Question

A certain weak acid, HA, has a Ka value of 1.4

Answer 1

Dissociation equation of the acid;

HA(aq) <-----> H+(aq) + A-(aq)
0.010-x M ........ x M ........ x M

Ka = [H+] [A-] / [HA]

1.4x10^-7 = (x)(x) / (0.010 - x)

Since the amount of x is very small compared to 0.010 M, it is neglected.
1.4x10^-7 = (x)(x) / (0.010)
x^2 = 1.4x10^-9
x = 3.6x10^-5 M

Since 3.6x10^-5 M is the amount dissociared (ionized)
and 0.010 M is the initial amount,
the percent ionization of the acid will be:
(3.6x10^-5 M / 0.010 M) x 100 = 0.36 %

If this amount is greater than 5%, we cannot apply the approximation of neglecting (x). In such cases, the roots of the quadratic equation must be found. (One of the roots will satisfy the condition).