Question

Calculate the percent ionization of formic acid (HCO2H) in a solution that is 0.139 M in formic acid. The Ka of formic acid is 1.77

Answer 1

HCO2H dissociates as:

HCO2H          ----->     H+   + HCO2-
0.139                 0         0
0.139-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)*0.139) = 4.96*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/(0.139-x)
2.46*10^-5 - 1.77*10^-4 *x = x^2
x^2 + 1.77*10^-4 *x-2.46*10^-5 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 1.77*10^-4
c = -2.46*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 = 9.844*10^-5

roots are :
x = 4.872*10^-3 and x = -5.049*10^-3

since x can't be negative, the possible value of x is
x = 4.872*10^-3

% dissociation = (x*100)/c
= 4.872*10^-3*100/0.139
= 3.505 %
Answer: 3.50 %