Question

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

Answer 1

Answer:-
`Ka=9*10^-^5`

Solution:- percent ionization =
`((x)/(c))100`

where,
`x` 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:

`3.0=((x)/(0.10))100`

`x=(3.0*0.10)/(100)`

`x=0.003`

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

`HA(aq)      \leftrightharpoons H^+(aq)   +A^-(aq)`

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.

`Ka=([H^+][A^-])/([HA])`

Let's plug in the values in it:

`Ka=((0.003)^2)/((0.1-0.003))`

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

`Ka=((0.003)^2)/((0.01))`

`Ka=9*10^-^5`

Second choice is the right one.