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Acidic Buffer Action

What will be ...

Question

What will be the new $p H$ if $0.01 m o l$ of $H C l$ is added to a $1 L$ buffer solution that is $0.02 M$ in propionic acid $(C H_{3} C H_{2} C O O H)$ and $0.015 M$ in sodium propionate $(C H_{3} C H_{2} C O O N a)$ ?
Dissociation constant $K_{a}$ of propionic acid at $25^{\circ} C$ is $1.34 \times 10^{- 5}$
(Take $l o g (1.34) = 0.13$ )

4.09

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4.95

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5.15

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5.87

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Solution

The correct option is A 4.09
$p K_{a} = - l o g (K_{a}) p K_{a} = - l o g (1.34 \times 10^{- 5}) p K_{a} = (5 - 0.13) = 4.87$
Finding initial $p H$ of the buffer
$p H$ for acidic buffer can be defined by :

$p H = p K_{a} + l o g (\frac{[C H_{3} C H_{2} C O O^{-}]}{[C H_{3} C H_{2} C O O H]}) p H = 4.87 + l o g (\frac{0.015}{0.02}) p H = 4.87 + l o g (0.75) p H = 4.87 - 0.12 = 4.75$

Since, Volume = $1 L$
Concentration = moles
$∴ [H C l] = 0.01 M$

On addition of acid externally, concentration of anion of salt decreases whereas concentration of acid increases by same amount.

So,
$[C H_{3} C H_{2} C O O^{-}] = 0.015 - 0.01 = 0.005 M [C H_{3} C H_{2} C O O H] = 0.02 + 0.01 = 0.03 M$

Finding the changed pH

$p H_{n e w} = p K_{a} + l o g (\frac{[C H_{3} C H_{2} C O O^{-}]}{[C H_{3} C H_{2} C O O H]}) p H_{n e w} = 4.87 + l o g (\frac{0.005}{0.03}) p H_{n e w} = 4.87 + l o g (\frac{1}{6}) p H_{n e w} = 4.87 - 0.78 p H_{n e w} = 4.09$

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Similar questions

Q. How many moles of sodium propionate

(C H_{3} C H_{2} C O O N a)

should be added to one litre of an aqueous solution containing 0.02 mole of propionic acid

(C H_{3} C H_{2} C O O H)

to obtain a buffer solution of

p H = 5.87

?
Dissociation constant of propionic acid

K_{a}

25^{\circ} C

1.34 \times 10^{- 5}

Take

l o g (1.34) = 0.13

Q. How many moles of sodium propionate

(C H_{3} C H_{2} C O O N a)

should be added to one litre of an aqueous solution containing 0.02 mole of propionic acid

(C H_{3} C H_{2} C O O H)

to obtain a buffer solution of

p H = 5.87

?
Dissociation constant of propionic acid

K_{a}

25^{\circ} C

1.34 \times 10^{- 5}

Take

l o g (1.34) = 0.13

Q. The

K_{a}

of propionic add is

1.34 \times 10 - 5

. What is the

p H

of a solution containing

0.5

M propionic acid and

0.5

M sodium propionate?

Q. I. How many moles of sodium propionate should be added to one litre of an aqueous solution contain

0.02

mole of propionic acid to obtain a buffer solution of

p H

4.7

?
II. What will be the

p H

0.0005

mole of hydrogen chloride is dissolved in the above buffer solution?
III. Compare the last

p H

value with the

p H

of a

0.01

molar

H C l

solution.
Assume that dissociation constant of propionic acid,

25^{o} C

1.00 \times 10^{- 5}

Q. A buffer solution was prepared by dissoving

0.02

mol propionic acid and

0.015

mol sodium propionate in enough water to make

1.00 L

of solution.

(K_{a}

for propoinic acid is

1.34 \times 10^{- 5}

). What is the

p H

of the buffer?

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What will be the new pH if 0.01 mol of HCl is added to a 1 L buffer solution that is 0.02 M in propionic acid (CH3CH2COOH) and 0.015 M in sodium propionate (CH3CH2COONa) ? Dissociation constant Ka of propionic acid at 25∘C is 1.34×10−5 (Take log(1.34)=0.13)