You visited us 1 times! Enjoying our articles? Unlock Full Access!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

4.95

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5.15

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5.87

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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

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

Suggest Corrections

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. 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?

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. 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. 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?

Join BYJU'S Learning Program

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)