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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 for all practical purpose, but since the ratio `(["Base"])/(["Salt"])` or `(["Salt"])/(["Acid"])` change, a slight decrease or increase in pH results in.
The amount of `(NH_(4))_(2)SO_(4)` to be added to 500 mL of `0.01M NH_(4)OH` solution `(pK_(a) for NH_(4)^(+) is 9.26)` prepare a buffer of `pH 8.26` is:

A

0.05 mole

B

0.025 mole

C

0.10 mole

D

0.005 mole

Text Solution

Verified by Experts

The correct Answer is:
B
`pH = pK_(a) + log.(["Base"])/(["Salt"]) rArr ["Base"] = (0.01 xx 500)/(500) = 0.01`
`[NH_(4)^(+)] = (a xx 2)/(500)`, Let a millimole of `(NH_(4))_(2)SO_(4)` are added.
`:. [Salt] = [NH_(4)^(+)]`.
`pH = 9.26 + log [(0.01)/(2a//500)]`
`8.26 = 9.26 + log.(0.01 xx 500)/(2a) :. a = 25`
`:.` Mole of `(NH_(4))_(2)SO_(4)` added `= 0.025`.
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