You visited us 2 times! Enjoying our articles? Unlock Full Access!

Acidic Buffer Action

A 30 mL solut...

Question

A $30 m L$ solution of $0.2 M C H_{3} C O O H$ is titrated with $0.2 M N a O H$ . Find the $p H$ of the solution when $10 m L$ of $N a O H$ has been added.
Given $p K_{a} (C H_{3} C O O H) = 4.74$

5.04

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

4.44

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

3.54

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

4.04

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

Open in App

Solution

The correct option is B 4.44
mmol of $C H_{3} C O O H = 30 \times 0.2 = 6$
mmol of $N a O H = 10 \times 0.2 = 2$

$C H_{3} C O O H (a q) + N a O H (a q) \to C H_{3} C O O N a (a q) + H_{2} O (l) I n i t i a l l y : 6200 F i n a l : 4022$

After titration we can see that $4 m m o l$ of $C H_{3} C O O H$ and $2 m m o l$ of $C H_{3} C O O N a$ are present.
$Final Concentration = \frac{Moles}{Total Volume}$
Total Volume $= 30 + 10 = 40 m L$
$[C H_{3} C O O H] = \frac{4}{40} M$
$[C H_{3} C O O N a] = \frac{2}{40} M$
An acidic buffer is formed. So,
$p H = p K_{a} + l o g (\frac{[C H_{3} C O O N a]}{[C H_{3} C O O H]}) p H = 4.74 + l o g ⎛ ⎝ \frac{\frac{2}{40}}{\frac{4}{40}} ⎞ ⎠ p H = 4.74 + l o g (\frac{1}{2}) = 4.74 - 0.3 p H = 4.44$

Suggest Corrections

Similar questions

Q. A

50 m L

solution of

0.2 M C H_{3} C O O H

is titrated with

0.2 M N a O H

. Find the

p H

of solution when:
(a)

0 m L N a O H

is added (

p H_{i n i t i a l}

)
(b)

10 m L

N a O H

has been added (

p H_{10 m L}

)

Given:

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

Q. A

30 m L

solution of

0.2 M C H_{3} C O O H

is titrated with

0.2 M N a O H

. Find the

p H

of the solution when

10 m L

N a O H

has been added.
Given

p K_{a} (C H_{3} C O O H) = 4.74

Q. A

30 m L

solution of

0.2 M N H_{4} O H

is titrated with

0.2 M H C l

. Find the

p H

of the solution when

10 m L

H C l

has been added.
Given

p K_{b} (N H_{4} O H) = 4.74

20 m L

0.2 M C H_{3} C O O N a

solution is titrated against

0.1 M H C l

. Calculate the

p H

of solution when

10 m L H C l

has been added.
Given:

p K_{a} (C H_{3} C O O H) = 4.74

Q. What is the pH when 25.0 mL of 0.2 M of

C H_{3} C O O H

has been titrated with 35.0 mL of 0.1 M

N a O H

p K_{a} (C H_{3} C O O H) = 4.752

l o g 2.333 = 0.368

Join BYJU'S Learning Program

Explore more

Acidic Buffer Action

Standard XII Chemistry

Join BYJU'S Learning Program

A 30 mL solution of 0.2 M CH3COOH is titrated with 0.2 M NaOH. Find the pH of the solution when 10 mL of NaOH has been added. Given pKa(CH3COOH)=4.74

A $30 m L$ solution of $0.2 M C H_{3} C O O H$ is titrated with $0.2 M N a O H$ . Find the $p H$ of the solution when $10 m L$ of $N a O H$ has been added.
Given $p K_{a} (C H_{3} C O O H) = 4.74$