Question

If the percentage ionization of an 0.10 M acid is 3.0%, what is its acid dissociation constant?

Answer 1

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

Solution:- percent ionization = $(\frac{x}{c})100$

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:

$3.0=(\frac{x}{0.10})100$

$x=\frac{3.0*0.10}{100}$

$x=0.003$

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.

$Ka=\frac{[H^+][A^-]}{[HA]}$

Let's plug in the values in it:

$Ka=\frac{(0.003)^2}{(0.1-0.003)}$

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

$Ka=\frac{(0.003)^2}{(0.01)}$

$Ka=9*10^-^5$

Second choice is the right one.