- + H+
thus the equilibrium expression of Ka is as follows:
Ka= [H+][CH3CH2CH2COO-] / [CH3CH2CH2COOH]= 1.5 X 10-5
since only a bit of the acid ionizes, we use x to denote how much has ionized...
based on the equation,
[CH3CH2CH2COOH] left at equilibrium= 0.006-x
[CH3CH2CH2COO-] formed at equilibrium= x
[H+] formed at equilibrium= x
thus the Ka expression becomes
Ka= (x)(x) / (0.006-x)= 1.5 X 10-5
solve for x..
u get x= 2.926 X 10
-4
thus the percent ionization= (x / 0.006) X 100%
= 4.88 %
hope it helps...
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