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Deduce Hendersons equation for an acidic buffer. |
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Answer» Solution :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])`. |
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