Deduce Handerson's equation for a basic buffer.

Explanation

Consider the basic buffer, a mixture of ammonium hydroxide and ammonium chloride,
`NH_(4)OHhArrNH_(4)^(+)+OH^(-)to(1)`
`NH_(4)ClhArrNH_(4)^(+)+Cl^(-)to(2)`
Applying law of mass action for (1) `K_(b)=([NH_(4)^(+)][OH^(-)])/([NH_(4)OH])to(3)`
where `K_(b)` = dissociation constant of weak base.
In this solution, `[NH_(4)^(+)]=["Salt"]to(4)`
`[NH_(4)OH]=["Base"]to(5)`
Substituting for (4) and (5) in (3), we get, `K_(b)=(["Salt"][OH^(-)])/(["base"]),[OH^(-)]=K_(b)(["base"])/(["salt"])`
Taking `-log_(10)` on both the sides, `-log_(10)[OH^(-)]=-log_(10)K_(b)+(-log_(10))(["base"])/(["salt"])`
`pOH=pK_(b)+"log"(["Salt"])/(["Base"])rArrthereforepOH=pK_(b)+"log"([S])/([B])`.