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Answer the following in brief : The pH of a weak monobasic acid is 3.2 in its 0.02 M solution. Calculate its dissociation constant.

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Question

Answer the following in brief :

The pH of a weak monobasic acid is 3.2 in its 0.02 M solution. Calculate its dissociation constant.

Sum
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Solution

Given:

pH = 3.2

Concentration Dissociation = 0.02 M

Dissociation constant (Ka) = ?

Calculation:

Ka = `(Calpha^2)/(1 - alpha)`

`alpha = {[H^+]}/C`

pH = −log10[H+]

3.2 = −log10[H+]

−3.2 = log10[H+]

`10^-3.2 = [H^+]`

`6.3 xx 10^-4 = [H^+]`

`alpha = {[H^+]}/C`

= `(6.3 xx 10^-4)/0.02`

= 0.0315

Ka = `(Calpha^2)/(1 - alpha)`

`(0.02 xx (0.0315)^2)/(1 - 0.0315)`

= `(0.02 xx 0.000992)/0.9685`

= `(0.02 xx 9.92 xx 10^-4)/(1)`

= `0.1984 xx 10^-4`

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Chapter 3: Ionic Equilibria - Exercises [Page 62]

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Balbharati Chemistry [English] Standard 12 Maharashtra State Board
Chapter 3 Ionic Equilibria
Exercises | Q 3. ix. | Page 62

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