Question

Draw the structure of the predominant form of ch3cooh (pka = 4.8) at ph = 10.

Answer 1

The Henderson-Hasselbalch equation is:

`pH = pK_a + log ([conjugate base, (A^(-))])/([weak acid, (HA)])`

From the above equation.,

• When concentration of both the forms that is the concentration of conjugate base and weak acid are equal then pH =
  `pK_a`.

`[conjugate base, (A^(-))] = {[weak acid, (HA)]}`

`pH = pK_a + log ([weak acid, (HA)])/([weak acid, (HA)])`

`pH = pK_a + log 1`

`pH = pK_a`

• When pH <
  `pK_a`, protonated species.
• When pH >
  `pK_a`, deprotonated species.

pH of
`CH_3COOH` is 10 and
`pK_a` is 4.8. (given).

Since, pH >
`pK_a` so, deprotonated form of
`CH_3COOH` will be predominant that is
`CH_3COO^{^(-)}`.

The structure of the predominant form of
`CH_3COOH` is shown in the image.