The `pK_(a)` of a weak acid `(HA)` is `4.5`. The `pOH` of an aqueous buffered solution of `HA` in which `50%` of the acid is ionized is:

Options

A. `7.0`
B. `4.5`
C. `2.5`
D. `9.5`

Answer

D

Explanation

For acidic buffer,
`pH = pK_(a) +log .[["Salt"]]/[["Acid"]] =pK_(a) + log .[[A^(-)]]/[[HA]]`
When acid is 50% ionised.
`[A^(-)] =[HA]`
`:. pH =pK_(a) + log .[[HA]]/[[HA]] = 4.5 + log 1 = 4.5`
`pOH =14 - 4.5 =9.5`