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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

For the buffer solution of `NH_(3)` & `NH_(4)^(+)`
`pH=pK_(a)+log (([NH_(3)])/([NH_(4)^(+)]))rArr 8.26=9.26+log ((500xx0.01)/("m.moles"of NH_(4)^(+)))`
rArr "m.moles" of `NH_(4)^(+)=50`:. moles of `(NH_(4))_(2)SO_(4)` required `=0.025`
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To prepare a buffer of pH 8.26 , amount of (NH_(4))_(2)SO_(4) to be added into 500mL of 0.01M NH_(4)OH solution [pK_(a)(NH_(4)^(+))=9.26] is:

What will be the degree of dissociation of 0.005M NH_(4)OH solution. If pK_(b) for NH_(4)OH is 4.7

Knowledge Check

  • What will be the amount of (NH_(4))_(2)SO_(4) (in g) which must be added to 500 mL of 0.2 M NH_(4)OH to yield a solution of pH 9.35? ["Given," pK_(a)"of "NH_(4)^(+)=9.26,pK_(b)NH_(4)OH=14-pK_(a)(NH_(4)^(+))]

    A
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    B
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    C
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    D
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