Question

The Ka of a monoprotic weak acid is 6.46 × 10-3. What is the percent ionization of a 0.193 M solution of this acid?

Answer 1

Lets write the dissociation equation of HA

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

0.193 0 0

0.193-x x x

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

Ka = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((6.46*10^-3)*0.193) = 3.531*10^-2

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Ka = x*x/(c-x)

6.46*10^-3 = x^2/(0.193-x)

1.247*10^-3 - 6.46*10^-3 *x = x^2

x^2 + 6.46*10^-3 *x-1.247*10^-3 = 0

Let's solve this quadratic equation

Comparing it with general form: (ax^2+bx+c=0)

a = 1

b = 6.46*10^-3

c = -1.247*10^-3

solution of quadratic equation is found by below formula

x = {-b + √(b^2-4*a*c)}/2a

x = {-b - √(b^2-4*a*c)}/2a

b^2-4*a*c = 5.029*10^-3

putting value of d, solution can be written as:

x = {-6.46*10^-3 + √(5.029*10^-3)}/2

x = {-6.46*10^-3 - √(5.029*10^-3)}/2

solutions are :

x = 3.223*10^-2 and x = -3.869*10^-2

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

x = 3.223*10^-2

% dissociation = (x*100)/c

= 3.223*10^-2*100/0.193

= 16.7 %

Answer: 16.7 %