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

Acidic Buffer Action

8g of NaOH wa...

Question

8g of $N a O H$ was dissolved in 1 L of a solution containing one mole of acetic acid $C H_{3} C O O H$ and one mole of sodium acetate $C H_{3} C O O N a$ . Find the pH of the resulting solution.
The dissociation constant for $C H_{3} C O O H$ is $1.8 \times 10^{- 5}$ .

6.88

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

3.68

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

5.76

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

Open in App

Solution

The correct option is B 4.92
Concentration of $N a O H = \frac{8}{40} = 0.2 m o l L^{- 1}$
$C H_{3} C O O H (a q) + N a O H (a q) \to C H_{3} C O O N a (a q) + H_{2} O (l) I n i t i a l : 1 0.2 10 E q u i l i b r i u m : (1 - 0.2) 01 + 0.2 0.2$
$p K_{a} = - l o g (K_{a}) p K_{a} = - l o g (1.8 \times 10^{- 5}) p K_{a} = (5 - 0.26) = 4.74$

$[C H_{3} C O O H]_{e q b} = 0.8 M [C H_{3} C O O N a]_{e q b .} = 1.2 M p H = p K_{a} + l o g \frac{[s a l t]}{[a c i d]} p H = p K_{a} + l o g (\frac{[C H_{3} C O O N a]}{[C H_{3} C O O H]}) p H = 4.74 + l o g (\frac{1.2}{0.8}) p H = 4.74 + l o g (1.5) = (4.74 + 0.18) p H = 4.92$

Suggest Corrections

Similar questions

N a O H

C H_{3} C O O H

C H_{3} C O O N a

C H_{3} C O O H

1.8 \times 10^{- 5}

p H

0.01 m o l

N a O H

1 L

0.1 M

(C H_{3} C O O H)

0.1 M

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

K_{a}

25^{\circ} C

1.8 \times 10^{- 5}

0.063

4.5

K_{a}

{C H}_{3} C O O H = 1.8 \times 10^{- 5}

p H = 5.74

C H_{3} C O O H

C H_{3} C O O N a

0.01 M

C H_{3} C O O H

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

l o g (1.8) = 0.26

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

C H_{3} C O O H + (C H_{3} C O O N a

C H_{3} C O O H

Join BYJU'S Learning Program