The Henderson-Hasselbalch equation is:
[tex]pH = pK_a + log \frac{[conjugate base, (A^{-})]}{[weak acid, (HA)]}[/tex]
From the above equation.,
[tex][conjugate base, (A^{-})] = {[weak acid, (HA)]}[/tex]
[tex]pH = pK_a + log \frac{[weak acid, (HA)]}{[weak acid, (HA)]}[/tex]
[tex]pH = pK_a + log 1[/tex]
[tex]pH = pK_a[/tex]
pH of [tex]CH_3COOH[/tex] is 10 and [tex]pK_a[/tex] is 4.8. (given).
Since, pH > [tex]pK_a[/tex] so, deprotonated form of [tex]CH_3COOH[/tex] will be predominant that is [tex]CH_3COO^{^{-}}[/tex].
The structure of the predominant form of [tex]CH_3COOH[/tex] is shown in the image.
