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

Answer

C

Explanation

Ionization of weak acid (HA) is represented as
`HA(aq.)hArr H^(+)(aq.)+A^(-)(aq.)`
When HA is `50%` ionized.
`C_(HA)=C_(A^(-))`
According to the Henderson equation,
`pH=pK_(a)+log.(C_(A^(-)))/(C_(HA))`
`=pK_(a)+log 1`
`=pK_(a)=4.5`
At `25^(@)C`,
`pH+pOH=pK_(w)=14`
Thus, `pOH=14-pH`
`=14-4.5`
`=9.5`