If the degree of dissociation of acetic acid CH3COOH in a 0-9 M solution is 4-5- 10-3- Then- its pKa value will be-

Given: degree of dissociation (α) of CH_3COOH=4.5× 10^−3=0.0045 So, according to given reaction: CH_3COOH (aq.) ⇌ H^+ (aq.)+CH_3COO^ (aq.) Initial: ...

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

Byju's Answer

Standard XIII

Chemistry

pH the Power of H

If the degree...

Question

If the degree of dissociation of acetic acid $(C H_{3} C O O H)$ in a $0.9 M$ solution is $4.5 \times 10^{- 3}$ . Then, its $p K_{a}$ value will be:

4.74

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

5.74

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

6.74

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

7.26

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

Open in App

Solution

The correct option is A 4.74
Given: degree of dissociation $(α)$ of $C H_{3} C O O H = 4.5 \times 10^{- 3} = 0.0045$

So, according to given reaction:

$C H_{3} C O O H (a q .) ⇌ H^{+} (a q .) + C H_{3} C O O^{-} (a q .) I n i t i a l : C 00 A t e q u i l i b r i u m : C (1 - α) C α C α$

$K_{a} = \frac{[C H_{3} C O O^{-}] [H^{+}]}{[C H_{3} C O O H]} = \frac{(C α)^{2}}{C (1 - α)} = \frac{C α^{2}}{1 - α} K_{a} = \frac{0.9 \times (0.0045)^{2}}{(1 - 0.0045)} \approx 1.8 \times 10^{- 5}$

$p K_{a} = - {log}_{10} (K_{a})$
putting values,
$p K_{a} = - {log}_{10} (1.8 \times 10^{- 5}) = (6 - l o g_{10} 18)$
$p K_{a} = 6 - 1.26$
$p K_{a} = 4.74$

Suggest Corrections

Similar questions

Q. If the degree of dissociation of acetic acid

(C H_{3} C O O H)

in a

0.9 M

solution is

4.5 \times 10^{- 3}

. Then, its

p K_{a}

value will be:

Q. The degree of dissociation of acetic acid in a 0.1 N solution is

1.32 \times 10^{- 2}

. At what concentration of nitrous its, degree of dissociation will be same as that of acetic acid?

{K_{a} (H N O_{2}) = 4 \times 10^{- 4}}

Q. The degree of dissociation of acetic acid in a

0.1 M

solution is

1.32 \times 10^{- 2}

, find out its

p K_{a}

value.
Take

l o g_{10} (1.76) = 0.25

Calculate the degree of ionization of 0.05M acetic acid if its pKa value is 4.74.
How is the degree of dissociation affected when its solution also contains
0.01 M

0.1 M in HCl?

Q. What will be the percentage degree of dissociation of

0.2 M

acetic acid

(C H_{3} C O O H)

solution with

K_{a} = 1.8 \times 10^{- 5}

Join BYJU'S Learning Program

Explore more

pH the Power of H

Standard XIII Chemistry

Join BYJU'S Learning Program

If the degree of dissociation of acetic acid (CH3COOH) in a 0.9 M solution is 4.5×10−3. Then, its pKa value will be:

If the degree of dissociation of acetic acid $(C H_{3} C O O H)$ in a $0.9 M$ solution is $4.5 \times 10^{- 3}$ . Then, its $p K_{a}$ value will be: