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

Explanation

Let the acid be HA.
`HA(aq)hArr A^(-)(aq)+H^(+)(aq),K_(a)=([H^(+)][A^(-)])/([HA])`
In solution, `[H^(+)]=[A^(-)]`
Given: pH=4.15.
So, `[H^(+)]=10^(-pH)=10^(-4.15)(M)=7.08xx10^(-5)(M)`
`therefore`In solution, `[A^(-)]=7.08xx10^(-5)(M)`
Again, for a monoprotic acid we know,
`[H^(+)]=sqrt(K_(a)xxc)`
Therefore, `7.08xx10^(-5)=sqrt(K_(a)xx0.01)&K_(a)=5.01xx10^(-7)`
`therefore pK_(a)=-log_(10)K_(a)=-log_(10)(5.01xx10^(-7))=6.30`.