If the degree of dissociation of acetic acid CH 3 COOH in a 0-9 M solution is 4-5 - 10-3- Then- its pK a value will be-A- 4-74B- 5-74C- 7-26D- 6-74

Name: JEE - Grade 11 - Chemistry - Ionic Equilibrium - Session 04 - D08
Uploaded: 2022-07-04T10:36:42+05:30
Description: The correct option is A 4.74Given: degree of dissociation (α) of CH_3COOH=4.5× 10^−3=0.0045 So, according to given reaction: CH_3COOH (aq.) ⇌ H^+ (aq.)+CH_3C ...

The correct option is A 4.74Given: degree of dissociation (α) of CH_3COOH=4.5× 10^−3=0.0045 So, according to given reaction: CH_3COOH (aq.) ⇌ H^+ (aq.)+CH_3C ...

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Byju's Answer

Standard XII

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

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5.74

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6.74

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7.26

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

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