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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 ratio of pH of solution (I) containing 1 mole to pH of solution (II) containing 1 mole of `CH_(3)COONa` and 1 mole of acetic in one litre is :

A

`1:2`

B

`2:1`

C

`1:3`

D

`3:1`

Text Solution

Verified by Experts

The correct Answer is:
A
`{:(CH_(3)COONa+,HCl rarr,CH_(3)COOH+,NaCl,),(" "1," "1," "0," "0,),(" "0," "0," "1," "1,):}`
`:. [CH_(3)COOH] = (1)/(2) =1`.
`:. [H^(+)] = C alpha = C sqrt((K_(a))/(C)) = sqrt(K_(a).C) = sqrt(K_(a))`
or `pH_(1) = -(1)/(2) log K_(a) = (1)/(2) pK_(a)`
`{:(CH_(3)COOH+,CH_(3)COONa),(" "1," "1):}`
`:. pH = pK_(a) + log.(1)/(1) rArr pH_(2) = pK_(a) :. (pH_(1))/(pH_(2)) = (1)/(2)`.
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