Question

Calculate the percent ionization of formic acid (HCO₂H) in a solution that is 0.219 M in formic acid. The K_a of formic acid is 1.77 x 10^-4.

Answer 1

HCO2H dissociates as:

HCO2H -----> H+ + HCO2-

0.219 0 0

0.219-x x x

Ka = [H+][HCO2-]/[HCO2H]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((1.77*10^-4)*0.219) = 6.226*10^-3

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Ka = x*x/(c-x)

1.77*10^-4 = x^2/(0.219-x)

3.876*10^-5 - 1.77*10^-4 *x = x^2

x^2 + 1.77*10^-4 *x-3.876*10^-5 = 0

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

a = 1

b = 1.77*10^-4

c = -3.876*10^-5

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 = 1.551*10^-4

roots are :

x = 6.138*10^-3 and x = -6.315*10^-3

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

x = 6.138*10^-3

% dissociation = (x*100)/c

= 0.0061*100/0.219

= 2.80 %

Answer: 2.80 %