The `pK_(a)` value of acetic acid is 4.7447 at `25^(@)C`. How would you obtain a buffer of acetic acid and sodium acetate with pH = 4?

Explanation

`pH=pK_(a)+"log"(["Salt"])/(["Acid"])rArr4=4.7447+"log"(["Salt"])/(["Acid"])`
`"log"(["Salt"])/(["Acid"])=4.000-4.7447=-0.7447=1.2553`
Taking antilog on both sides, `therefore(["Salt"])/(["Acid"])="anti"log(-1.2553)rArr(["Salt"])/(["Acid"])=0.18`
So by mixing sodium acetate and acetic acid in the ratio 0.18 : 1 mole per `dm^(3)`, the buffer of pH = 4 is obtainetd.