Question

A certain weak acid, HA, has a Ka value of 2.6×10−7. Calculate the percent ionization of HA in a 0.10 M solution.

Answer 1

Answer:

The percent ionization is 0,16%

Explanation:

The percent ionization is defined as the number of ions that exist in a substance.

$PI=\frac{[A-]}{[HA]} x100$

First, we find the [A-] using the ka equation

HA ⇄ $H^{+} + A^{-}$

[H+] = [A-]

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

since the ionization constant is very small we can assume that the final concentration of [HA] is the same

$Ka=\frac{[H+]^{2} }{[HA]}$

$[H+]=\sqrt[2]{Ka.[HA]}$

$[H+] =\sqrt{(2,610^{-7} )(0,1)} = 1,61210^{-4}$

Now we calculate the percent ionization using these values

$PI=\frac{1,61210^{-4} }{0,1} X100$

PI=0,16%