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The degree of dissociation of a 0.01 M w...

The degree of dissociation of a 0.01 M weak acid is ` 10^(-3)` . Its pOH is

A

5

B

3

C

9

D

11

Text Solution

Verified by Experts

The correct Answer is:
C
`[H^(+)]=sqrt(K_(a)*C)`
`or = sqrt(alpha^(2)* C^(2))" " [ :. K_(a) = alpha^(2)C]`
` =sqrt((10^(-3))^(2)*(0.01)^(2))`
=`10^(-5)`
`pH=-log[H^(+)]=-log10^(-5)` =5
` pOH= 14-5=9`
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