Deduce Hendersons equation for an acidic buffer.

Explanation

Consider the acidic buffer, a mixture of acetic acid and sodium acetate.
`CH_(3)COOHhArrCH_(3)COO^(-)+H^(+)to(1)`
`CH_(3)COONatoCH_(3)COO^(-)+Na^(+)to(2)`
Applying law of mass action to (1), `K_(a)=([CH_(3)COO^(-)][H^(+)])/([CH_(3)COOH])to(3)`
where, `K_(a)` is dissociation constant of weak acid.
In this solution, `[CH_(3)COO^(-)]=["Salt"]to(4)`
`[CH_(3)COOH^(-)]=["Acid"]to(5)`
Substituting for (4) and (5) in (3), `K_(a)=(["Salt"][H^(+)])/(["Acid"])rArr[H^(+)]=K_(a)(["Acid"])/(["Salt"])`
Taking `-log_(10)` on both the sides, `-log_(10)[H^(+)]=-log_(10)K_(a)+(-log_(10))(["Acid"])/(["Salt"])`
`pH=pK_(a)+"log"(["Salt"])/(["Acid"])rArr" "thereforepH=pK_(a)+"log"([S])/([A])`.