The ratio of volumes of `CH_(3)COOH`0.1 (N) to `CH_(3)COONa` 0.1 (N) required to prepare a buffer solution of pH 5.74 is
(given : `pK_(a) " of " CH_(3) COOH " is " 4.74)`

Options

A. `10:1`
B. `5:1`
C. `1:5`
D. `1:10`

Answer

D

Explanation

`pH = pK_(a) ++log. (["Salt"])/(["acid"])`
`5.74 = 4.74 + log . (["salt"])/(["acid"])`
`log .(["salt"])/(["acid"]) = 1`
`(["salt"])/(["acid"]) = 10/1 `
If volume of acid is `V_(a)` and volume of salt is `V_(s)`and molarity of acid= molarity of salt=0.1, then
`((0.1 xxV_(s))//(V_(a)+V_(s)))/((0.1 xxV_(a))//(V_(a)+V_(s)))=10/1 `
`(V_(s))/(V_(a)) =10/1or (V_(a))/(V_(s))=1/10`