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The amount of (NH(4))(2)SO(4) to be adde...

The amount of `(NH_(4))_(2)SO_(4)` to be added to `500 mL` of `0.01 M NH_(4)OH` solution `(pK` 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 =pKa + log. (["Base"])/(["Salt"])`
`["Base"] = (0.01 xx 500 )/500 = 0.01`
Let .a. milli moles of `(NH_(4))_(2)SO_(4)`
`[NH_(4)^(+)] = (a xx2)/500 " " :.[ "Salt"] = [ NH_(4)^(+)] `
`pH = 9.26 + log. [(0.01)/(2a//500)] `
` 8.26 = 9.26 + log [ (0.01xx 500)/(2a)] `
` a = 25`
` :. ` Moles of `(NH_(4))_(2)SO_(4) ` added `= 0.025`
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