The pH of basic buffer mixtures is given by : `pH=pK_(a)+log((["Base"])/(["Salt"]))` , whereas pH of acidic buffer mixtures is given by: `pH= pK_(a)+log((["Salt"])/(["Acid"]))`. Addition of little acid or base although shows no appreciable change for all practical purpose, but since the ratio `(["Base"])/(["Salt"])` or `(["Salt"])/(["Acid"])` change, a slight decrease or increase in pH results in.
The volume of `0.2M NaOH` needed to prepare a buffer of `pH 4.74` with 50 mL of `0.2M` acetic acid is: `(pK_(a) of CH_(3)COO^(-)= 9.26)`

Options

A. `50 mL`
B. `25 mL`
C. `20 mL`
D. `10 mL`

Answer

B

Explanation

Let V mL of NaOH be neede to given `CH_(3)COONa`.
`{:(NaOH+,CH_(3)COOHrarr,CH_(3)COONa+,H_(2)O),(0.2xxV,50xx0.2,0,0),(-,[10-0.2V],0.2V,0.2V):}`
`:. pH = pK_(a)+log ((["Salt"])/(["Acid"]))`
`=pK_(w)-pK_(b)+log((["Salt"])/(["Acid"]))`
`= 14-9.26+log ([(0.2V)/(50+V)])/([(10-0.2V)/(50+V)])`
`4.74=4.74+log [(0.2V)/(10-0.2V)]`
`:. v= (10)/(0.4)= 25 mL`