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

Acidic Buffer Action

20 mL of 0.2 ...

Question

$20 m L$ of $0.2 M$ sodium hydroxide is added to $50 m L$ of $0.2 M$ acetic acid to give $70 m L$ of the solution. Calculate the $p H$ of the solution.
The dissociation constant for $C H_{3} C O O H$ is $1.8 \times 10^{- 5}$ .

3.56

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

4.56

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

5.56

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

6.56

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

Open in App

Solution

The correct option is B 4.56
We know,
millimoles = molarity $\times$ volume (mL)
millimoles of $N a O H = 0.2 \times 20 = 4$
millimoles of $C H_{3} C O O H = 0.2 \times 50 = 10$

$C H_{3} C O O H (a q) + N a O H (a q) \to C H_{3} C O O N a (a q) + H_{2} O (l) I n i t i a l : 10400 E q u i l i b r i u m : (10 - 4) 044$

$[C H_{3} C O O N a] = \frac{4}{70} M [C H_{3} C O O H] = \frac{6}{70} M$
$p K_{a} = - l o g (K_{a}) p K_{a} = - l o g (1.8 \times 10^{- 5}) p K_{a} = (5 - 0.26) = 4.74$
$p H = p K_{a} + l o g \frac{[s a l t]}{[a c i d]} p H = p K_{a} + l o g (\frac{[C H_{3} C O O N a]}{[C H_{3} C O O H]}) p H = 4.74 + l o g ⎛ ⎝ \frac{\frac{4}{70}}{\frac{6}{70}} ⎞ ⎠ p H = 4.74 + l o g (0.67) p H = 4.74 - 0.18 = 4.56 p H = 4.56$

Suggest Corrections

Similar questions

20 m L

0.2 M

sodium hydroxide is added to

50 m L

0.2 M

acetic acid to give

70 m L

of the solution. Calculate the

p H

of the solution.
The dissociation constant for

C H_{3} C O O H

1.8 \times 10^{- 5}

Q. 20 mL of 0.2 M sodium hydroxide is added to 50 mL of 0.2 M acetic acid to give 70 mL of the solution. What is the pH of this solution. Calculate the additional volume of 0.2 M NaOH required to make the pH of the solution 4.74. (Ionisation constant of

C H_{3} C O O H

1.8 \times 10^{- 5})

Q. 20 ml of 0.2 M NaOH is added to 50 ml of 0.20 M acetic acid to give 70 ml of the solution. What is the pH of the solution?

The ionisation constant of acetic acid is

1.8 \times 10^{- 5}

20 mL of 0.2 M NaOH is added to 50 mL of 0.2 M acetic acid to give 70 mL of the solution. What is the $p H$ of the solution? Calculate the additional volume $x$ (in mL) of 0.2 M NaOH required to make pH of the solution 4.74. The ionisation constant of acetic acid is $1.8 \times 10^{- 5}$ .

Q. 20 ml of 0.2 M NaOH is added to 50 ml of 0.2 M acetic acid. What is the pH of the resulting solution? Calculate the additional volume of 0.2 M NaOH required for making the solution of pH 4.74. The ionization constant of acetic acid is

1.8 \times 10^{- 5} .

(l o g 1.8 = 0.2552, l o g 0.66 = - 0.18)

Join BYJU'S Learning Program

Explore more

Acidic Buffer Action

Standard XIII Chemistry

Join BYJU'S Learning Program

20 mL of 0.2 M sodium hydroxide is added to 50 mL of 0.2 M acetic acid to give 70 mL of the solution. Calculate the pH of the solution. The dissociation constant for CH3COOH is 1.8×10−5.

$20 m L$ of $0.2 M$ sodium hydroxide is added to $50 m L$ of $0.2 M$ acetic acid to give $70 m L$ of the solution. Calculate the $p H$ of the solution.
The dissociation constant for $C H_{3} C O O H$ is $1.8 \times 10^{- 5}$ .