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

Acidic Buffer Action

2 g of acetat...

Question

$2 g$ of acetate ions in the form of sodium acetate are added to $500 mL$ of a $0.10 M$ acetic acid solution of $p K_{a} (C H_{3} C O O H) = 4.74$ . What is the $p H$ of the final solution?

4.56

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

4.74

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

5.01

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

5.70

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

Open in App

Solution

The correct option is A $4.56$
No of mmoles of acetate ion $= \frac{2 \times 1000}{60} = 33.33 m m o l e s$ also, 50 mmoles of $C H_{3} C O O H$ are present in the solution.
$p H = p K_{a} + l o g \frac{[C H_{3} C O O^{-}]}{C H_{3} C O O H]}$
$p H = 4.74 + l o g \frac{33.33}{50}$
$p H = 4.74 + l o g 2 - l o g 3 = 4.564$

Suggest Corrections

Similar questions

2 g

500 mL

0.10 M

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

p H

10 m L

0.10 M

20 m L

0.1 M

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

500 m L

0.2 M

500 m L

0.30 M

1.5 \times 10^{- 5}

p H

p K_{a}

4.74

p H 5.74

p H

0.1 M

0.01 M

(p K_{a} = 4.74)

Join BYJU'S Learning Program