Question

Calculate the percent ionization of benzoic acid at the following concentrations a).30 b) .00030

Answer 1

ka for benzoic acid = 6.3*10^-5

a)

C6H5COOH dissociates as:

C6H5COOH -----> H+ + C6H5COO-

0.3 0 0

0.3-x x x

Ka = [H+][C6H5COO-]/[C6H5COOH]

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 ((6.3*10^-5)*0.3) = 4.347*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)

6.3*10^-5 = x^2/(0.3-x)

1.89*10^-5 - 6.3*10^-5 *x = x^2

x^2 + 6.3*10^-5 *x-1.89*10^-5 = 0

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

a = 1

b = 6.3*10^-5

c = -1.89*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 = 7.56*10^-5

roots are :

x = 4.316*10^-3 and x = -4.379*10^-3

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

x = 4.316*10^-3

% dissociation = (x*100)/c

= 4.316*10^-3*100/0.3

= 1.4387 %

Answer: 1.44 %

B)

C6H5COOH dissociates as:

C6H5COOH -----> H+ + C6H5COO-

3*10^-4 0 0

3*10^-4-x x x

Ka = [H+][C6H5COO-]/[C6H5COOH]

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 ((6.3*10^-5)*3*10^-4) = 1.375*10^-4

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Ka = x*x/(c-x)

6.3*10^-5 = x^2/(3*10^-4-x)

1.89*10^-8 - 6.3*10^-5 *x = x^2

x^2 + 6.3*10^-5 *x-1.89*10^-8 = 0

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

a = 1

b = 6.3*10^-5

c = -1.89*10^-8

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 = 7.957*10^-8

roots are :

x = 1.095*10^-4 and x = -1.725*10^-4

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

x = 1.095*10^-4

% dissociation = (x*100)/c

= 1.095*10^-4*100/0.0003

= 36.5133 %

Answer: 36.5 %