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 $N a O H$ 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 $25^{\circ} C$ is $1.8 \times 10^{- 5}$

5.52

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

4.92

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

4.1

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

6.13

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

Open in App

Solution

The correct option is B 4.92
$p K_{a} = - log (K_{a}) p K_{a} = - log (1.8 \times 10^{- 5}) p K_{a} = (5 - 0.26) = 4.74$
Finding initial $p H$ of the buffer
$p H$ for acidic buffer can be defined by :

$p H = p K_{a} + log (\frac{[C H_{3} C O O^{-}]}{[C H_{3} C O O H]}) p H = 4.74 + log (\frac{0.1}{0.1}) p H = 4.74 + log (1) p H = 4.74$

Since, Volume = $1 L$
Concentration = moles
$∴ [N a O H] = 0.02 M$

On addition of small amount of base ( $N a O H$ ) externally there have a reaction like,
$C H_{3} C O O H + N a O H ⇌ C H_{3} C O O N a + H_{2} O$ .
So, the concentration of conjugate base increases whereas concentration of acid decreases by same amount.

So,
$[C H_{3} C O O^{-}] = 0.1 + 0.02 = 0.12 M [C H_{3} C O O H] = 0.1 - 0.02 = 0.08 M$

Finding the changed $p H$ :

$p H_{n e w} = p K_{a} + log (\frac{[C H_{3} C O O^{-}]}{[C H_{3} C O O H]}) p H_{n e w} = 4.74 + log (\frac{0.12}{0.08}) p H_{n e w} = 4.74 + log (\frac{3}{2}) p H_{n e w} = 4.74 + log (3) - log (2) p H_{n e w} = 4.74 + 0.48 - 0.30 = 4.92$

Suggest Corrections

Similar questions

Q. A buffer solution of

p H = 5.74

is to be prepared using

C H_{3} C O O H

and

C H_{3} C O O N a

. Find the concentration of sodium acetate to be added to

0.01 M

solution of

C H_{3} C O O H

K_{a} (C H_{3} C O O H) = 1.8 \times 10^{- 5}

Take

l o g (1.8) = 0.26

Q. A buffer solution contains 0.1

M

of acetic acid and 0.1

M

of sodium acetate. What will be its

p H

, if

p K_{a}

of acetic acid is 4.75

?

Q. What amount (in milli mole) of sodium acetate

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

must be added to

2 L

0.1 M

acetic acid solution to given a solution of pH = 3 [Given:

K_{a (C H_{3} C O O H)} = 10^{- 5}]

Q. A buffer solution of

p H = 4

is to be prepared, using

C H_{3} C O O H

and

C H_{3} C O O N a .

The concentration of acetic acid and sodium acetate are

\frac{1}{10} M

and

x M

. How much will the

p H

be if

1.5 L

H_{2} O

is added to above buffer ?

K_{a} (C H_{3} C O O H) = 4.7447

5.0 m L

0.1 M N a O H

solution is added to

50 m L

of the

0.1 M

acetic acid solution. Calculate the

p H

of the resulting acetic acid solution.

(K a = 1.8 \times 10^{- 5})

Join BYJU'S Learning Program

Explore more

Acidic Buffer Action

Standard XI Chemistry

Join BYJU'S Learning Program

What will be the new pH if 0.01 mol of NaOH is added to a 1 L buffer solution that is 0.1 M in acetic acid (CH3COOH) and 0.1 M in sodium acetate (CH3COONa) ? Dissociation constant Ka of acetic acid at 25∘C is 1.8×10−5

What will be the new $p H$ if $0.01 m o l$ of $N a O H$ 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 $25^{\circ} C$ is $1.8 \times 10^{- 5}$