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Standard XII

Chemistry

Van't Hoff Factor

To prepare a ...

Question

To prepare a buffer of pH = 8.26, amount of $(N H_{4})_{2} S O_{4}$ to be added into $500$ ml. of M $N H_{4} O H$ solution $[P K_{a} N H_{4} = 9.26]$ is:

0.05

mol

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0.025

mol

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0.10

mol

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0.005

mol

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Solution

The correct option is D $0.005$ mol
$0.005 m o l$
Using henderson equation
$p H = p k a + log \frac{b a s e}{a c i d}$
now,
$500 m l (\frac{1 l}{1000 m l}) (\frac{0.01 m l}{l})$
$= 0.005 m o l$

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Similar questions

Q. The amount of

(N H_{4})_{2} S O_{4}

to be added to

500

mL of

0.01

N H_{4} O H

solution (

p K_{a}

for

N H_{4}^{+}

9.26

) prepare a buffer of

p H

= 8.26

is?

Q. What will be the amount of

(N H_{4})_{2} S O_{4}

(in g) which must be added to 500 mL of 0.2 M

N H_{4} O H

to yield a solution of pH 9.35? [Given,

p K_{a}

N H_{4}^{+}

= 9.26,

p K_{b} (N H_{3})

= 14 -

p K_{a} (N H_{4}^{+})

]

Q. If

p K_{a} (N H_{4}^{+}) = 9.2

, then the amount of

(N H_{4})_{2} S O_{4}

to be added in 500 mL of 0.1 M

N H_{4} O H

for the preparation of a buffer of pH=8.5 would be:
(Given: log5=0.7)

Q. To prepare a buffer solution of

p H = 4.04

, amount of Barium acetate to be added to

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[p K_{b} (C H_{3} C O O^{-}) = 9.26]

is:
Given

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Q. In order to prepare a buffer of

p H = 8.26

, the amount of

(N H_{4})_{2} S O_{4}

required to be mixed with one litre of

0.1 (M) N H_{3} (a q),

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To prepare a buffer of pH = 8.26, amount of (NH4)2SO4 to be added into 500 ml. of M NH4OH solution [PKaNH4=9.26] is:

The correct option is D 0.005 mol0.005molUsing henderson equationpH=pka+logbaseacidnow,500ml(1l1000ml)(0.01mll)=0.005mol

To prepare a buffer of pH = 8.26, amount of $(N H_{4})_{2} S O_{4}$ to be added into $500$ ml. of M $N H_{4} O H$ solution $[P K_{a} N H_{4} = 9.26]$ is:

The correct option is D $0.005$ mol
$0.005 m o l$
Using henderson equation
$p H = p k a + log \frac{b a s e}{a c i d}$
now,
$500 m l (\frac{1 l}{1000 m l}) (\frac{0.01 m l}{l})$
$= 0.005 m o l$