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)`.

Explanation

pH=4.15
`-log [H_3O^+]`=4.15
or log `[H_3O^+]=-4.15 =bar(5).85`
`therefore [H_3O^+]` = antilog `(bar(5).85)`
`=7.08xx10^(-5)`
Now `[H_3O^+]=[A^-]=7.08xx10^(-5)` M
(i)Concentration of anion = `7.08xx10^(-5)` M
(ii) Ionization constant of acid
`HA(aq)+H_2O(l) hArr H_3O^(+)(aq) + A^(-)(aq)`
`K_a=([H_3O^+][A^-])/"[HA]"`
Concentration of undissociated acid =0.01-0.000071
=0.009929
`therefore K_a=((7.08xx10^(-5))xx(7.08xx10^(-5)))/0.009929`
`=5.05xx10^(-7)`
(iii)`pK_a=-log K_a`
`=-(5.05xx10^(-7))` =6.26