Answer: 2.75%
Explanation:
[tex]pH=-log [H+][/tex]
[tex]3.26 = -log [H+][/tex]
[tex][H+] = 5.495\times 10^{-4} M[/tex]
[tex]HA\rightleftharpoons H^++A^-[/tex]
initial 0.020 0 0
eqm 0.020 -x x x
[tex]K_a=\frac{[H+][A-]}{[HA]}[/tex]
[tex]K_a=\frac{[x][x]}{[0.020-x]}[/tex]
[tex]x=5.495\times 10^{-4}[/tex]
[tex]K_a=\frac{[5.495\times 10^{-4}]^2}{[0.020-5.495\times 10^{-4}]}[/tex]
[tex]K_a =1.553\times 10^{-5}[/tex]
percent dissociation = [tex]\frac{[H^+_eqm]}{[Acid_{initial}]}\times 100[/tex]
percent dissociation=[tex]\frac{5.495\times 10^{-4}}{0.020}\times 100[/tex]
Thus percent dissociation= 2.75 %
