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^(-)]=7.08xx10^(-5)M, K_(a)=5.08xx10^(-7), pK_(a) = 6.29`

Explanation

pH = 4.15 means `- log [H^(+)] = 4.15 or log [H^(+)] = - 4.15 = bar(5) . 85 or [H^(+)] = 7.08 xx 10^(-5)M`
`HA hArr H^(+) + A^(-)`
Hence, at equilibrium `[H^(+)]=[A^(-)] = 7.08 xx 10^(-5)M ~= 7.1 xx 10^(-5)M`
`[HA] = (0.01-7.1xx10^(-5))=(0.01-0.000071)M = 0.009929 M`
`K_(a) = ([H^(+)][A^(-)])/([HA])=((7.1xx10^(-5))^(2))/(9.929xx10^(-3))=5.08 xx 10^(-7)`
`pK_(a) = - log K_(a) = - log (5.08 xx 10^(-7))= 6.29`.