Question

The Ka of a monoprotic weak acid is 6.90 × 10-3. What is the percent ionization of a 0.176 M solution of this acid?

Answer 1

For simplicity lets write the monoprotic acid as HA

HA dissociates as:

HA -----> H+ + A-

0.176 0 0

0.176-x x x

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

Ka = x*x/(c-x)

6.90*10^-3 = x^2 / (0.176-x)

1.2144*10^-3 - 6.90*10^-3*x = x^2

x^2 + 6.90*10^-3*x - 1.2144*10^-3 = 0

This is quadratic equation (ax^2+bx+c=0)

a = 1.0

b = 0.0069

c = -0.0012144

Roots can be found by

x = {-b + sqrt(b^2-4*a*c)}/2a

x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 0.00491

roots are :

x = 0.03157 and x = -0.03847

since x can't be negative, the possible value of x is

x = 0.03157

% dissociation = (x*100)/c

= 0.03157*100/0.176

= 17.9 %

Answer: 17.9 %