Question

Calculate the percent ionization of acetic acid solutions having the following concentrations.

A. 1.20M

B. 0.550

C. 0.100

D. 4.60*10^-2

Answer 1

A)

CH3COOH dissociates as:

CH3COOH -----> H+ + CH3COO-

1.2 0 0

1.2-x x x

Ka = [H+][CH3COO-]/[CH3COOH]

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 ((1.8*10^-5)*1.2) = 4.648*10^-3

since c is much greater than x, our assumption is correct

so, x = 4.648*10^-3 M

% dissociation = (x*100)/c

= 4.648*10^-3*100/1.2

= 0.3873 %

Answer: 0.387 %

B)

CH3COOH dissociates as:

CH3COOH -----> H+ + CH3COO-

0.55 0 0

0.55-x x x

Ka = [H+][CH3COO-]/[CH3COOH]

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 ((1.8*10^-5)*0.55) = 3.146*10^-3

since c is much greater than x, our assumption is correct

so, x = 3.146*10^-3 M

% dissociation = (x*100)/c

= 3.146*10^-3*100/0.55

= 0.5721 %

Answer: 0.572 %

C)

CH3COOH dissociates as:

CH3COOH -----> H+ + CH3COO-

0.1 0 0

0.1-x x x

Ka = [H+][CH3COO-]/[CH3COOH]

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 ((1.8*10^-5)*0.1) = 1.342*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)

1.8*10^-5 = x^2/(0.1-x)

1.8*10^-6 - 1.8*10^-5 *x = x^2

x^2 + 1.8*10^-5 *x-1.8*10^-6 = 0

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

a = 1

b = 1.8*10^-5

c = -1.8*10^-6

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 = 7.2*10^-6

roots are :

x = 1.333*10^-3 and x = -1.351*10^-3

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

x = 1.333*10^-3

% dissociation = (x*100)/c

= 1.333*10^-3*100/0.1

= 1.3327 %

Answer: 1.33 %

D)

CH3COOH dissociates as:

CH3COOH -----> H+ + CH3COO-

4.6*10^-2 0 0

4.6*10^-2-x x x

Ka = [H+][CH3COO-]/[CH3COOH]

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 ((1.8*10^-5)*4.6*10^-2) = 9.099*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)

1.8*10^-5 = x^2/(4.6*10^-2-x)

8.28*10^-7 - 1.8*10^-5 *x = x^2

x^2 + 1.8*10^-5 *x-8.28*10^-7 = 0

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

a = 1

b = 1.8*10^-5

c = -8.28*10^-7

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 = 3.312*10^-6

roots are :

x = 9.01*10^-4 and x = -9.19*10^-4

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

x = 9.01*10^-4

% dissociation = (x*100)/c

= 9.01*10^-4*100/0.046

= 1.9587 %

Answer: 1.96 %