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The Ph of basic buffer mixtures is given...

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 in Ph for all practical purposes, but sicne the ratio `(["Base"])/(["Salt"])` or `(["Salt"])/(["Acid"])` changes, a slight decrease or increase in pH results.
The volume of 0.2 m NaOH needed to prepare a buffer of pH 4.74 with 50 mL of 0.2 m acetic acid `pH_(b)` of `CH_(3)COO^(-)=9.26` is :

A

`50mL`

B

`25mL`

C

`20mL`

D

`10mL`

Text Solution

Verified by Experts

Let `VmL` of `NaOH` be needed to give `CH_(3)COONa`
`{:(NaOH,+,CH_(3)COOH,rarr,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)(["Salt"])/(["Acid"])`
`=14-9.26+(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)/(10-2V)]:.V=(10)/(0.4)=25mL`.
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