It has been found that the `pH` of a `0.01 M` solution of an organic acid is `4.15`. Calculate the concentration of the anion, the ionization constant of the acid and its `pK_(a)`.

Answer

`[A^(-)] = 7.08 xx 10^(-5)M, K_(a) = 5.08 xx 10^(-7), pK_(a) = 6.29`

Explanation

Let the organic acid be HA.
`rArr HA harr H^(+) - A^(-)`
Concentration of HA `= 0.01 M`
`pH = 4.15`
`-"log" [H^(+)] = 4.15`
`[H^(+)] = 7.08 xx 10^(-5)`
`K_(a) = ([H^(+)][A^(-)])/([HA])`
Now,
`[H^(+)] = [A^(-)] = 7.08 xx 10^(-5)`
`[HA] = 0.01`
Then,
`K_(a) = ((7.08 xx 10^(-5))(7.08 xx 10^(-5)))/(0.01)`
`K_(a) = -"log" K_(a)`
`= - "log" (5.01 xx 10^(-1))`
`pK_(a) = 6.3001`