A certain weak acid, HA, has a Ka value of 6.7?

weak-acid

#1

A certain weak acid, HA, has a Ka value of 6.7.

Answer:

Dissociation equation of the acid;

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

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

6.7x10^-7 = (x)(x) / (0.010 - x)

Since the amount of x is very small compared to 0.010 M, it is neglected.
6.7x10^-7 = (x)(x) / (0.010)
x^2 = 6.7x10^-9
x = 2.5x10^-5 M

Since 2.5x10^-5 M is the amount dissociared (ionized)
and 0.010 M is the initial amount,
the percent ionization of the acid will be:
(2.5x10^-5 M / 0.010 M) x 100 = 0.25 %

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).