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To prepare a buffer of pH 8.26, amount o...

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:

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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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 amount of (NH_(4))_(2)SO_(4) to be added to 500mL of 0.01 M NH_(4)OH solution (pH_(a)NH_(4)^(+) is 9.26) to prepare a buffer of pH 8.26 is :

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:

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Amount of (NH_(4))_(2)SO_(4) which must be added to 50mL of 0.2 M NH_(4)OH solution to yield a solution of pH 9.26 is ( pK_(b) of NH_(4)OH=4.74 )

For the preparation of a buffer of pH = 8.26 , the amount of (NH_4)_(2)SO_(4) required to be mixed with one litre of 0.1 (M) NH_3(aq), pK_b = 4.74 is ?

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)^(+))]

Knowledge Check

  • 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 amount of (NH_(4))_(2)SO_(4) to be added to 500mL of 0.01 M NH_(4)OH solution (pH_(a)NH_(4)^(+) is 9.26) to prepare a buffer of pH 8.26 is :

    A
    0.05 mole
    B
    0.025 mole
    C
    0.10 mole
    D
    0.005 mole

  • 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

  • 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_(1) + log.(["Salt"])/(["Acid"]) . Addition of little acid or base although shows no change in pH for all practical purposes , but since the ratio (["Base"])/(["Salt"]) for (["Salt"])/(["Acid"]) changes, a slight decrease or increase in pH results. The amount of (NH_(4))_(2)SO_(4) to be added to 500 mL of 0.01 M NH_(4)OH solution ( pK_(a) for NH_(4)^(+) is 9.26) to prepare a buffer of pH 8.26 is

    A
    0.05 mole
    B
    0.025 mole
    C
    0.10 mole
    D
    0.005 mole