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Standard XII

Chemistry

Common Ion Effect

Suppose it is...

Question

Suppose it is required to make a buffer solution of pH = 4, using acetic acid and sodium acetate.
How many grams of sodium acetate is to be added to one litre of $\frac{N}{10}$ acetic acid? $K_{a} = 1.8 \times 10^{- 5}$

2.567

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3.257

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1.476

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4.235

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Solution

The correct option is C 1.476
Using Henderson’s equation
$p H = l o g \frac{[S a l t]}{[A c i d]} - l o g K_{a}$
$4 = l o g [S a l t] - l o g (0.1) - l o g 1.8 \times 10^{- 5}$
$L o g [S a l t] = (4 - 1 - 5 + 0.2552)$
$∴ [S a l t] = 0.018$ g mole/litre

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Similar questions

Q. Suppose it is required to make a buffer solution of

p H = 4

, using acetic acid and sodium acetate. How much of sodium acetate is to be added to

1

litre of

N / 10

acetic acid?
Dissociation constant of acetic acid

1.8 \times 10^{- 5}

Q. How many moles of acetic acid and sodium acetate each should be dissolved to prepare one litre of

0.063

molar buffer solution of pH

4.5

?
(

K_{a}

for

{C H}_{3} C O O H = 1.8 \times 10^{- 5}

)

Q. Consider an acidic buffer solution of acetic acid and sodium acetate at

25^{o} C

. What should be the ratio of concentration of sodium acetate and acetic acid so that

p H = 4

is obtained ?

Given :

K_{a}

value of acetic acid is

1.8 \times 10^{- 5}

and 10^{(- 0.75)} = 0.18

Q. What will be the new

p H

0.01 m o l

N a O H

is added to a

1 L

buffer solution that is

0.1 M

in acetic acid

(C H_{3} C O O H)

and

0.1 M

in sodium acetate

(C H_{3} C O O N a)

?
Dissociation constant

K_{a}

of acetic acid at

25^{\circ} C

1.8 \times 10^{- 5}

Q. One litre of a buffer solution has

0.13833

mole of acetic acid and

0.1951

mole of sodium acetate. Calculate

p H

(nearest integer) of solution. Neglect volume change.

K_{a} = 1.8 \times 10^{- 5}

. (log1.8 = 0.255)

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Suppose it is required to make a buffer solution of pH = 4, using acetic acid and sodium acetate. How many grams of sodium acetate is to be added to one litre of N10 acetic acid? Ka=1.8×10−5

The correct option is C 1.476Using Henderson’s equation pH=log[Salt][Acid]−log Ka 4=log[Salt]−log(0.1)−log 1.8×10−5 Log[Salt]=(4−1−5+0.2552) ∴[Salt]=0.018 g mole/litre

Suppose it is required to make a buffer solution of pH = 4, using acetic acid and sodium acetate.
How many grams of sodium acetate is to be added to one litre of $\frac{N}{10}$ acetic acid? $K_{a} = 1.8 \times 10^{- 5}$

The correct option is C 1.476
Using Henderson’s equation
$p H = l o g \frac{[S a l t]}{[A c i d]} - l o g K_{a}$
$4 = l o g [S a l t] - l o g (0.1) - l o g 1.8 \times 10^{- 5}$
$L o g [S a l t] = (4 - 1 - 5 + 0.2552)$
$∴ [S a l t] = 0.018$ g mole/litre