What is the percent ionization of a weak monoprotic acid?

Question

The concentration of the monoprotic acid = 0.116 M and the Ka = 6.21x10^-3

Answer 1

We can see if we can use the "5% rule". This is where we calculate the dissociation and % ionization without using the quadratic and if it is less than 5% of the original concentration, we can neglect it and not use the quadratic.

Let HA be the weak monoptoric acid. Then, HA <==> H⁺ + A^-

Ka = [H+]² / [HA] 

6.21x10^-3 = [H⁺]² / 0.116 

[H⁺]² = ( 6.21x10^-3) x0.116 

[H⁺]² = 7.20*10^-4 

[H⁺] = 2.68x10^-2 M 

% dissociation = (2.68x10^-2) /0.116 x 100 = 23.14% 

This is clearly greater than 5% so this method cannot be used legitimately, so we must now go back and use the full method including the quadratic equation.

Ka = [H⁺] [A^-] / [HA] 

Let x be the [H⁺]

6.21x10^-3 = x² / ( 0.116 - x) 

x² = (6.21x10^-3) ( 0.116 - x) 

x² = 7.20x10^-4 - 6.20x10^-3x 

x² + 6.20x10^-3x - 7.20x10^-4 = 0 

x = 0.0239 M

Percent dissociation = 0.0239/0.116 (x100%) = 20.6 %