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Deduce Handerson's equation for a basic buffer. |
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Answer» Solution :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])`. |
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