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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

`{:(CH_(3)COONa,+,HCl,rarr,CH_(3)COOH,+,NaCl),(1,,1,,0,,0),(0,,0,,1,,1):}`
`:.[CH_(3)COOH]=(1)/(1)=1`
`:.[H^(+)]=Calpha=Csqrt((K_(a))/(C))=sqrt(K_(a).C)=sqrt(K_(a))`
or `pH_(1)=-(1)/(2)logK_(a)=(1)/(2)pK_(a)`
`underset(1)(CH_(3)COOH)+underset(1)(CH_(3)COONa)`
`:.pH=pK_(a)+(log)(1)/(1)`
`pH_(2)=pK_(a)`
`(pH_(1))/(pH_(2))=(1)/(2)`
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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. Mole of HCI required to prepare a buffer solution of pH=8.5 with 0.1 mole of NaCN in one litre solution is: (pK_(a) for CN^(-)=4.61)

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)