Question

Ionization constant of CH3COOH is 1.8X10 5, degree of ionization of 0.01 M CH3COOH
is​

Answer 1

Step-by-step explanation:

When acetic acid is dissolved in water, it partially dissociates (2%).

Ionisation constant K

a

=1.87×10

−5

K

a

=

C(1−α)

Cα×Cα

We assume α∼0⇒1−α∼1

⇒1.87×10

−5

=

1

Cα

2

=C×0.02×0.02

⇒C=

0.02

2

1.87×10

−5

=0.0467M

Correct answer is option-B.