If the percentage ionization of an 0.10 m acid is 3.0%, what is its acid dissociation constant? 0.003 9 x 10-5 .0009 .009

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Solution:- percent ionization =

where, is the equilibrium concentration of product ion and c is the initial concentration of the acid.

Let's plug in the values in the formula:

In general, the acid is represented as HA, the ice table is shown as:

I            0.10                                             0             0

C           -X                                               +X             +X

E         (0.10 - X)                                         X               X

where X is the change in concentration that we already have find out using percentage ionization formula.

Let's plug in the values in it:

0.003 is almost negligible as compared to 0.10, so (0.10 - 0.003) could be taken as 0.10.

Second choice is the right one.

it is a solid

explanation: